Eigenvalues and eigenvectors are fundamental concepts in linear algebra where for a square matrix A and non-zero vector X, if AX = λX, then X is called an eigenvector and λ is the corresponding eigenvalue. The characteristic equation det(A - λI) = 0 yields the eigenvalues, and the characteristic polynomial is det(A - λI). For an n×n matrix, there are at most n distinct eigenvalues. The algebraic multiplicity is the number of times an eigenvalue appears as a root of the characteristic equation, while the geometric multiplicity is the number of linearly independent eigenvectors corresponding to that eigenvalue. The trace of a matrix equals the sum of its eigenvalues, and the determinant equals the product of its eigenvalues. For real matrices, complex eigenvalues appear in conjugate pairs.
Eigen Values & Eigen Vectors (Part 1) || Engineering Mathematics || PrepFusionAjouté :
Yeah. Hi everyone, welcome back. So we have been studying about linear algebra, right? So we have covered a lot of topics. In the last lecture what we were studying, we were studying about null space. So just like null space, there is another concept that is igon value and igon vector. And it is very very very important in your gate examination. It is very important. So in null space, what was the scenario? Let's say there would be a matrix A. If I multiply my vector X with the matrix A, then what this matrix A was doing? This matrix A was mapping this vector X to zero.
Then X is known to be in the null space of A. X belongs to the null space of A.
So vector X being multiplied with A is going towards the null or you can say A is mapping the vector X to null or you can say X belongs to the null space of A. So that was the concept of null space. Here what will happen again I'm having a matrix A. Now vector X will be multiply with matrix A only in the output you will receive again vector X.
So matrix A multiply with vector X again in the output you are receiving vector X only when this condition is possible when sir A is identity matrix right because identity matrix multiply with X will give you X only but here I'm not expecting the output to be X here I'm expecting output to be X multiply with some factor lambda what I mean to say is that let's just say I give my input as 1 2 3 Then there was this matrix A which I don't know whatever this matrix A was there. Then at the output what I received 1 2 3 multiply with 4. That means what I actually received? I actually received 4 812 4 812. So when I saw this 482 I got to know that if I take 4 common if I take 4 common from here if I take four common from here what I will be getting?
I will be getting 1 2 3. So when I applied input X this was the system or this was a machine eventually it gave me 4x right. So if this is the scenario then X is known to be the igon vector of matrix A what it is known to be on vector X is the igon vector and what is lambda?
Lambda is known to be the igon value.
Okay, so lambda is the igon value and x is the igon vector. Now there could be multiple igon values. There could be multiple igon vectors. Are you able to answer this point? Let's just say there was this matrix A. I multiplied I multiplied here X1. So what I received first I multiplied X1. There was this matrix A. I received some lambda 1 multiply with X1. Then what I did to the same matrix I multiply X2. What I received X2 again multiplied with some factor which is lambda 2 or what I did again there was a matrix A then I sent some X3 what I received X3 again multiply with some factor lambda 3 right so there could be multiple vectors like that there could be multiple igon values as well but with respect to each vector there will be one igon value right understandable I hope it is clear to you what is the meaning of igon vector and igon Is that clear?
So matrix A was multiplied with vector X. Why I'm calling it vector X? Because it will be a column vector, right? Just in the null space as well, this was a column vector. No. So I'm saying that this is my N cross N matrix. Okay, this is my N cross N matrix. So this will be a column vector of N cross 1. So A will be multiplied with X. You will receive the vector X again multiplied with some factor lambda. The value of lambda could be anything. It could be 1x2. It could be three. It could be four. It could be minus one, it could be minus j, it could be 1 - 1 + j, it could be anything.
Okay.
So this is the theory that is written that let a be a square matrix. This will be a square matrix only. In the whole lecture, a will be a square matrix.
Okay. The concept of value and vector is defined only for the square matrix. So let a be a square matrix and x be a non-zero vector x. So and lambda is a scalar such that a x equ= to lambda x.
Then X is called igon vector of matrix A. X is the igon vector and lambda is known to be the igon value of matrix A corresponding to the igon vector X or other language what you can say there is other language that lambda is the igon value of matrix A. Okay. And X is the igon vector corresponding to the igon value lambda of matrix A. What I'm saying X is the on vector corresponding to corresponding to on value lambda because you can see here that with respect to X3 there is a different igon value lambda 3 with respect to X2 there is a different igon value lambda 2 with respect to x1 there is a different igon value lambda 1 okay so all these things you don't need to care much basically what you need to care you need to care only one thing that I sent x1 in the output I received x1 only multiply with some factor I sent x2 in the output I received x2 only multiply with some factor I sent x3 in the output I received x3 only multiplied with some factor is that clear to you okay so x1 x2 x3 will be known as the igon vectors and lambda 1 lambda 2 lambda 3 will be known as the igon value or you can say x1 is the igon vector corresponding to lambda 1. x2 is the igon vector corresponding to lambda 2. x3 is the ig vector corresponding to lambda 3.
Understandable okay let's sol this matrix A is given now with respect to this matrix A which alligs are there what the what this question is saying which of the following vectors are the igon vectors of A and for each valid igon vector you need to tell the corresponding igon value as well. So that means basically what we need to do is that we need to pick up this matrix A then multiplied this matrix A with the igon vector. This could be possibly the igon vector. This could be also possible on vector. Then we need to multiply our matrix with these vectors and we need to see what results we are getting. If we are getting the results as the same vector I receiving multiply with some factor then yes X is igon vector.
X is a igon vector and if we are not receiving the result like that that means X will not be a igon vector. Is that clear to you? So let's check example by example. Let's take the first example. So this this is our matrix A which is being multiplied with 1 0 0. So this is our matrix A which is being multiplied with 1 0 0. So 2 0 0 2 0 03 is multiply with 1 0 0. What you will receive? Sir 2 multiply with 1 then 0 0.
So sir this will be two. Then sir this will also be two.
Oh sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry sorry 2 mult* with 1 will be 2 0 0 then 0 multiply with 1 0 0 and 0 answer 0 0 and zero so this is what you are receiving this is your result when I saw this vector what I thought let's take two common so when I take two common what I'm getting so what you saw here this was your matrix A right and this was your vector x what you are getting in the output this is the final output again vector x multiplied with constant value. So I'm saying that there was a matrix A. I multiply with X1.
Let's call this as X1. So what I'm getting? I'm getting the same vector X1 multiply with some constant value which is lambda 1. So here what is your lambda 1? Lambda 1 is 2. And what is your X1?
X1 is nothing but 1 0 0.
So 2 is a value of matrix A. 2 is a value of matrix A. 2 is a value of matrix A.
and x1 is a igon vector corresponding to lambda 1.
Is that clear to similarly what was our next question? So yes, this is a vector for sure. Next was 0 1 0. So 0 1 and 0.
Do the multiplication. So 2 mult*ly with 0 0 multiply with 0 0. So 0 multiply with uh 0 will be 0. 2 multiply with 1 will be 2 and then 0 0 then zero zero here and zero here when I saw this vector carefully I got to know that if I take 2 common what I will be getting this is what I will be getting again you can sense that this is nothing but 2 into the same vector let's call this as x1 and let's call this as x2 okay so this is your x2 so it is the same thing it is same x2 is appearing here again 2 into x2 okay so what is is your igon value.
This is your igon value which is lambda 2 into x2. So x2 is the igon vector and lambda 2 is the igon value.
Understandable say yes or no. So this is also one of the igon vector. So this was the matrix a. I multiply this matrix a with 0 1 0.
What I received 0 2 0.
Okay. Next I multiply with 5 0 0 5 0.
So what I will receive 5 into 2 10 and then rest everything would be zero. When I saw it carefully what I got to know that I can take five common sorry I can take yes I can take two common.
What I can take? I can take two common.
So I'll be getting 5 0 0. Now someone would say sir why why don't you take 10 common? Sir if you take 10 common then this is something you will get. But this is the form that I don't want to make.
No I want to make this form.
that a being multiplied with x3 a being multiplied with x3 it is giving me x3 again with some constant factor so this is your x3 this is your x3 this is also your x3 and this is also your x3 so I want to make this this kind of form no so that's why I'm taking two as a common not 10 as a common because if I take 10 as a common then I will be getting 1 0 0 I will not be getting anything fruitful right so what I saw here is that what I'm getting I'm getting nothing but 2 into x3 or you can call it lambda 3 into x3.
So shall I call it lambda 3?
What is this? lambda 1 equ= to 2. What is this? This is also two. What is this?
This is also true. So shall I write down lambda 2 here or lambda 1 only? What do you say? Let's write down lambda 1 only.
Understandable?
So till this point what you have seen with respect to one igon value with respect to igon value two you have received this igon vector you have received this igon vector and you have received this igon vector with respect to igon value two you are having 1 2 three kind of igon vector till this point. Now if you think closely what you would understand that sir if 1 0 0 is a igon vector then for sure 5 0 will also be on vector because what I'm saying is that if a is being multiplied with x1 and you are receiving some lambda 1 into x1 now if you multiply a with some 5 x1 so what is five? Five is a constant value.
Five is a constant value. You can take five out.
And what is five? What is a1 into x1? So a1 into not a1. What is a into x1? 5 is a value that I took out a into x1 I'm getting. So what is a into x1? A into x1 is nothing but lambda 1 into x1. Can I write down this as lambda 1 into 5 x1.
So what you saw here that's why I multiplied my matrix A with 5 X1 what I received lambda 1 5 X1. So this was your matrix you multiplied this matrix with A you are receiving your matrix again multiply with the constant value lambda 1. So it is pretty obvious know that if I multiplying 1 0 0 with the matrix I'm getting 1 0 0 as the igon vector then it's a linear combination linear combination of x1 right that could be 2 x1 3x1 - x1 3x2 x1 that will surely be one of the igon vector for sure it is pretty obvious the same kind of concept we studied in the null space as well that if x1 and x2 are two of the igon values if x1 sorry if x1 and x2 are two of the igon vector Or I said if x1 and x2 belongs to the null space of a then linear combination of x1 and x2 will surely belong to the null space of a. Is that clear to you or not? Do you remember that if x1 is one of the igon vector x2 is one of the igon vector then the linear combination will also be one of the igon vector for sure. So it is the linear combination only you know it is simply x1 multiply with five. Okay.
So from this particular question only we are going to make some conclusions. But as of now whatever the conclusion that I'm writing down I hope that is clear to you.
Understandable. Started with 1 0 0 we got the same result. It was multiply with some factor 2. Started then 0 1 0 got the same result multiply with some factor 2. Start then took 5 0 got the same result multiply with some factor 2.
So as of now we have got to know that x1 x2 x3 all these are igon vectors with their corresponding igon values. For all these three examples the corresponding igon value came out to be two only. Can you let me know what will be the igon value of this particular vector and will it be a IG vector or not? Sir, if 0 1 0 is igon vector then this particular thing is nothing but 0 1 0 multiply with 5. 0 1 0 multiply with 5. So sir if x2 is igon vector then s x2 multiply with some constant factor alpha. If x2 is igon vector then alpha into x2 will surely be one of the igon vector for sure. There is no doubt about it sir. So 050 will also be igon vector. And sir I can already comment that icon value will come out to be two only. Why is that? So let's check the let's let's check the let's check this out. So 0 5 0. So this will become zero. This is 10 and this is zero. Someone will say let's say let's take 10 common. No I will take five common.
I will take two common not five. Why is that? So because I need to make 0 5 zero here. So what I'm getting some constant value which I'm calling as lambda 1 only which I'm calling as lambda 1 only multiplied our vector x4.
So this is your igon vector and this is your igon value. This I'm calling as x4 again I'm getting x4 here.
Understandable. So it was pretty obvious that I already knew what about 05. So sir 1 0 0 you have taken 0 1 0 you have taken sir even if you take their linear combination then also you will not get 0 05. So sir this is a different igon vector. 05 is a different igon vector.
Sir even if you take their linear combination what will be the linear combination of 1 0 0 and 0 1 0. Sir multiply this with alpha and multiply this with beta and add both of them. So that will be alpha alpha into 1 will be alpha. Beta into 0 will be zero. Alpha into 0 will be 0.
Beta into 1 will be beta and zero zero.
So this will be the linear combination.
So here this vector should be zero but this vector is 5 A. So this is a different vector. So let's check that out if it is IG vector or not. And if it is on vector then igon value will be different for sure. Why is that? So I value might be different. Not for sure on value might be different. Why value might be different? Let's check that out. So 0 0 0 you are getting 0 0 again 0 you are getting and 0 multiply 0 0 multiply by 0 here you are getting 50.
So what common I will take? I will take common as three.
Right? So if I'm taking three as common what I'm getting here so this was my x5 and in the output again I'm receiving x5 only but this on value is different this time I'm calling this as lambda 2. So what you are getting lambda 2 x5. So this is your igon vector and this is your igon value which is different than the previous value. Right? So yeah that means yes sir 0005 is also one of the igon vector and the ig value was three.
So with respect to the igon value three the igon vector was x5 understandable let's check the next example sir if 005 is one of the igon vector then I'm pretty sure that s 0 01 will also be on vector why is that so because it is simply multiply with 1x 5 if you multiply this vector with 1x 5 what you will be getting 0 0 1 so if x5 is a on vector then alpha into x5 will surely be one of the igon vector for sure right so alpha into x5 Five will also be one of the igon vector for sure.
That means 0 0 1. So this will become zero. This will also become zero. And this will become three. And this is 0 0 1. What common you are taking? Three.
Again the same on value and vector is coming out to be x6. If I take this as x6, this is also x6. And this is my lambda 2. Okay. So what is the igon value? Ig value is also lambda 2.
Okay. So igon value is lambda 2. So with respect to igon value 3. With respect to igon value three with respect to igon value three we are getting the ig vectors 0 05 and 0 0 1. 03 will also be one of the igon vector. Yes sir. 0 0 2 will also be one of the igon vector. Yes sir.
Right.
Now what about the next example? Sir 1 1 1.
Now someone would say that sir 1 0 0 was the igon vector.
Then sir 0 1 0 was the ig vector and sir 0 0 1 was also the igon vector. sir if I take the linear combination let's say I simply add them what you will be getting 1 + 0 + 0 will be 1. 0 + 1 + 0 will be 1. 0 + 0 + 1 will be 1. So sir this is you will be getting that mean their linear combination should also be one of the igon vector for sure. So sir I will say that this is also one of the igon vector wrong. Why is that so? They were the igon vector with respect to lambda 1. No and this is the igon vector with respect to lambda 2. What I mean to say is that a into x1 gave you lambda 1 into x1. Right? And a into x5 and the other vector let's call this as x2. So that also give you lambda 1 into x2. Right? A into x1. This was your x1.
This I'm calling as x1. And this I'm calling as x2. Okay? This I'm calling as x2. This was your x2. This is your x1.
This is your x2. And this is your x6. So a into x1 gave you lambda 1 into x1. A into x2 gave you lambda 1 into x2. And a into x6 gave you lambda 2 into x 6. Is that understandable to you?
Right. Now let's do one thing. Let's add up equation one and two. Let's add up equation 1 and two. What I'm doing? I adding up equation 1 and two. So what I will be getting? So a into x1 + x2 then lambda 1 I can take common from here. Then x1 + x2. This is what I will be getting. That means linear combination of x1 + x2 it was. So matrix A multiply with this vector. Matrix A multiply this vector Y is giving you lambda 1 Y only. That means yes Y is igon vector. This is igon vector corresponding to value corresponding to IG value lambda 1. Is that clear to you?
Y is vector corresponding to IG on value lambda 1. It is absolutely fine. It is the linear combination of X1 and X2.
Both had the same igon value. Both had same on value.
Same igon value. But what if I add equation 1 + 2 + 3? What I will be getting? If I add equation 1 + 2 + 3, I will be getting a x1 plus a x2 + a x3. So what I can do is that I can take a as common. So here I will be getting a into x1 + x2 + x3.
But on the other side what I will be getting? So lambda 1 x1 plus lambda 1 x1 plus lambda 1 x2. So lambda 1 you can take common from here plus lambda 2 x 6 it is not 3 it is 6 what you sense here sir let's call this as vector y a into y am I getting some constant value y sir y you cannot even take common s lambda 1 was 2 this was three right so here y was being common no x1 plus x2 plus x3 I can take separate and lambda 1 I I could have taken separate but can I take lambda 1 separate here.
If both were same then I could have taken it but if both are different then I cannot take it. So yes linear combination of igon vector will also be one of the igon vector but for the same on value not for the different igon values here it looks like that s 1 0 is one of the igon vector 0 1 0 is one of the igon vector and s 0 0 1 is also one of the igon vector. So let's take their linear combination. We are getting the option G which is on which is X7. So sir X7 will also be one of the igon vector.
But the question would be what is the igon value? So the question you will ask is that that sir what is the igon value with respect to x1? If I'm having value lambda 1 with respect to x2 as well I'm having a value lambda 1 only and with respect to x6 as well if I'm having value lambda 1 then for sure their linear combination will be a igon vector. But sir if they are having different igon values then it might not be a igon vector.
Is that understandable to you or not? So they need the same on values here lambda 1 and lambda 1. Then only it would have been the igon vector otherwise it is not. Okay. And you can check as well 111. So sir why you took these examples?
These examples are not your not the examples which which will be asked in the examination but they will build up your concept. Okay. So these examples are taken to build up your concept not to test your understanding. So here we are building up the concept right. So when I multiply with 11 one what I will be getting? So two multiply with 0 will be 0 2 multiply with 1 will be two then 0 and 0 then zero 2 multiply with 1 will be two and then this will be three. Can I take anything common?
If I take two common let's just say if I take two common what I will be getting 1 1 and 3x2 right? So if I call this as X 7 right. So is is this X7? No this is not X7 right? That means not a on vector not on vector.
Not on vector.
Is that understandable to you? So a into X7 cannot be written in some constant value. X7 cannot be written in this form. That means it is not a vector. I hope that is understandable to you.
Okay. And the linear combination thing is also the linear combination thing is also understandable to you. Why is that so? Because we got to know that they are having different on values. X1, X2 and X6. X1, X2 are having the same igon values but X6 is having different values than A and B. So from this as well we got to know that if they are having the different igon values then sir this lambda 1 I'm not able to take common.
But if they were having the same igon values then their linear combination of igon vectors was giving me another possible igon vector. Is that understandable to you? What about 555?
If x7 is not igon vector then its linear combination will also be not the igon vector. 111 is not the igon vector. Then 555 will also not be the igon vector. It is pretty obvious. No because 111 is not giving me it not it 111 is not giving me the results. Then 555 will also not give me the results.
5 into 2 10 10 and then 15 what I will do? Let's take two guys common. If I take two common what I will be getting five five but here I will get 15 by2 again not on vector not igon vector and what is the last one? Last one is also anything random 554 anything random I have taken 54. It is not the linear combination nothing any random veh any random vector I have taken. So let's check that out what I will be getting. 5 into 2 10 sir 2 into 5 s are 10 and here you will be getting 12. What common I should take? Let's take two common what you will be getting five five and six. So are they matching?
No they are not matching. What does that mean? Not igon vector.
So anything random cannot be vector right here. What allig vector you got?
Sir, we got 1 0 0 0 1 0 5 0 0 0 5 0 05 0 0 1 these six igon vector we got. So can I write down what all vectors I got. So corresponding to our igon value lambda 2 corresponding to igon value lambda sorry corresponding to igon value lambda equals to 2 these igon vectors I got 1 0 0 0 1 0 5 0 0 05 can you let me know that I will have other on vectors as well what that would be sir that would be 03 0 as well why is that so because they are the linear combination along with that sir you will be having 2 0 as well why is that so because they are the linear combination along with that sir addition of these two as well. Addition of these two as well. Why is that? So because of the linear combination. So what will be the addition of these two? 2 3 0.
So addition of these three as well.
Because linear combination. So 2 + 2 4 3 + 3 6 0 that will also be one of the igon vector. So there will be infinite igon vectors here. 1 0 0 is an igon vector. 0 1 0 is an igon vector. 5 0 0 0 5 0 03 0 2 0 0 2 3 0 4 6 0 all these will be the ig vectors. But in the complete linear algebra I have said one thing that if infinite vectors are possible or if infinite elements are possible this is not a matter of concern for me. What is the matter of concern?
Independent elements. So how many independent vectors you are having?
Two independent vectors you are having.
Are they independent of each other?
Yes sir. Mega matrix 1 0 0 0 1 0. Is it in row estone form? Yes sir. Row estone form. What is the rank you are getting?
Two. That means here I'm having two independent rows. So yes they are independent of each other. Rest everything is dependent. 0 5 0 is depending on 1 0 0. 050 is depending on 1 0 1 0. 03 0 depending on 0 1 0. 2 0 depending on 1 0 0. 2 3 0 depending on the linear combination of these two. And these are depending on these two guys.
460 depending on linear combination of these three and these three are depending on these two guys. So everything is dependent only two independent vectors we are having two independent igon vectors. So in the examination they will not ask you that with respect to on value two how many igon vectors you are having. That will be the wrong question. You will say sir infinite vectors I will have with respect to lambda with respect to IG value two. How many independent vectors you are having? Sir two independent vectors I will have rest all will be dependent. Sir how will I find it out that we will see but as of now you have understood one thing that my matter of concern would be the independent igon vectors and with respect to all these on vectors I was having the same on value.
If I ask you one question, okay, X1 is a vector corresponding to value lambda 1. My question is will 5 X1 be be a vector? My question is will 5 X1 be a igon vector? Yes sir, it will be on vector. What will be its corresponding on value?
Five lambda 1.
Wrong. These kind of simple question could be asked in the examination and seven out of 10 people will get it wrong. If x1 is a igon value sorry if x1 is a igon vector then corresponding igon value is lambda 1. If 5 x1 is ig vector yes sir it is ig vector. If it is vector then what will be the corresponding igon value? That will also be lambda 1 only.
No, here what I have written you have seen here that 1 0 0 was on vector corresponding to value lambda equ= to 2. 5 0 0 was vector corresponding to on value lambda equ= to 2 only. This was your x1. This was your x3. What is x3? x3 is nothing nothing but 5 x1. This is also 5 x1.
So with respect to 5 x1 you are having a value two. With respect to x1 you are having value two only. And it is pretty obvious. How will you think that sir a multiply with x1 will give me lambda 1 into x1.
Now sir a multiply with 5 x1 5 is a common thing sir I will get 5 into lambda 1 into x2 sorry 5 into lambda 1 into x1 but my matrix is 5 x1 sir. So if I need to make the same matrix no so lambda 1 will be outward and here I need to make the same matrix which will be 5 into x1 only a into x1 is giving me lambda 1 into x1. Then I multiplied five here. So here also I will multiply with five. But this is my matrix. So my matrix multiply with matrix A is giving me my matrix multiply with some factor. So what is that constant factor that is lambda 1 only. So this kind of simple equation could be as in the examination and you will be messing it up. Okay. So with respect to a matrix A we are having a value lambda 1. So corresp and the corresponding igon vector will be X1. So for the igon vector 5 X1 what will be my igon value? people will click on five lambda 1. Eight out of 10 people will click on five lambda 1. But the answer would be lambda 1. You either you can do by this thing or you can remember the example. Okay. Better will be this particular scenario that in the examination you see this condition based on that you write it down. Then you see what is your vector and based on that you see that what is my igon value.
Understandable?
Yeah. So these kind of things in short somewhere you can write down. Next on vector corresponding to lambda equals to 3. So corresponding to lambda equ= to 3 we got this igon vector which was 0 05 0 0 1 that's it. So here we saw sir 001 we saw then we saw s 0 0 5.
Can anyone let me know that 0 0 1 by 2 will also be igon vector or not 0 0us 2 will also be the vector or not and so on. But only one independent igon vector we will have. Only one independent igon vector we will have.
Okay just one second. Yeah. So what I was saying I was saying that here with respect to lambda 3 we will have only one one independent igon vector. So what I'm saying with respect to lambda equals to 2 we are having two independent on vectors.
Right? And with respect to lambda equals to 3 we are having one independent on vector right?
Is that clear?
So total three independent vectors we are getting for matrix for this matrix A. Right. And one more thing are they independent of each other?
Yes, it is pretty obvious that they are independent of each other. And this is also one of the independent vectors. So they will be independent of each other.
Basically, let's just say for this matrix A, I was getting X1 as X1 as a vector, X2 as vector and X3 as vectors.
Three independent vectors I'm getting.
So they will be independent of each other. this guy and this guy is independent and this guy is independent of of both of these guys. What I'm saying this is independent of both of these people, right? That means all three will be independent of each other.
Pretty obvious, right? Are you able to understand this point? So all these points will again come up. But you you need to keep on building up the concept, right? You need to keep on building up the concept. From this question what we have understood that yes you can get different igon vectors but with respect to one igon value with respect to each igon value whatever the igon vectors that you are getting the linear combination of those igon vectors will also be the new igon vector for sure but with respect to different igon values with respect to different igon values if you are getting two different igon vectors then it is not necessary that their linear combination will also be the igon vector Okay. So different igon value is a different concept. Different igon values you find no relation but same igon values two different igon vectors then their linear combination will surely be one of the vectors. Here it is proven as well. This particular portion is also very important. So when you revise you will revise this portion as well. This is also very very very important. So here I will write down y is not igon vector not on not vector.
Okay. So this portion is also very important. All these conclusion will also be written. No need to worry about it. This is this question is like your examination problem. I guess this is a gateway only. So how will you find out the how will you get this question correct in the examination? That matrix A is given and which of them will be will be an igon vector. This is the question they are asking. So what I will do is that sir I in the examination quickly I will take this matrix and then sir I will multiply with the first option 3a 3. So what I will be getting sir - 12 + 6 - 6 and the answer sir 12 + 6 18 okay so sir here I can sense that no matter whatever I do I will not be able to get three three here if I need to get three here I need to take minus2 common but if I'm taking minus2 common then here I will get minus 6. So it is not possible for me to get this kind of structure. So this is not a vector. Then sir again next option I will check - 42 in quicker manner you will be able to check right it's not it will not take much time 4 3. So directly you will be doing the you will be doing the calculation in a quicker manner. So -6 + 6 right - 16 + 6. So you'll be getting - 10 here. So and here you will be getting 16 + 9 which will be 25. So pretty obvious again you will not be able to get it right. Again you will not be able to get a 43 because because if you need four here here you you would need 2.5 which will not give you the results.
Right?
Next will be 2 - 1.
So what you will get - 8 and -2 - 10 12 and -3 12 and - 3 9 - 8 - 2 will be - 10 8 - 3 will be 5 okay now I can see one thing that sir if I take -5 common here here I will be getting two and here I will be getting minus So yes, this is the igon value and this is the igon vector. So this option is correct. This also you can check. This will also not be correct. Okay, this also you can check -4 2 4 and 3. So -1 and 2 what you are getting? 4 + 2 into 2 8 - 4 + 6 will give me 2. Okay. So no way this is also not a vector. Okay. So 2 into minus 1 will be on vector with the corresponding value minus 5. And let me tell you one thing in your competitive examination no matter whichever the examination you are writing down. If there are four option A, B, C, D, A, B, C, D are the options and you got to know that sir I will have to check the options. To get the answer of this question, I will have to check the options. There is a question having four options. You got to know that to check the answer of this question. To get the answer of this question, I will have to check the options. without checking the option I will not get the answer. So always first check option C because A is in the beginning D is in the beginning. What some people will do that either they will start from A or they will start from B.
So first check C and most of the people will start from the beginning. No so C will come at the third place right? So first check C then check B and then any of A or D. Okay. So this should be the order. I always check from C. Even if you take even if you see the question paper of J even if you see the question paper of G examination I have seen one thing that I professor set the option C only when you need to check the options okay most of the times I have seen that option C would be the would be the correct answer so always check by options option C then go to B and then any of A or D okay always in that order we we should check here here as well I did not know I did not know that answer would be C okay I did not But it came out to be C only. If I was there in the examination, I would have started from 2 - 1 only. Okay. And I would have get the answer. If it was MCQ, then without even checking all these options, I would have clicked on C. Right? So this is how we can save our time. Let's go to next concept that how would I find out the igon value of matrix A. That sir, here you solve these questions, you got to know that minus 5 is igon value. Here you got to know that 2 is igon value, 3 is on value. But sir, how will I find out these values? If matrix A is given, if matrix A is given, how will I find out the IG values? Will I be thinking that 10 0 might be igon vector then no right? How will I find out the igon values and how will I find out the igon vector as well? That also we will see but let's first talk about the igon values. So first we need to find out the igon values. So how we will find it out. So let's let's assume that lambda is a value corresponding to vector x. Okay. So what does that mean?
that if I multiply my vector x with a so the result I will be getting is vector x only multiply with some constant value lambda is that clear to so a x will give me x only multiplies with some constant vector constant value lambda so this is a constant value lambda is a scalar don't forget it is not a matrix it is a scalar and a and x are the matrix they are the matrix right now what if can I make somewhere like this a x minus lambda x equals to zero lambda x is going there right if I take x common what I'm doing I'm taking x common I'm taking x common what I will write down here lambda let's just say a is 2 3 4 5 and lambda is a constant value right what I have done a x minus lambda x let's just say lambda is 3 so previously what I was doing a x - 3x so a was a matrix x was also a matrix minus plus 3 X was also a matrix. So yes, it is absolutely fine to do like that. But when I take X common, what will come here?
I will come. No, A minus lambda I I will do. No, because this is a matrix. This is also a matrix.
This I need to make a matrix. Is that clear to you?
X I'm taking common. So here I would come. No, this is a square matrix. This also I need to make a square matrix. Now lambda is a number. Lambda is a number. This is a scalar number or constant number.
Scalar or a constant number. But from the matrix I need to subtract the matrix only. So what will come here? Identity matrix would come lambda into I. So what does that mean? I'm subtracting I is 1 0 0 1 lambda value as 3. So I'm subtracting 3 0 3. You can take any random example where a x equ= to lambda x. Then you take a minus lambda i into x you do and the result will be zero only.
Take any random example you will be able to understand. If you are not able to sat satisfy that sir how this I came this I came because this is a matrix and this is a constant value. I need to make it a matrix. So I will multiply with the identity matrix. Is that clear to you?
Okay. So eventually what I have got is that I have got this condition that a minus lambda i into x is coming out to be zero. So if lambda is your igon value corresponding or you can say if lambda is one of the igon value of matrix a and the corresponding igon vector is x then a minus lambda i multiply with x will be zero. I don't think it is very tough for you to understand it is easy only what you will remember you will remember this step a x= to lambda x and the very next step you will be able to write down that will be this particular step how is that so you will take lambda here so a minus lambda i into x would be equals to zero.
Now have we seen this kind of equation?
Can I write down this equation as a minus lambda is a matrix only. A minus lambda i will also be a matrix.
This will also be a matrix. So I'm calling this as matrix cx= to0. Can I write down this as cx= to0?
Yes sir. Have you seen the solution of this equation cx= to0? Yes sir. We have seen two kind of solutions. One is unique solution and other one is infinite solution.
What is the condition of unique solution? Condition of unique unique solution is that it should be a full rank matrix, right? It should be a full rank matrix or you can say that or you can say let's say A is a N cross N matrix. Okay, A is a N cross N matrix.
So C will also be N cross N. C will also be N cross N, right? because a minus lambda I I will also be n cross n so c will also be n cross n matrix. So what was the condition that rank of a minus lambda or rank of c rank of c should be equals to full rank and here rank of c which is a minus lambda i should be less than n. It should not be full rank. If it is full rank then what would be the solution vector? What would be vector x?
So vector x has to be the null vector.
And if it is not a full rank then vector X could be anything. Here elements would come and these elements would belong to R. Here some elements would come. Here some elements would come. These elements would be it is not necessary that they will be real. They could be complex as well. I don't care about it whether they are real or complex. I don't care about it. But in this case only one vector is possible which is 0000. Do you remember the last le lecture of null space? In null space also I said that 00 0 will never be my matter of concern. Why is that? So take any random matrix. No here I would have taken this matrix. Someone would say that sir I need to get a x equ= to lambda x. Then one intelligent guy would come that no matter what whatever the matrix you take if I multiply this matrix with 00 0 eventually I will end up getting 0 0 only. And here the constant value lambda equals to 1 I can take. So what will what will happen? Someone would come and say that x is my igon vector corresponding to igon value lambda equals to 1.
Someone would say I will take two here.
2 * 0 0 0 will also be zero. So my igon value will be two as well. Someone would say I will take 3x2. Someone would say I will take minus one. So with respect to all possible values 0 0 will be the igon vector. No and no matter whatever the matrix is there I don't even know what is my matrix. So any random matrix you take multiply that with 0 0 0 eventually you're going to get 0 0 0 only and here you can multiply any constant vector. So in that manner you will have infinite values and and only one vector which is 0 0 0. So 0000 is not considered as a vector. If you remember the definition what we have written non zero vector x.
So 0000 is not igon vector. Even in the null space also I said that 00 0 we will ignore we will not care about 00 0. So here you can clearly see that nonzero vector we can't. So if we are studying the concept of igon vector we are only talking about the nonzero vectors. What we are talking about we are only talking about non zero on vectors.
Is that clear to you or not?
Understandable?
Okay. So 00 0 will never be my matter of concern. What does that mean? This condition I don't want. Which condition?
This condition I don't want. So what I will say not desired.
This is not the desired condition. But which is the desired condition? This is the desired condition.
And uh if you want this condition, what is the scenario? That means it should not be a full rank matrix. And if it is not a full rank matrix, can I say one thing about it? That determinant of matrix C which is A minus lambda I.
Determinant of a matrix C which is A minus lambda I. That should be equals to zero. There you go. So matrix A would be given. You will find out matrix A minus lambda I. You will calculate its determinant and put that equals to zero.
You will get the value of lambda from here.
And that will be your on values. Is that understandable to you or not? Do you remember this thing? I hope you remember it. Like at least 50 times we have written that if it is not a full rank matrix then it determined will be zero.
It will be irreversible. All those things we have written. Right? So if it is not a full rank matrix A minus lambda is not a full rank matrix. That means it will have some nullity. What is the meaning of nullity? That means it will have all zero row. If it is having a all zero row what does that mean? Its determinant will be zero. So if it is not a full rank matrix then it statement will be zero. So in the examination if they are asking how to find out the igon values what you will do matrix A would be given you will find out the matrix A minus lambda I then you will calculate its determinant and you will equate it to zero you will get the values of lambda that's it we will sol example and it will be clear to you so let's take this example in this example a lot of things are asked what is the things they asking that first they asking characteristics equation characteristic polomial s I don't even know second question they're asking on values yes that I know sir I will try to find it out next question what they asking arithmetic multiplicity Geometric multiplicity I don't know sir vectors they're asking that also I can find out then a lot of things I don't know right so with respect to this question we will learn okay what is the meaning of algebraic multiplicity or not algebraic multiplicity it is arithmetic multiplicity okay this name is algebraic also you can say but arithmetic multiplicity then geometric multiplicity what is the meaning of that what is the meaning of characteristic polomial what is the meaning of characteristic equation all those things we will learn and how we can comment on the nullity of a minus plus lambda I that we need to see. So what is our matrix A? This is our matrix A. What we need to find out?
Let's take question B. Sir, we need to find out the igon values.
Ig values we need to find out. So I told you one thing for finding out the values. What you do?
Calculate a minus lambda. Calculate this matrix. Take its determinant and equate it to zero.
Okay. So this is our matrix A which is this one which is 2 0 0 0 2 0 0 and 0 0 5. What I will calculate A minus lambda I will calculate A minus lambda I. What I will calculate? I will calculate a matrix C which is A minus lambda I. So matrix A is given minus lambda I 1 0 0 1 0 and 0 0 1. So can this lambda go inside? Lambda multiply with 1 will be lambda only.
Here also this is lambda. Here also this is lambda. This is also lambda. This is also lambda. So what you will getting? A minus lambda. What that will be? 2 - lambda.
Then 0 - 0. Then 0 - 0. then 0 2 - lambda and that is 0 then 0 0 and then 5 - lambda this is your a minus lambda i.
From now onwards if some matrix I have given a b c d e f g h i and I ask you to calculate a minus lambda i. Will you follow this process? No. You know that sir identity matrix will have only these elements as lambda. Rest all will be 0 0 0 only. And then I need to subtract. So what I will do direct directly I will subtract minus lambda minus lambda and minus lambda I will get my a minus lambda i.
Is that clear too? Because I know that's a lambda i will have only these elements as lambda lambda lambda lambda. No rest all element will be zero. So a minus lambda e minus lambda i minus lambda.
That's it.
So 2 - lambda 2 - lambda 5 - lambda there you okay. So I don't need to follow all these processes that this is how I'm subtracting directly. I will be able to write down this particular matrix. Now what I need to do I need to equate the determinant equals to zero.
Okay 5 minus lambda. Can you calculate it determinant? Yes sir. I will calculate with respect to I will calculate with respect to first row. So what I'm getting 2 minus lambda multiply with 2 - lambda multiply with 5 - lambda coming out to be zero. Or I can write down in a better sense. I want to write down lambda - 2. I'm taking minus common. Okay. Lambda minus 2 and lambda minus 5. So 3 minus common I have taken.
So this is the negative sign I will be getting that is equated to zero.
Understandable.
Okay.
Now this particular can you solve the these equations?
Yes, I can solve this equation. But this particular thing this thing which is on the left hand side is it a function of lambda? Is it a function of lambda? Yes sir. What is the function of lambda? You are getting lambda - 2 and lambda minus 5.
Right? This is the function of lambda you are getting. And what is your what is the equation that you are getting?
The equation that we are getting is p lambda equals to0. This is the equation we are getting and this is the function we are getting. Right? So this is known as characteristics polomial. What it is known as? Characteristics.
Characteristics polomial.
And what this is known as? This is known as characteristics equation.
Characterics equation.
So eventually this is your characteristics equation. This is your characteristics equation and this is your characteristics polomial. Polomial means it is a function not equated to zero. It is a function and this is your characteristics equation. Is that clear to you? I hope it is understandable.
Very straightforward thing. So this is my characteristic equation. This is my characteristic polomial. And if I solve this equation, what is my characteristic equation? So your characteristics equation is lambda minus 2 multiply with lambda minus 2 multiply with lambda minus 5 equated to zero. If I solve this equation, the value of lambda I'm getting is value of lambda here you will get two. Here also again you will get two and here also again you will get five. So these are the igen values I'm getting. Someone would say that siron value is repeated. Yes valu is repeated.
So on values that you are getting is 225. What values you are getting? Siron values that I'm getting is 225. So 22 to5 are the igon values that you are getting. Okay. So on value 2 is repeated twice. So on value two repeated twice and on value five only on value five only came once.
So this is known as algebraic or arithmetic multiplicity.
This is known as arithmetic multiplicity. So if I ask you what is the arithmetic multiplicity of value lambda equals to 2.
What is that? So you will write down s arithmetic multiplicity of lambda= to 2.
Arithmetic multiplicity of lambda= to 2 is nothing but two. It is repeated twice. Okay. And what is the arithmetic multiplicity of value lambda= to 5? That is 1. So how many times on value is repeated that is known as arithmetic multiplicity. Okay. How many times value is repeated? What that is known as? That is known as arithmetic multiplicity. So arithmetic multiplicity of lambda= to 2 is 2 and arithmetic multiplicity of lambda= to 5 is 1.
Understandable? So we have seen the meaning of arithmetic multiplicity. We have seen the meaning of characteristic polomial and characteristic equation.
What is this?
Isn't it characteristic equation only sir? When you solve this particular side then only you got this equation.
Are you able to answer this point? when I solve this side determinant of a minus lambda. When I solve determinant of a minus lambda, then I got this particular equation. No, that means it is nothing but characteristic equation only. This is your characteristic equation.
Characteristics equation. And what do you mean by characteristics polomial? This particular side is characteristics polomial. Which side? This side is characteristics polomial. Okay. So what is your characteristics polomial?
This is your characteristics polomial characteristics polomial.
Okay. So in textbook this would be written that characteristic polomial is determinant of a minus m. Characteristic equation is determinant of a minus m i equals to zero. That is pretty understandable to you. Another thing if on values are mentioned okay if on values are mentioned values are mentioned which are 2 2 and five can I write down the characteristics poly can I write down the characteristics equation these values are mentioned let's call these values as lambda 1 lambda 2 and lambda 3 where value of lambda 1 is 2 value of lambda 2 2 and value of lambda 3 is five. Okay, someone would say they are same. Yes, they are same. If on values are mentioned lambda 1, lambda 2 and lambda 3. Can I write down the characteristic equation?
Characteristics equation.
Can I write down this equation? Yes sir, you will be able to write down the equation. It will be lambda - 2 or lambda - lambda 1. lambda - 2 lambda minus lambda 2 lambda minus 5 which is lambda minus lambda 3 that would be equated to zero. So if I know the ig values I will be able to write down the characteristic equation as well and from here I would get to know that this will be my characteristic polomial as well.
Okay although characteristic equation will be the matter of concern characteristic polinomial will not be a matter of concern but still we will be able to write it down.
Okay well and good so this much we have seen right. So all these points you need to understand what was the order of this matrix 3 + 3.
If you solve this correctly equation what will be the maximum power of lambda you will get. So lambda minus lambda 1 will be multiply with lambda minus lambda 2 multiply with lambda minus lambda 3. So lambda will be multiply with lambda will be multiply with lambda. So maximum power of lambda here maximum power of lambda maximum power of lambda will be equals to three only or you can call it n. If the order was if the order was n cross n if the order was n + n the maximum power of lambda would have been n. What does that mean? That means there could be three roots of this equation. Whatever the maximum power is there, let's just say I give you one equation a x square + b x + c equals to zero. So how many roots it will have? Two roots it will have alpha and beta. Now it might be possible that both of the roots are same. It might be possible no that both of the roots are same but maximum two roots it can have. So if there is a matrix of order n cross n. So what is the maximum number of different on values it can have? If there is a matrix of order n cross n then it can have maximum n different on value.
Is that understandable to you or not? A x² + b x + c will have two different roots. Similarly, if I'm having three gross three matrix, then I can have maximum three different on values because its characteristic equation will be of the form lambda minus lambda 1 multiply with lambda minus lambda 2 multiply with lambda minus lambda 3 equals to 0. So for a matrix of n cross n for a matrix of n cross n the maximum number of igon values maximum number of different on values on values will be equals to n. Say yes or no.
Or you can say it will surely have n on values but some values might be same that here it is having 22 2.
It is having three igon values which are 225 but they are same but surely it will have three values. Two of them are same.
2 and two are same.
Is that understandable to you? So if there is a matrix of order n cross n then it will surely have n igon values.
Okay it will surely have n igon values but some of them might be same understandable.
So these things we have seen on values we have find out on values we have find out and we have find out the characteristic equation we have find out the characteristics polomial and all these things. Next what we need to find out we need to find out the igon vectors and then we will understand the meaning of this geometric multiplicity as well.
Okay. So let's find out the igon vectors. Our next target is finding out on vectors.
So what equation we are having? The equation we are having is ex= to lambda x. Okay. And we got to know that in lambda we are having these elements 2 and three. Sorry 2 and five. Sorry guys.
2 and five on values are two and five.
So here we are having two and five when we write down the vectors no sorry when we write down the set when we write down the set we don't write down the repeated value in set repeated value will not come so this set should be written as two and five so we are having two different on values which is two and five so let's find out the on vector on vector corresponding to lambda equals to 2. This is what we are doing. So let's just say my igon vector is x only. My igon vector is x. So what it will give me? A x will give me 2x.
Right? Because on value is 2. So when a is multiply with x, it will give me 2x.
What is my a? So a is your 3 + 3 matrix which is given as 2 0 0 2 0 and 0 05.
Now this will be a vector x. In this vector x, x1 element would be there. X2 element would be there. X3 element would be there. That should give you two X1 element, X2 element and X3 element. Say yes. So this will be the scenario.
This vector X would be some this kind of vector. No, X1 element would be there, X2 element would be there, X3 element would be there.
Understandable. A X would give you 2X.
From here can I write down the equation?
2 into X1 multiply with 0. So 2x1 is coming out to be 2x1 only. 0 2x2 multiply 2x2 is coming out to be 2x2 only. And 5 x3 is coming out to be 5 into x3 is coming out to be 2x3 only.
Isn't it obvious that x1 will always be equals to x1 always.
What does that mean? It is a free variable.
CS X 2 X1 is coming out to be equals to 2 X1. 2 X1 is coming out to be equals to 2 X1. So X1 - X1 will be equ= to 0. 2 and 2 is getting cancel out. X1 - 0 equ= to 0. It is always true. No. So it is a free variable. What about the condition here? X2 is coming out to be equals to X2. It is also always possible. So what does that mean? It is also a free variable.
Right? And what about this guy? Is it always true 5 X3 = to 2X3? No. No, let's just say x3 is 1. So 5 = to 2. Is it true? No. What does that mean? I need to solve it. 5 x3 = to 2 x3. So this x3 2 x3 comes here. So what I will get? 5 x3 - 2 x3 = to0. So 3 x3 = to 0. So from here x3 = to 0 we are getting. So value of x3 is 0. Value of x1 and x2 are 3. So what will be your solution vector? Let's call x = to x1 as c1 and x = to x2 as c2. Okay, they are the free variables.
They are the free variables that we are having. What does that mean? That means our igon vector is coming out to be somewhat like this.
Ong vector is coming out to be somewhat like this.
X1, X2 and X3. This is C1. This is C2 and this is zero.
C1 and this is C2. I hope you remember this particular part in null space. We understood this part that how I will write down this. So C1, C2, C3. So this will be 1 0 and 0 because C1 multiply with 1 will give you C1. C1 multiply with 0 will give you 0. C1 multiply with 0 will give you 0. Now what do you need here? Here you need C1. So C1 multiply with 1 will be C1. C2 multiply with 0 will be zero. So C1 + 0 will be C1 only.
Here what you need? Here you need C2. So C1 multiply by 0 will be 0. C2 multiply by 1 will be C2. So this is how I can write down C2. And this value is zero.
That means in both of the vectors you would know that X3 will be fixed. But we are having these two different vectors which is 1 0 0 and 0 1 0 two independent on vectors we will have.
So with respect to lambda equals to 2 the igon vectors are let's call this igon vector as v_sub_1 and this igon vector as v_sub_2. So on vectors would be v_sub_1 v_sub_2 alpha v_sub1 beta v v_sub_2 and their linear combination but only two independent on vectors we will have two independent on vectors we will have Right? Two independent on vectors we will have.
Is that clear to you? So with respect to lambda equals to 2, two independent vectors we will have.
Now what about with respect to lambda equals to 5. So find out that this is the process of finding the vectors.
Okay. But what if sir in the examination we will do this much process. It will take a lot of time. So for that we took this question. In the examination these kind of question can come when it comes to IG vector. So you check by the options basically if someone would say that sir I need to find out the igon vector. So what I will do first I will find out the ig value then he will do this this process a minus lambda equals to zero. He will get the value of lambda and with respect to each igon value he will he will find out the ig vector and then he will match the option which is not desirable right because I know the concept of vector. No igon vector means simply matrix multiply with that vector will give me that vector only multiply with some factor. So if I know the concept then I will check the options right but here just for the theory purpose who knows in the examination it might be used. So just for the theory purpose we are doing it and we will make some conclusion here. Okay as of now you can see that with respect to lambda equals to 2 we are having two independent vectors and you know that lambda was repeated two was repeated. So some conclusion we are going to make okay have some patience.
So what do you mean by geometric multiplicity? If someone would ask what is the okay let we will write down geometric multiplicity later on but let's first find out the vector corresponding to lambda equ= to 5 and I will have to add the pages as well so I will add some page so what we are finding out on vector corresponding to lambda = to 5 vector corresponding to lambda equ= to 5 this is what I am supposed to find out understandable so let's find it out what I will do sir a multiply with vector x will give me 5x so matrix a is given 2 0 0 2 0 and 0 0 5 multiply with vector X which will be X1 X2 X3. It should give me 5X which should be 5 X1 X2 and X3.
Can you solve this equation sir? Here also you took X1, X2, X3. Ideally I should have taken X4 X5 X6. But I hope you are not getting confused. This is a separate part. This is also a separate part. All I need to find out is this vector. That is my matter of concern.
Right? So 2x1 2x2 5 x3. So first I will get 2x1. And from this equation I will get 2x1 should be equ= to 5 x1. This is the equation I'm getting. Then I'm getting 2x2 should be equals to 5 x2.
2x2 should be equ= to 5 x2. And then the equation that I'm getting is 5 x3 should be equals to 5 x3. 5 x3 should be equals to 5 x3. This is always true sir.
Always. This is always true that 5 x3 will be equ= to 5 x3. Take any value of x3. x3= to 2. 10 will be equ= to 10. x3= to 5. 25 will be equ= to 25. So this is s always true. But s these are not always true. No, that I have to solve.
So here I will get - 3x1= to0. Here I will get - 3x2= to0. So if I solve the equation this is always true which is free variable right but here I'm getting sir x1= to0 x2= to0 and x3= to some free variable I'm taking which is c3. So what is the solution vector that I'm getting?
solution vector x1 x2 x3 that I'm getting is s 0 0 which are fixed but some constant value c3 that could be written as c3 0 0 1 so this is the igon vector we are getting this is the igon vector we are getting right so on vector corresponding to lambda equals to 5. Right? What is the vector you are getting? So on vector corresponding to lambda= to 5 how we made it corresponding to lambda= to 5 we were getting as let's call this as v3 v_ub3 some 2 v3 and some gamma v3 you will get but there is only one independent igon vector here there is only one independent igon vector one independent vector.
Understandable?
Is that clear to you? Now let's understand the meaning of a geometric multiplicity.
Geometric multiplicity.
What is the meaning of geometric multiplicity which is GM?
The meaning of geometric multiplicity is that number of independent igon vectors.
Number of independent on vectors on vectors corresponding to each on value.
each on value. That is the meaning of geometric multiplicity. If I ask you what is the geometric multiplicity of what is the geometric multiplicity of value two? What you will say? Two. Why is that? So because we are having two independent on vectors. What we are having? We are having two independent on vectors with respect to lambda equals to 2. With respect to lambda equals to 2.
We are having two independent vectors.
two independent igon vectors on vectors which are v_sub_1 and v_sub_2 and what is geometric multiplicity of lambda equals to 5 that is one. Why is that? So because we are having one independent on vector which is v_3.
Is that clear to you?
So v_sub_1 and v_sub_2 are independent and v_sub_3 is also one independent vector we are having. Right?
One more thing v_sub1 v_sub2 v_sub3 will be independent of each other as well. Look about v1 v2 I have already said that v1 and v2 are independent of each other. That's why I am saying two independent vectors are there. But the doubt would be that sir let us say I was having a matrix A.
Okay.
Through this matrix A I'm getting these kind of igon values lambda 1 again lambda 1 then lambda 2 and lambda 3. These kind of values I'm getting with respect to these values.
I'm getting two independent igon vectors. Two independent on two linearly independent igon vectors I'm getting.
And here with respect to lambda 2 I will get one independent igon vector and here also I will get one one independent igon vector let's call this indep this here we were getting two independent vectors v1 v2 here I'm getting independent vector v3 here I'm getting independent vector v4 first of all one thing is for sure that if there is a igon value if there is igon value which is which is repeated only once. This igon value has been repeated only once then there could be only one independent igon vectors. There could be only one independent igon vector. If it is repeated only once then there could be only one independent igon vectors.
Let's just say sir one igon value is repeated twice.
So how many independent vectors it will have? Sir, it it may have one as well or it may have two as well. But maximum it can have two. One value is repeated twice. Lambda 1 lambda 2 it is repeated twice. So maximum it can have two independent vectors. It cannot have more than two igon vectors. Here this was repeated twice. How many vectors it had?
Two. It might be possible that it will have only one independent vectors. It might be possible. We will see the example. Okay? Don't worry. We will see the example. There it will have only one independent vector. But the maximum it can have two.
This is repeated only once. Lambda 2 is coming only once. So it will have one independent vector. It is coming once it will have one independent vector. It is coming twice. It may have two or more than two. And whatever the igon vectors you are getting they will be all independent of each other. What does that mean? That means these igon vectors that I will get this igon vector this igon vector this igon vector this igon vector all will be independent of each other. V_sub_1 v_sub_2 v3 v4 all will be independent of each other.
All are independent of each other.
independent of each other.
Is that understandable to you or not?
Okay, well and good. Now let's talk about the nullity. So till this point we have understood the meaning of geometric multiplicity as well. We have understood the meaning of algebraic multiplicity as well. So let's write down the values of geometric and algebraic multiplicity. So what all values you are getting?
Let's do this thing. Yes. So what is the algebraic multiplicity we are getting?
What is the meaning of algebraic multiplicity of lambda= to 2. What does that mean?
That means how many times lambda= to 2 has been repeated twice. And what was the algebraic multiplicity of lambda= to 5? How many times lambda= to 5 has been repeated? That is 1. Right? How many times lambda= to 5 has been repeated?
That means how many times you have got the value lambda equ= to 5. That is one.
And what is the geometric multiplicity here in the last what I have written here? I hope I did not write down five.
Okay. Algebraic multiplicity I explained where I explained algebraic multiplicity.
Yeah. One only. Okay. So in lang in written sometimes I can write down five here but I hope you understand how many times lambda equals to 5 has been repeated only once. No right. Oh, one more thing before this I will write down that the igon values that we were having was 22 2 5 they were 225 and with respect to 22 we got igon values which were v_sub_1 and v_sub_2 and with respect to five we got on value which is on vector which is v_sub3 with respect to 22 we got two independent vectors which were v_sub1 and v2 that means what was the geometric multiplicity of lambda 2 that was two and what was the geometric multiplicity of lambda= to 5 that was 1 because with respect to lambda equ= to 5 we got only one independent on vector and with respect to lambda equals to 2 we got two independent vector so I hope the meaning of algebraic multiplicity and geometric multiplicity is clear to you am gm I will call them understandable let's understand them meaning of nullity of a minus lambda i what we are understanding nullity of a minus lambda i so we are having this equation for vectors we are having this equation that ax equ= to lambda x from here we are getting the equation a minus lambda i into x= to0 and one thing also I know that if I'm considering the vector I can take only one condition that rank of a minus lambda i should be less than n or can I say nullity of a minus lambda i would be greater than or equals to 1 or I can say greater than zero is Is that understandable to you?
Nity of A minus lambda I would be greater than zero. Can I say so? Quickly think about it.
This is your 3 + 3 matrix. If it is having rank less than n, let's say rank is two. So 1 2 3 0 2 4 rank is two. That means this has to be all zero. So what is the nullity? Nity would be the number of columns minus the rank. So what is the nullity? Nity is one. So nullity has to be greater than zero.
Now what does nullity represent?
Nullity represents the number of free variables.
Number of free variables.
And this number of free variables. What does it represent? Number of independent vectors. In the last lecture of null space, we saw number of free variables.
Here how many number of free variables were there? Only one free variable was there. So how many independent vector I got? One. How many number of free variables were there here? How many number of free variables were there?
Those two were the free variables. Two free variables. Two free variables.
Two free variables.
If I am having two free variables, then what I got? I got two independent on vectors. The same thing we saw with respect to null space as well. Do you remember dimension of null space? So we calculated the nullity and what we were saying nullity of a would represent the number of free variables. The number of free variables would represent the number of independent solution. The number of independent solution would represent the number of independent vectors in the null space of a.
Similarly here nity of a minus lambda would represent the number of free variables and it will represent the number of independent vectors or you can say number of independent igon vectors with respect to the igon value lambda.
What it will represent? Number of independent igon vectors.
number of independent on vectors with respect to igon value lambda.
Let's understand by the example. This is this was the concept. If you are following the concept right from the beginning from the system of homogeneous equation then you will be able to understand this concept. Otherwise you can take the example. What you can take?
Let lambda equ= to 2. Let lambda equ= to 2. So what will be my matrix? A minus 2 i right? A minus lambda. A minus lambda.
So what I'm supposed to find out? I'm supposed to find out the nullity of A minus lambda. So can I find out A minus lambda? Yes sir. What is my what was my A? A was 20 0 0 2 0 0 5.
What was A? 2 0 0 2 0 0 5 minus lambda 2 I 2 0 0 2 0 0 2 What you will get? 2 - 2 will be 0. 0 - 0 will be 0. 0 - 0 will be 0. 0 - 0 will be 0. 2 - 2 will be 0. 0 - 0 will be 0.
0 - 0 will be 0. 0 - 0 will be 0. And 5 - 2 will be 3. You can interchange the rows and column. This is the matrix that you are getting. Can you comment on the nity? Nity of a minus 2 i. What you are getting? That you are getting as two.
This is the nullity you are getting.
What you can comment it will have two independent on vectors.
Can you comment on the nity of a - 5 I? Yes sir. A - 5 I A 2 0 0 0 2 0 0 05 - 5 I 5 5 0 0 0 5 0 0 0 5 What you will end up getting so I will end up getting 2 - 5 will be minus 3 0 0 - 0 will be 0 2 - 5 will be - 3 0 and 0 0 there you go is it in the form absolutely sir it is in the ro form only what is the nullity you are getting so nity of a minus - 5 I getting as one.
What does that mean? It will have one independent vector one independent on vector.
So what does nullity of A minus lambda 1 I a minus lambda 1 I represents?
What does this represent?
Lambda 1 basically here lambda 1 is 2 here lambda 1 here lambda 1 is five. So a minus lambda i would represent the number of independent igon vectors. The number of independent on vectors with respect to value lambda= to lambda 1. There you go. Are you mugging it up? Are you mugging it up? Are you just seeing some example and then you are making the conclusion? No, with the concept we are making the conclusion. If you feel like that s we need to see the example as well, then here is the example. So from this particular question, what all things you have understood? You have understood the meaning of characteristics polomial.
What is characteristics polomial?
Determinant of a minus lambda. That is your characteristics polomial. Have you understood the meaning of characteristics equation? Yes sir.
Determinant of a minus lambda equated to zero is the characteristics equation.
Have you understood the meaning of finding out the igon values of a? Yes sir. How will you find out the igon values of a? Determinant of a minus lambda i should be equated to zero. Then you will be able to find out the igon values of a. Sir, I equated that to zero. I was able to find out the igon values of a which were 2 to 5. With respect to igon values can you comment on the arithmetic multiplicity? Yes sir.
Arithmetic multiplicity simply means how many times the igon value has repeated.
Lambda equals to 2 has been repeated twice. Lambda equ= to 5 has been repeated once. So that is known as the arithmetic multiplicity. Then can I find out the igon vectors? Yes sir. I can find out the igon vectors. Along with that I got to know one thing. If there is a matrix of n cross n then maximum number of igon values I can have.
Maximum number of different values I can have is n. There could be some same igon values. There could be some different on values as well. So but maximum number of different values I can have is n. Then can I find out the ig vector? Yes sir. I know the equation a x= to lambda x. I need to get lambda equ= to 2. So that I put simply I will solve the equation. I will get the ig vector. In this case, I was getting two free variables. Two independent vectors I was having. Here I was getting one free variable, one independent vector I was having. Then I understood the meaning of geometric multiplicity. What is the meaning of geometric multiplicity? The number of independent vectors you are getting with respect to a particular value. So with respect to value two, I was getting two independent vector and with respect to value five, I was getting one independent vector. Right? That I understood. Along with that, I understood one more thing. Let's just say there is a Yeah. Can you let me know what what will be the order of matrix A?
That will be 4 + 4. Why is that? So because I am having four on values. So if I'm having four igon values okay lambda 1, lambda 1, lambda 2, lambda 2.
I am saying that from my side it is given that lambda 1 and lambda 1 these are giving me two independent igon vectors which is v_sub1 and v_sub2. And this is giving me one independent igon vector and this is also giving me one independent igon vector. Right? So if we are getting v1, v2, v3, v4, they will be surely independent of each other. So all these vectors will be independent of each other. But here this is not the shorty. Okay. This is not the shorty that for sure you will get one independent vector. This is not sure not one I should say two. With respect to this repeated on value you will get two independent vector. That is not sure. But these are not repeated. No. So one vector you will get here. Here also one vector you will get. But here it is coming twice. So you may get two or you may get one if one value is repeated four times. If one value is repeated four times. This is repeated four times.
So maximum how many independent vectors I can have. So maximum independent vectors I can have four or I can have three as well or I can have two as well or minimum I will have one. So with respect to lambda 1 I can have one independent vector as well or two or three or four. I cannot have five independent vector.
Is that understandable to you? I hope it is understandable. It is absolutely clear to you. Right? So here if lambda 1 is repeated twice then you can have two independent vectors and that is the case here. Here we are having geometric multiplicity of two and arithmetic multiplicity of two as well.
Understandable then we would show that how will I find out what will be the geometric multiplicity. So basically what this is representing nity of a minus lambda i. Basically what is nity of a minus lambda i is representing nity of a minus lambda 1 i. What it is representing? It is representing nothing but geometric multiplicity of lambda 1 that how many independent vectors it will have. If I need to find out the with respect to lambda equals to 2 how many independent vectors you are having.
So what I would do I would calculate the nity of a minus 2 i. So nity of a minus 2 i came out to be two. So with respect to lambda equals to two I having two independent vectors. And why is that so?
Because nity would represent the number of free variables. Number of free variables means number of independent solution. Number of independent solution means number of independent igon vectors.
So these many independent vectors I will have. So nity of a minus lambda i represents the geometric multiplicity of lambda or you can say how many times or how many on vectors independent vectors you are having with respect to a on value lambda equals to lambda 1. Is that understandable to you or not? So this is a complete revision question which is telling you most of the things about igon values and on vector. I hope it is completely understandable to you. Again we will solve one more problem there. We will understand more concepts. What do we need to find out? IG values you find out. Okay. So make the matrix calculate a minus md. It will be a tough calculation. Okay. It will be a tough calculation. This will be a tough calculation. In the examination this kind of calculation might not be asked but who knows it may be asked as well.
So calculate it on your own. Sir, calculating the determinant of 3 + 3 is fine but sir 4 + 4 it will be a bit tough sir. Yes it will be a tough task but you need to do it on your own. Try it on your own and then what equation you will get? This equation you will get which I have already written down. Okay.
So you try doing it you try doing it on your own and find out the ig values.
Once you have find out the igon values can you find out the algebraic multiplicity and geometric multip multiplicity? Yes sir. That also we can find out and then you will be able to comment on the relation of am and gm.
Okay, you will be able to comment on the relation of algebraic multiplicity and geometric multiplicity. When you solve it, you will understand the relation.
When you solve it, you will understand the relation. Okay, and then you will tell me the total number of independent vectors. And after that, you will calculate the trace and determinant of this matrix A and you will comment something comment on the relation to the values of A. So four questions you can solve and then you can calculate the trace and determinant. What will be the relation that we will see that you leave up to me. I will take a break and then I will come back. But I ask you to solve this question on your own. Will you do it for me? Will you do it for yourself?
Yes sir, you will. We will do it. We will do it everything on on our own. We will find out the values as well. Then we can comment on the algebraic multiplicity from here. Then for finding out the geome geometric multiplicity what I need to do? I need to calculate the nullity sir. Then I will calculate the nity. Then I will see what is the relation I'm getting and then I will be able to tell the total number of independent vectors as well. And then I will calculate the determinant and trace and I will comment on the relation. As of now, if you don't know the relation, I will tell you the relation.
Understandable? So, do it on your own and I will come back. Okay. Yeah, welcome back. So, did you solve this question? So, our matrix A was given, we are supposed to find out the on values, right? So, what we need to do if we need to find out the igon values, what we will do is that a minus lambda i, we will equate that to zero. And when we solve this particular equation, what what kind of equation we will be getting? We'll be getting this kind of equation. lambda - lambda 1 multiply with lambda - lambda 2 multiply with lambda minus lambda 3 multiply with lambda - lambda 4. Why is that? So because sir this is the characteristics equation from this equation only I will get four kind of roots. Why four kind of roots? Because this is a 4 + 4 matrix that means I can get four values. Now these values might be same they might be different. I don't know that. So this kind of equation I will be getting and uh after solving this equation sir I will be get I will be able to get to know my igon values.
Right? So a matrix is given minus lambda i. So from here I will subtract lambda.
Here I will subtract lambda. Here I will subtract lambda. Here I will subtract lambda and then I will calculate the determinant and then I will equate it to zero. So when I equated this is the equation I was getting. And from here I got to know that I'm getting the values that 22 because lambda minus 2 square is there. No that means on value 2 is repeated twice. Three is repeated once and five is repeated once only. What does that mean? That means sir your algebraic multiplicity algebraic multiplicity of lambda= to 2 is 2. Your algebraic multiplicity of on value 3 is 1. And your algebraic multiplicity of igon value 5 is also one.
Understandable. So this we have seen.
Now what we need to find out? We need to find out the igon vectors or we need to comment on the geometric multiplicity.
We are not supposed to find out the igon vectors. You need to comment on the geometric multiplicity. So let's comment on the geometric multiplicity of lambda equals to 5. Can you let me know what will be the geometric multiplicity of lambda equals to five? For first let's talk about the geometric multiplicity of lambda equals to 3. So that will be equals to the nity of a minus lambda i which is a minus 3 i. Can you let me know the answer without even doing the calculation? Sir, it will be one only.
Why is that? So because this is repeated only once the value the igon value is coming only one time. That means it will have maximum one independent igon vector and minimum as well one independent igon vector. Right? I hope you understand that if lambda equals to 3 is coming.
This is one of the igon value. So with respect to one igon value you will have one independent ig vector or we can check let's calculate the nity of a minus 3 i. What is your matrix a? And from that you need to subtract three. So you need to check the nity. So let's check the nity. Okay it will not be difficult.
We can simply see 4 - 2 1 0. So I need to subtract three here. So 1 - 2 1 0 1 - 2 1 0. Then 2 - 3 1 0 then zero.
Here again zero then here I need to subtract three and this will become one. Right? And then I need to subtract three. So 1 - 1 0 2. This is your a - 3 i. I have calculated a - 3 i. What is the meaning of a - 3 i? That means all the diagonal element will be subtracted from 3. So this will become 1. This will become - 3. This will become 0 and this will become 2. 1 - 3 0 2. Rest all the elements are same. - 2 1 0 2 1 0 0 1.
And then 1 - 1 0. Yes. So this is your a - 3 i. Now what I need to do I need to find out the nullity. So let's uh try finding out the nullity.
So I need to find out the rank. I need to get the roon form. But if I check something can I get the answer in a quicker manner or I will have to follow the complete procedure. I will have to follow the complete procedure only. So how do we find out the rorow echelon form? So in row two I need to do row 2 - 2 r1. Right? And in row three I need to in row four. In row four, I need to do row four minus row 1. Then I will get the matrix and I will be able to comment on the rank.
Okay, this process you should not be watching. You should be doing it on your own because we have studied rank way before, right? So you should be able to do all these things on your own. Twice I am subtracting here. So zero twice I'm subtracting here. 1 - 1 0 then this is I'm not doing anything about it. And then here I'm directly subtracting. So this will become zero. This will become 1. This will become minus one. And this will become two.
Right? So more operation I can do what I can do is that I can add up row two, row four and row two.
Right? Basically what I'm going to do now is that row from row four I will subtract row two.
Okay. Okay. So what matrix I will be getting in a quicker manner I'm writing down. So matrix that I will be getting in 1 - 2 1 0 from row four I'm subtracting something now. So let's keep this as it is and this will be 0 0 0 1 only. And the last one would be 0 0 because it is being subtracted zero and this will become two. And next what you are going to do? Next you are going to do in row four.
Next this came here.
Okay. Next what you are going to do is that from row four you are going to subtract row twice of row three. Twice of row three. So what you will be able to see?
You will be able to see one thing that sir it is 1 2 minor 1 2 1 zero. Nothing is happening with the first three rows.
So first three rows will be as it is sir.
They will be as it is. And now sir twice of this is subtracted. So directly I can sense that so it is a row as shown form rank of a minus 3 i is coming out to be three. That's why nullity of a minus 3 i is coming out to be number of columns 4 minus 3 that is equals to 1. So nullity is 1. What does that mean? What does that mean? That means my geometric multiplicity is also one. My geometric multiplicity that means number of independent vectors with respect to lambda equ= to 3 I will have only one.
And here as well I don't need to calculate. I know that geometric multiplicity with respect to lambda= to 5 will also be one only. Is that understandable to you or not? Absolutely it is understandable. But now I need to comment on lambda= to 2. I need to find out that geometric multiplicity of lambda equals to 2. What does that mean?
I need to find out how many independent vectors I am having with respect to lambda equals to 2. That means I need to calculate the nity of a minus 2 i. So then I will able to understand how many independent vectors I'm having with respect to value two. Okay. So how many independent vectors I'm having. So what I will do is that I will simply copy this one right. So here I subtracted three but I'm supposed to subtract two. So what I need to do is that in the diagonal element I need to add one. So this will become two. 2 this will become minus2 this will become 1 and this will become three basically what I'm doing is that I'm subtracting two from the diagonal element so here you will get two here you will get minus 2 here you will get 1 and here you will get three 2 - 2 1 3 and rest all the elements are same - 2 1 0 2 1 0 rest everything will be same now what I need to do I need to calculate the nity directly you can see that these two rows are the same right and then I need to get the ro echelon form. So let's get the ro echelon form. So how will you get the ro echelon form? In R4 what you do? R4 is being subtracted with half of R1, right? And R2 what you do? R2 - R1 you.
So what you are going to get?
2 - 2 1 0.
Right? So here 0 0 0 1 1 and here half of that you are subtracting. So this will become 0 0 only. This will be 0 only and this will be three. So will it be in the form? You just interchange both of them. Yes, they will be in the form and this is the old zero row. What is the rank you are getting?
So rank of a minus 2 i is coming out to be nothing but three. So what will be the nullity of a minus 2 i? So that will come out to be rank minus sorry number of columns minus the rank that is coming out to be one. So nity of a minus 2 i is coming out to be one. What does that mean? Geometric multiplicity of lambda= to 2 is coming out to be one only. What do you understand from here? I will have only one independent igon vector. Only one independent igon vector.
Independent on vector.
Is that understandable to you or not?
So geometric multiplicity of lambda equ= to 2. Although it is being repeated twice but only one independent vector you will have. Can you find out that vector? Yes sir. Simply multiply x1 x2 x3= to x1 x2 x3 x4 equ= to 0ero and then you will get the igon vector you will get to know that 3 x4 will come out to be zero that means x4 will be equals to zero from here x1 x2 value also you will be able to get are you able to understand what I'm saying what I mean to say is that if I'm supposed to find out the vector what I will do a minus 2 i multiply with vector x that should be equated to zero what is a minus 2 i this is a minus 2 i but I told you after doing the transformation after doing the after applying transformations whatever the matrix that you are getting that also you can write down here and then you can multiply with your variables x1 x2 x3 and x4 and when you solve those variables you will get the answer basically 2 - 2 1 0 here 0 0 0 1 1 and 0 03 so from here you can get to know that 3 x4 is coming out to be zero and uh x3 + x4 is coming out to be 0.
Then this is 0 equals to 0 only. And in the first equation you are getting 2x1 - x2 + x3 = to0. You will be able to understand all those things if you have followed the previous lectures. Right?
So from here what you will be able to get x4 is coming out to be zero. From here x3 is also coming out to be zero.
Now if x3 is coming out to be zero then equation from here you are getting is 2 x1= to x2. So one of the free variables could be there. Let's just say if x2 is or let us just say if x1 is c then what will be x2? x2 will be 2 c. So what is the solution vector you are getting? The solution vector that we are getting is x1 x2 x3 x4. So the solution vector that we are getting is this is c this is 2 c and this is 0 0.
So c you take common 1 2 0 0 only one free variable we are getting or you can say one independent solution we are getting or you can say one independent igon vector we are getting which is 1 2 0 0. So this is the independent vector we are getting. So what is your geometric multiplicity here? So let's copy this particular thing.
Let's copy this particular thing here and we can comment on the geometric multiplicity as well. That geometric multiplicity is coming out to be geometric multiplicity is coming out to be one. Here what is the geometric multiplicity? One here also one from here you can comment that algebraic multiplicity will always be greater than geometric multiplicity. And it is pretty obvious as well. Why is that so? Let's just say I am having a matrix 7 + 7 matrix. Okay. And its igon values are lambda 1, lambda 2, lambda 3, lambda 3, lambda 4, lambda 4 and lambda 4.
Okay. So seven values it should have. So 3 to 5 to 7. Seven value it should have.
So now it will have one independent ig vector, one on vector. This will also have one independent igon vector. One independent igon vector.
These are repeated twice. It may have one or two independent vector. So what is the algebraic multiplicity of lambda 1? Algebraic multiplicity of lambda 1 that is one. And what will be the geometric multiplicity here? For sure you can say that s one independent vector I will have. What is the algebraic multiplicity of lambda 2? That is also one. And for sure I can say that I will have one independent vector. But here what is the algebraic multiplicity of lambda 3?
So that is two. Lambda 3 is repeated twice. Now how many independent vectors you will have sir? Either I will have one or two. I cannot comment. And what about these these guys algebraic multiplicity? That means how many times lambda 4 has been repeated that is repeated thrice. But how many independent vectors you will have? So sir I can have independent vectors either 1 O2 or 3.
Is that understandable to you? Either 1 O2 or 3.
Is that clear? Here also 1 2 or 3. And whatever the independent vectors you are having they will be also independent of each other. Let us say here I'm having V_sub1 vector. Here I'm having V2 vector. And considering here only one independent vector is there and here two independent vectors are there or only one independent vector is there. This is what I'm considering. So four independent vectors I'm getting V1,1 V2, V3, V4. So they will also be independent of each other. Is that clear to you?
These guys will also be independent of each other. So this is the relation we are getting that arithmetic multiplicity will always be greater than the geometric multiplicity. AM greater than GM. Is that understandable to you or not? So what all questions we have answered? First one was find the values of A. Then for each on value find out the algebraic multiplicity. Comment on the relation. Find out the total linearly independent vectors. How many total linearly independent vectors you are having. So sir for matrix A we have find out that we are having the on values 2 2 3 and 5. So basically it was a 4 + 4 matrix. So we were having four igon values. With respect to this guy I was having only one independent vectors.
Here also only one independent igon vector and here also one independent igon vector. So they are independent of each other. They are independent of each other. So total three independent igon vectors.
Total three independent on vectors.
Is that true or not? So total three independent vectors we are having. So this is answer. Find out the trace and determinant of A. What is the trace of A?
Trace. Trace means addition of all the main diagonal elements. So that will come out to be 12. Right? 4 + 3 7 7 + 5 12. And what what is the det of A? Can you find out the determinant? Yes sir. You do the what I will do? I will do my expansion with respect to second row or I can do the expansion with respect to sorry this is your second column. I can do the expansion with respect to third column as well. So third row as well or I can do the expansion with respect to fourth column because here there are two zero element here also two zero element here also two zero element. So how you will do it? You can do the expansion with respect to fourth column.
Okay. So plus minus plus minus this will be minus this will be plus then minus and then plus. So directly I will go to this one. So one then that is gone. So what you are left with 4 - 2 1 2 0 1 1 - 1 1 - 1 0. So + - - plus - plus + -. So minus sign would come up here. Then again next plus 5. So I will block the row and column with respect to five. Block the row and column with respect to five. What I will be left with? 4 - 2 1 2 0 1 0 0 3. So what I will be getting here this is three and what will be the determinant of this one that will be 3 multiply with 4 into 0 will be 0 0 - 4 that will be 4 so this will become 12 so this determinant was 12 right so 5 into 12 here I was getting 5 into 12 so this was 60 and what is the determinant of this one so let's find out the determinant here so minus 1 is already there I need to find out the detent of this matrix So what I would do I would plus minus plus I would do it with respect to last row.
So last row minus 2 then 1 you should do it all on your own right I'm not supposed to do the calculations and all but again this is a recorded video so I don't know if you are able to calculate the determinant or not but at least you should be able to calculate the determinant till this point if you are studying in this lecture value vector which is nearly the end of linear algebra then you should be able to calculate the determinant what you will do pick up what we were doing we were doing it with respect to fourth column Right? Fourth column. Right? Fourth column. So this was your fourth column.
Then I did it with respect to zero. It will be zero only. Then zero zero. And you you remember the concept of co-actor. Right? Plus, - plus - plus minus in the alternative order. So this will be minus one. Then block the row and column. This is the matrix that you are getting. You need to find out the determinant of this matrix that is coming out to be zero only.
Okay. The the uh determinant is coming out to be zero. So here it was 60. So it is coming out to be 60. Okay. the determinant is coming out to be 60.
So this is the condition we are getting.
We have calculated the determinant as well and we have calculated the trace as well. So determinant is coming out to be 60. You should be able to do it on your own. Understandable? So this is the scenario.
What all are your values sir? Values are 235. Right? 235. If you write down that sir my values are 235. This is wrong.
Your values are 22 235. What are what are your igon values? So on values are 2 2 3 5. If I add up all the igon values, let us say this is lambda 1. This is lambda 2. This is lambda 3 and lambda 4.
If I add up all the igon values, what it will come out to be? Lambda 1 + lambda 2 plus lambda 3 + lambda 4. So 2 + 2 4 + 3 7 7 + 5 12. It is coming out to be trace of a. And if I multiply all the values even if it is repeated then also you need to take in action 2 into 2 4 4 into 3 12 12 into 5 60 which is determinant of A. So what do you understand?
Addition of values addition of igon values will represent the trace of matrix A and the multiplication of igon values will represent the determinant. Will you be able to remember this? Yes sir. So whatever that we have studied conclusion of everything we are going to write down.
Yeah, but one more page I'm having.
After that, we will write down the conclusion. We will solve this question and after that we will write down the conclusion. But from this particular question, what is the conclusion that we are making? We have understood the meaning of algebraic multiplicity. We have understood the meaning of geometric multiplicity. The nullity of a minus lambda will represent the geometric multiplicity of lambda. And arithmetic multiplicity will always be greater than the geometric multiplicity. It is pretty obvious. Now this particular matrix was having four values. Two of them were same. Two of them were same, right? Two of them were same. They were having one independent igon vector. This was also having one independent igon vector and this was also having one independent ig vector. So total three independent igon vectors they were having. Although it is a 4 + 4 matrix but it is having total three independent igon vectors. Maximum it can have four independent igon values. Sorry four different igon values and four independent vectors. But here with respect to the repeated igon value I was having only one independent igon vector. But in the previous example with respect to the repeated igon value I was having three two independent igon vectors. So it was repeated twice it was repeated twice and two independent vector I was getting but here I'm getting only one independent vector which is pretty obvious because nullity may be lesser right but with respect to the repeated value you can never get three independent vector. with respect to two independent two different igon values. With respect to two different igon values, I can only get maximum two independent ig vector. With respect to two different igon values, I can maximum get only two independent vectors which is pretty obvious and very easy to understand. Right? So all these things we have seen and then we got to know that trace of a trace of matrix a is simply the addition of all the igon values. trace of matrix A which is 12 which is simply the addition of all the igon values lambda 1 plus lambda 2 plus lambda 3 plus lambda 4 and the determinant of matrix A determinant of matrix A is simply the multiplication of all the igon values understandable so this also you will remember let's see this question you need to think of this question and need to give me the answer okay we are not commenting on the nullity of a minus lambda we are commenting on the nullity of a so which of the following statements are true matrix A is there which is of order N cross N.
Can you let me know what will be the meaning of nullity of A minus lambda I?
That means it will comment on the it will comment on the geometric multiplicity of what value geometric multiplicity of value or let's call it lambda 1. So it will comment on that geometric multiplicity of value lambda equals to lambda 1.
Understandable now they're talking about the nullity of a.
If I want to make if I want to make this function as nullity of a, what I need to do? Let lambda 1 equals to0 is one of the igon value of is one of the on values of a. If it is one of the on values of a then what will happen? Nity of here you will put lambda 1 equ= to z. What you will get? Nality of A equals to geometric multiplicity of lambda equals to 0. There you go. What it is representing? What nity of A is representing? Nality of A is simply representing the number of independent on vectors.
on vectors corresponding to corresponding to lambda equals to zero.
So with what nullity of a is representing? nullity of a is representing the number of igon vectors corresponding to lambda equals to zero.
So if nullity of a is greater than equals to 1 that means if nity of a or you can say geometric multiplicity of a is greater than or equals to 1. So this is what they are saying that geometric multiplicity of lambda equ= to0 is greater than equals to 1. One thing you already know that algebraic multiplicity is already greater than the geometric multiplicity right and this value is already greater than equal to 1. This value is already greater than equals to 1. What does that mean? That means algebraic multiplicity will also be greater than or equals to 1. What does that mean? The number of times lambda equals to0 you are getting as an value that is more than 1 or= to 1. That means for sure at least once you will be getting the zero on value and it is pretty obvious. No, you are saying that here I am having more than one independent on vector with respect to igon value lambda equals to0. So someone is saying that with respect to lambda equ= to0 I having more than one vector that means zero would have come at least once. No let us say there is a matrix whose values are 0 0 0 and four. These are the igon values. So with respect to 0 0 0 you will have at least one igon vector at least one igon vector you will have.
So if I am saying that I am having one igon vector with respect to 0 0 0 that means 0 is surely value. No is it always necessary that zero will be igon value?
No it is not always necessary. Why is that so? If you take a matrix A and multiply with a vector X then you need to get vector X only. It should be multiply with a constant vector. It could be three, it could be two, it could be one, it could be zero as well.
If it is zero then 0 into 0 will be zero.
So x will be on vector corresponding to lambda equ= to0 only if a x equals to 0 and a x equ= to0 is always not possible.
a x equ= to0 is always not possible. Can you say that sir I will take a random matrix a and I will take a random column vector a and I will multiply both of them and I will always get zero. No, there has to be some specific x. No, there has to be some specific a. Always it is not possible. Understandable?
Always it is possible if x equals to zero but we are talking about the non-zero vectors right so for non-zero vector x always it is not possible that you will get a x equals to zero right what is the condition is that rank of a should be less than equals to n so it should not be a full rank matrix if it is not a full rank matrix that means it is its nity would be greater than equals to 1 so this is what first option is saying so all the things you can relate we have studied about null space as well we have studied about a x equ= to0 as well we have studed about a x equ= to b as you are able to relate everything. What we are saying is that if nity is greater than or equals to 1, nity is greater than or equal to 1. What does that mean sir? That means nity of a minus lambda is greater than equals to 1. Lambda is zero. So it is representing the number of independent vectors with respect to value lambda equals to0. You are having greater than 1 or equals to 1.
What does that mean? That means one value I will have which is lambda equals to0 for sure. Is that clear to you?
Nality of A represents the number of linearly independent vectors corresponding to value lambda equals to 0. Absolutely correct. Nality of A represents that algebraic multiplicity of value zero means the number of times zero has repeat repeated zero has appeared as an igon value of a. This is wrong. Why is that? So what I have said is that let us say nity of a has come out to be zero. So the nity of a has come out to be two. What does that mean?
That means geometric multip geometric multiplicity of lambda equals to zero is 2. Say yes or no.
Right? Nity of a means nity of a could be written as nity of a minus lambda i.
A minus 0 i that is equals to 2. So nity of a is equals to 2. That means geometric multiplicity of lambda equ= to 0 is 2. And one thing you know that am is greater than or equals to gm. That means algebraic multiplicity of lambda= to 2 is greater than equals to two. So what you can write down from this statement what you can write down the number of times the number of times on value zero has appeared is two. Is that statement correct? No.
What will be the correct statement? Is at least two.
Is at least two. Say yes or no. That means on value zero would have been repeated at least twice. So here the number of times zero has appeared. No.
The number of times minimum number of times zero has appeared. This would have been the correct statement. No. So it is not representing the arithmetic multiplicity. It is representing the geometric multiplicity and if geometric multiplicity is two that means arithmetic multiplicity would be greater than equals to two. If I am saying that with respect to this matrix with respect to this matrix I am having some particular igon values okay I'm having some particular igon values and here with respect to a particular igon value I'm having two independent igon vectors two independent igon vectors then you will say that if you are having two independent ig vectors that means at least twice that igon value would have been repeated at least twice you that igon value would have been repeated even if it is repeated three times then also you can have two independent igon vectors and one value could be different. So this is the similar thing we have written that if it is having two independent vectors. If it is having two independent vectors then it might be possible that the value is repeated thrice.
Right? So it is not representing the algebraic multiplicity. It is telling the minimum number of times that a value has been repeated. Minimum number of time. Okay. And if zero is an igon value then dependent of A will surely be equals to zero. Why is that so? Let's just say there is a matrix A which is having a values of lambda 1, lambda 2, lambda 3 and zero. Right? So what will be the determinant of A? So determinant of A will be simply lambda 1 multiplication of all the values lambda 1, lambda 2, lambda 3 multiply by 0 that will be equals to zero only.
And it is pretty obvious if 0 is one of the values then nity of a would surely be greater than equals to 1 and if nity is greater than equals to 1 then the determinant will also be equals to zero.
So from here also you can tell. So all the points are interrelated. The only thing is if you don't want to think much what you will understand that what is the meaning of geometric multiplicity.
Geometric multiplicity means are simply nullity of a minus lambda. What does it represent? The number of independent vectors with respect to lambda equals to lambda 1. If I'm saying that nullity of a nullity of a means lambda 1 equals to zero I need to put that means it will represent the number of independent vectors with respect to the on value lambda equals to zero. So if I'm saying that nity of a is greater than equals to 1. that means at least one independent igon vector I'm having with respect to lambda 1 equ= to0. So if I'm having at least one independent vector that means at least once the value zero has been repeated as a on value. So zero would have come as an igon value at least one time. It could be two time, it could be three time, it could be four time.
Whatever it is, I don't know. But if I'm having one independent vector, that means at least once the value has come.
So this has been written here. Nity A represents the number of linearly independent vectors with respect to value lambda equals to zero. It represents the number of times the value has been repeated. Absolutely not. It represents the number of it represents the minimum number of times the value has been repeated. If nity of a is one, that means minimum one value I will get zero. If nity of a is two then minimum two igon values I will get zero. It could be three as well it could be four as well that I don't know but minimum two I will get. If nity of a is four then minimum four zero values I will get. If nity of a is seven then minimum seven zero values I will get. Is that clear to you? And if zero is one of the values then determine would be zero only.
Say yes or no. So very very deep points but in a very easy manner we have understood everything. Only thing you need to remember is this one. That's it.
If you remember this particular thing, your job is done. If you remember this particular thing, your job is done.
And what is the actual meaning of geometric multiplicity? That also you need to know. What is the meaning of geometric multiplicity? The number of independent vectors that you are having with respect to one on value. Well and good. I hope it is understandable to you. That's it. Okay.
So this is these are the properties that we are going to write down. Here we have written all the properties that we will see later on. But one more small property is there. So this is a diagonal and a triangal matrices. This is a diagonal matrix right?
This is a upper triangular matrix and this is a lower triangular matrix.
This we have understood in the basics.
Right? What is the trace here? A + B + C. And what is the determinant? A into B + C. What is the trace here? A + B + C.
And what is the determinant? A b c. What is the trace here? A plus b + c. And what is the determinant? A b c only. Can you find out its value? What you will do for finding out the values? sir minus lambda minus lambda minus lambda. And then I will calculate the determinant and equate it to zero. So when I calculate equated to zero, the igon values will come out to be ABC only. And it is pretty obvious as well. I values will come out to be ABC only. And it is pretty obvious as well because addition of igon values will represent the trace and the multiplication of igon values will represent a minute. Is that understandable to you? If I find out the igon values here a minus lambda a minus lambda a minus lambda I will be doing and do the expansion with respect to first row sorry first column do the expansion with respect to first column then what you will get? A minus lambda multiply with b minus lambda multiply with c minus lambda. Okay you will get this answer only. Then you will understand that sir on values are ABC only and here as well values are ABC only and it is pretty obvious as well.
Why is that so? Because addition of igon values will represent the trace and multiplication of values will represent the determinant.
Understandable? So a plus b plus c is the trace. A into b into c is the determinant. So what I am saying is that these are the diagonal elements. No, these are the diagonal elements. These are the diagonal elements.
So for upper triangular matrix and for a lower triangular matrix the igon values are their diagonal elements only in a triangular or diagonal matrix the igon values are simply the elements of the main diagonal elements of the main diagonal and this example we have seen previously as well the very first example that we picked up now what were the on values when you solved it on value this is a diagonal matrix on values were 225 225 main diagonal element. Same goes for lower triangular as well and same goes for upper triangular as well. Is that clear to you? Not a very tough thing to remember. Absolutely very easy thing to remember. Okay, let's write down all the properties. What is the basic definition? This is the basic definition.
Ax= to lambda x. X is on vector of A.
Lambda is the corresponding value. What do you mean by diagonal? What do you mean by the values of diagonal and trional matrix? They will be equals to the diagonal element. Right? Then characteristic equation. What is the meaning of characteristic polinomial?
Determinant of a minus lambda i. This is known as determinant of a minus lambda i. Okay. And what is the meaning of characteristic equation? Determinant of a minus lambda i equated to zero. That is the meaning of uh characteristics equation. And roots of this equation will represent the igon values. This is also the basic thing. What is the meaning of trace and determinant? So trace of matrix A will be equals to the addition of all the igon values. Let's just say this is our matrix A which is having lambda 1 lambda 2. Matrix A of N cross N it is having N on values. So addition of lambda 1 plus lambda 2 plus dot lambda n will represent the trace of a. And similarly multiplication of lambda 1 lambda 2 lambda 3 dot lambda n will represent the determinant. Okay. Nity and value zero. If nity is greater than equals to 1 then zero will be igon value for sure and nity of a would represent the number of independent vectors with respect to lambda equals to 0. This we have proven and we have given roughly around 20 to 25 minutes of discussion here. Right? If one of the igon value is lambda equals to zero then nity of a would be greater than or equals to 0 or you can say the number of independent vectors with respect to lambda equals to 0 will be at least one. Okay. And it of a would represent the number of independent vectors of lambda equals to0 or you can call it what you can call it geometric multiplicity of lambda equals to 0. And you would also know one thing that algebraic multiplicity of lambda equals to0 would be greater than the geometric multiplicity here. That means at least one at least zero will be one of the igon value for sure. This is what it will say.
Geometric multiplicity. What is the meaning of geometric multiplicity? That means nullity of a minus lambda. What it represents? It represents the number of linearly independent vectors corresponding to lambda. Right? What is the meaning of algebraic multiplicity?
Number of time value has repeated has been repeated or it has appeared as a root of characteristic equation. Right?
What is the relation that arithmetic multiplicity would be greater than this is written as arithmetic multiplicity would be greater than or equals to geometric multiplicity or it would be greater than equals to one. It is pretty obvious. Now greater than equals to 1, right?
Arithmetic multiplicity should be greater than equals to 1. That means at least once it will be repeated. Now if I'm saying that lambda equals to 4 is one of the value that means it is coming once. If 4 is the value that means it is coming twice. If 44 is the value then that means it is coming thrice. So it has to be greater than equals to one.
No. If it is coming once then at least once it it is coming. No. So arithmetic multiplicity would be at least one and it will be surely greater than or equals to geometric multiplicity as well.
Right? Total number of independent vectors what that would be? That would be the summation of geometric multiplicity of each value. What does that mean? Let's just say that means simply we are having values lambda 1, lambda 1, okay, lambda 2 and lambda 3. So what I will find out? I will find out the geometric multiplicity of a minus lambda 1 i. Okay. Then then I will find out the geometric multiplicity of a minus lambda 2 I and then I will find out the geometric multiplicity of a minus lambda 3 I. So let's just say here I got the geometric multiplicity 1 1. So here one independent vector I'm having here also one independent vector I'm having here also one independent vector I'm having.
So total three independent vectors I will have. So simply I'm adding up the geometric multiplicity of each igon value. geometric multiplicity of each igon value. Okay. And all of them will be on vector corresponding to distinct igon values are always linearly independent.
All of them will be linearly independent. And maximum how many independent vectors I can have? Maximum I can have n independent vectors. Why is that? So because with respect to lambda 1 let's just say lambda 1 is repeated twice. So maximum I can get two independent igon vectors. Here one independent igon vectors and here one independent igon vectors. So total I can have four independent vectors. That means basically if there is a matrix of order n cross n then you can get maximum n different on values different on values and let us say if all are different then maximum n different vectors say yes or no it is pretty obvious. and very easy to understand.
Maximum and different values I can get and maximum and different vectors I can get. Understandable let's solve this question. So this is a new concept now okay it is a new concept new concept we are learning. So previously we have studied 10 properties and we have proved all the 10 10 properties intuitively as well with respect to the example as well. So everything is crystal clear. So this 2 point uh this two hour of discussion is giving us all the knowledge or maximum knowledge about the igon values and igon vectors.
Understandable let's see this question.
A is a invertible matrix and suppose lambda is an ig value of a with the corresponding ig vector x then you need to determine the ig values of the following matrices in terms of lambda and also you need to write the igon vectors as well. What they are saying is that what is given?
Given is a x= to lambda x. This is given.
Okay. Now what you are supposed to find out? Find out the on value and on vector of new matrix which is b a².
Okay. So basically they are saying that by y would be equals to some lambda y.
Here if I'm calling this as lambda so let's call it sum alpha. So by y should be equals to alpha y. So what will be the value of alpha in terms of lambda and what will be the value of y in terms of x. This is what you are supposed to find out. Okay. So what I will do?
I need to make a square. No I need to make a square. So what I will do here multiply with a multiply premultiplication by a on both of the side. Premultiplication pre multilication by a on both of the side. So a multiply with a multiply with x multiply with a multiply with x. Right? Now a into a will become a square into x. Now why I have taken lambda outside? Basically someone would say that's a lambda into x. You need to multiply premultiplication. No but lambda is a constant value. So that can be taken outside.
Now what is x? I know that a x is nothing but lambda x only. A x is nothing but lambda x. So this is lambda.
A x is nothing but lambda x. So again lambda x. So what this will become lambda square x. So what you you can see here is that a² x a² * with x is coming out to be lambda square * with x. This is what you are getting. A² * with x you are getting equals to lambda square multiply with x.
What does that mean?
There was a matrix B. You multiply that matrix B with some with some vector X.
You got the vector X again with some constant value. So this is the igon value and this is a on vector.
Say yes or no.
So what you can conclude if a was having the igon value of lambda and igon vector of x. A was having the igon value of lambda and igon vector of x. Then a² will have the igon value of lambda square and igon vector of x only. Do you understand this point or not? And this can be generalized for any n belongs to integer. Any n belongs to integer. Why it can be generalized? So let's check that out.
for a to the power 3. For a to the power 3, what will be the scenario? So what is given? Given is a x= to lambda x what I need to make? I need to make a the^ 3.
So what I will do? I will multiply with a square here. What I will do? I will multiply with a square here. So what I will get? This will become a ^ 3x. Let's check about this. So this is a square.
So a square could be written as a into ax, right? A square can be written as a into a into x. a into a into x. So what will be this a x? A X will be nothing but lambda X only. Now lambda is constant. So that can be taken outside.
So lambda square X. Now again a X could be written as lambda X only. This is already lambda square. A X could be written as lambda X only. So again lambda could be taken outside. So this will become lambda cube X. So a cube X ax given as lambda x. So a cub x I hope you understand. Multiply with a square on both of the side premultiplication. A square a square. So what is the meaning of a square? A multiply with a a multiply with a could be written as lambda x. Lambda x is taken outside lambda square a x. A x is written as lambda x. So lambda taken outside lambda qx. So a cube x is given as lambda qx.
What does that mean? This was some matrix b or this was some matrix c multiply with some matrix x. What you got? You got some constant value x. So what is this?
This is the on value. And what is this?
This is the igon value.
And what is this? This is known as vector.
Okay. So what you have understood that if there is matrix A which is having IG value of lambda and vector of X then A² will have a value of lambda and igon vector of value of lambda square and vector of X. So similarly a to the power n will have a value of lambda to the power n and vector of x only on vector will not change but on value will keep on changing just by the square. A is lambda A square lambda square A to the power N lambda to the power N.
If someone ask you for A to the power minus 1 then A to the power A is having a vector of sorry value of lambda and vector of X. Then A to the power minus 1 will have a value of lambda to the power minus 1 and vector of X only. What is lambda to the power minus 1? Lambda to the power minus 1 could be written as 1 by lambda. So if I'm saying that matrix A is having IG values of 2 2 3. So what will be the igon values of A inverse?
You will say sir 1x2 1x2 and 1x3 just the inverse of it. There you go.
Understandable.
Okay, we can prove it as well. Let's prove it. Let's prove this one. Let's prove it. So what I will do? I will add up a page. Do we have the page? This page we will need we will need to take.
So what is even? Ax equals to lambda x is given. Ax equals to lambda x is given. Right? I need to make something about a inverse. So what I do on both of the side I multiply with a inverse premultiplication of a inverse premultiplication of a inverse.
Now what this will become a inverse into a will be I. So this will become x lambda into a inverse x. Let's take this lambda here. So what I will be getting?
So from this equation what I will be getting is a inverse x I am keeping as it is. A inverse x. This a inverse x I'm keeping as it is. lambda goes downward.
So what this will become 1 by lambda into x. So what you can sense here there was a matrix a inverse I multiplied some vector x I got vector x in return multiply with some constant value which is 1x lambda. What does that mean? If a was having a lambda and x then a inverse is having 1 by lambda and x. Okay. So this goes on for a inverse as well. What about adjoint A? What about adjoint A? So, so before adjint A, let's understand another thing that again what is given given is that AX is having a value of lambda and vector of X. A is having a igon value of lambda and vector of X.
Now I need to comment about K A.
Okay, I need to comment about K A where K is a scalar where K is a scalar or constant value. K is a scalar or constant value. Understandable? So a x= to lambda x is given. What I do? K is a scalar. No. So I can multiply with k.
K into lambda. K into lambda into x directly. You can see that you are having a matrix K. You were having a matrix K which was multiply with the vector X. What you got? You got vector X in return. What you got? You got vector X in return and then K into lambda only.
You are having a matrix K multiply with vector X. You got vector X in return which is being multiplied with the factor of K into lambda. What does that mean? If A was having a IG value of lambda and vector of X then K A will have on value of K lambda and vector of X only. There you go.
So A is having a value of lambda. then k will have a value of k lambda and igon vector of x.
Understandable? So this also we have proved. What about a + i?
What about a plus i? So next again what is given? A x= to cx is given. What I need to comment on a plus some constant value ci. This is what they're asking.
No a + ci. Yes. So I need to comment on a plus ci. This is my target.
Understandable right? So what do I do?
Yes. So what do I do is that on both of the side I'm adding CX.
On both of the side what I'm doing I'm adding CX. Not CX I'm adding A plus CI is the matrix I need to make.
A x= to lambda x is given. What I'm doing? A x= to lambda x is given right.
A x= to lambda x because lambda is the igon value and x is the igon vector. Now I need to comment on a a plus ci. Right?
This is what I need to comment on where c is the scalar. Where c is the scalar.
Understandable. So what I'm doing I'm adding cx on both of the side. If I add cx on both of the side what I'm getting here I can take x as a common x as common. So will I write down a plus c?
No. A is a matrix and c is a constant value. So what I will write down a plus c. Right? And here what I'm doing here I'm taking x as common. So what I will write down here? Here I will write down lambda plus c only. Why is that? So why I'm writing down lambda plus c? Because sir why you are writing down lambda plus c sir? Because lambda is also a scalar value and C is also a scalar value.
Lambda is a scalar value only. No lambda equ= to 2, lambda equ= to 3 and c is also a scalar value. So what you can understand there was a matrix a + ci there was a matrix a + ci which was multiply with vector x. In return you got vector x only that was multiply with vector x. In return you got vector x only but that was multiply with the constant value of lambda plus c. So what you understood here if a is having a value lambda with igon vector x then a plus ci will have a on value of lambda plus c and vector is not changing which is x only.
Is that understandable to you? So this point also we have proven. So all of these things we have seen. Let's talk about adjint a. What is what is adjint a? Do you remember this thing? A into adjint a is nothing but this. So adjoint a would be nothing but so it is nothing but determinant of a a inverse.
If a inverse is having a igon value of 1 by lambda and vector of x why it will have value of 1 by lambda and vector of x that you already know because if a is having value of lambda and vector of x then a inverse will have value of 1x lambda and ig vector of x only. Then if I multiply this a inverse with a constant value I multiply this a inverse with a constant value.
What is the constant value? Determinant of a dependent of a will be a constant value. No 2 3 - 1 whatever it is it will be a constant value. Then what will happen to the vector? Have we seen it?
K. If a was the if a was the igon value a if a was having value of lambda then k was having value of k into lambda. So if a inverse is having IG value of 1 by lambda then determinant of a multiply with a inverse will have value of this and vectors remains the same. So here what you can understand if I write down the results of everyone what will be the scenario on value will be lambda square on vector will be x. What will be the scenario here? I value will be lambda to the power on vector will be x. What will be the scenario here? Ig value will be lambda to the power minus1 and igon vector would be x. What will be the scenario here? I on value is coming out to be determinant of a divided by lambda. Do you need to mug it up? Absolutely not.
Determinant of a divided by or multiply with lambda inverse. Multiply with lambda inverse and on vector of x only.
And what will be the scenario here? Ig value is coming out to be k multiply with value of lambda and vector of x.
And here it is coming out to be on value of A was lambda plus value of C was CI was C only. And this is the scenario. Is it tough to remember? Absolutely not.
Absolutely not. It is not tough to remember at all.
Is that clear to you?
Okay. So let's solve this question from this property only. We are solving this question that matrix B is given as 3 A square + 2 A - 4 I. We need to find out the determinant of B.
A is a 3 +3 matrix. Then what will be the order of B? A square means it will be a 3 + 3 matrix. This will also be a 3 + 3 matrix and I is also a 3 + 3 matrix.
Right? So determinant of sorry order of B will be 3 + 3. That means it will have three on values. Lambda 1, lambda 2 and lambda 3. It will have three igon values. Understandable. And what will be the determinant of B? That will be simply the multiplication of these values. So what I'm supposed to find out? I'm supposed to find out that on values of B. A is having a value of lambda with I vector X. A square will have on value of lambda square.
A is already given. And what will be the igon value of I? Identity matrix have value unity. What is identity matrix?
This is identity matrix. So all the igon values of identity matrix will be one only. Okay. And let's not talk about the igon vector. They are in the addition.
So forget about all those things. So basically what is your matrix B is given? B is given as 3 A² + 12 A - 4 I.
So on value of B will be equals to th of value of A² plus TW of value of A minus 4. What is value of I? That is one only.
That's it.
I value of B will be thrice of lambda A square + 12 * of lambda A - 4 into 1 value of I is 1 only. Is that clear to you? This is the thing that we did here. No, this was a square. A square will have lambda square. A inverse will have lambda^ - 1.
A + C will have lambda plus C. So, similar thing we are doing here. A square is having value of lambda square.
12 a will have the value of 12 A minus 4 I. And whatever the operation is there in between plus operation is there. plus operation I output.
Here there is minus operation. So minus operation I output.
Understandable very straightforward thing. So if lambda a equals to what are the igon values? lambda equals to 2. So if lambda a equals to 2 then what will be lambda b? So this will be 2 into 2 4 12 + 24 - 4. What answer we are getting?
32. And if lambda a= to 3. If lambda equals to 3. So what will be the lambda b? 3 into 3 9 27 + 36 - 4. So 32 + 27 that will be 59.
So what allig values you are having for matrix B? What allig values you will have? What allig values you will have?
You will have 32 as igon value and 59 as I value. Is that correct? No. If it is repeated twice then 32 is also repeated twice. So 32 multiply with 32 multiply with 9 multiply with 59. So these are the on values. So what will be the determinant of B? You can do the multiplication on your own. 32 * with 32 multiply with 59.
So what answer you are getting? 6416.
This is the answer you will get.
60 41 6.
Okay, you can do the multiplication.
This is the answer you will be getting.
Let's check back the calculation. 2 4 12 24 4 into 12 24 + 12 36 36 - 4 32 and here 3 if I put 9 27 36 27 + 36 - 4 32 32 5 27 59. Yes, do the multiplication you will get the remain of B. Very simple thing. No, very straightforward. There is nothing in this question that you don't know.
If I know the values of A, then I know the values of B as well. That will be simply 3 lambda square + 2 lambda minus 4. So what keep on putting up the value of lambda understandable? And whatever the igon vector you are having with respect to lambda let's just say with respect to lambda 2 lambda equ= to 2 we are having igon vector v_sub_1 and here we are having vector v_sub_2. So what will be the igon vector of b? Ig vector of b will also be v1 and v2 only. So this you will get with respect to igon value 32 and this you will get with respect to igon value 59. So in matrix B with respect to IG value 59 you are having IG vector B2 which is same as matrix A is having a IG value three and with respect to IG value 3 you are having vector B2.
So igon vectors are not changing. Ig vector remains the same. Ig values are simply becoming square here square and here square here simply the value n minus 4 vector are not changing. 4 a to the power n. These are the cases of a to the power n or k a multiply with some constant value. Have I talked about a transpose here?
Have I talked about a transpose?
No, I did not talk about a transpose.
Can anyone comment on the igon value of a transpose? If a is having a value of lambda, what will be the igon value of a transpose? That will also be lambda only. You know 2 3 4 5 you for finding out the ig value what you will do?
Multiply minus lambda and then you calculate the determinant. If I take the transpose of this so what if I take the transpose what will happen? Same determin I will get. No. What is the determinant of A? That is equals to the det of A transpose. Say yes or no.
Right.
So on value of A is lambda. Then on value of A transpose will also be lambda. But what about the igon vector?
What about the igon vector? This may change. This may change. Ig vector may change. Okay. So we will see this point.
We will see this point. Don't worry. But for with respect to igon value you can understand on your own. A is having a IG vector of IG value of lambda. So A transpose will also have the IG value of lambda only. There is no point in thinking too much. Right? A is having value of lambda. Then A transpose you are simply taking the transpose. It will not change the determinant. Determinant will be same. Characteristic equation will also be same. Roots of characteristic equation will also be same. That means value will be same. But on vector may change. Ig vector may change. Why is that so? Because a X equ= to lambda X will be there. No. Now if I take transpose on both of the side. If I need to make transpose then I will have to take transpose here I will have to take transpose. So if I'm taking transpose on both of the side what I will get xrpose into a transpose lambda as it is multiply with xrpose. So a transpose multiply with some xrpose is giving you xrpose multiply with the lambda. Is that statement correct? No that statement is wrong. Why it is wrong? Because a transpose a transpose should have been multiply with some constant matrix x here but x transpose is coming beforehand.
Are you able to understand what I'm saying?
Say yes or no. A transpose XRpose is coming here. It should have come here.
If it was coming later on, then I could have said that A transpose is being multiplied with some matrix which is Xrpose and some constant value Xrpose I'm getting. What does that mean? This is my igon vector. This is my igon vector and this is my igon value. But it is wrong. X transpose is coming before end. So you can you cannot comment on the vector of a transpose. Are you able to understand this point? Basically conclusion is here only. I will not teach it later on. The only conclusion is that if a is having a value of lambda and igon vector of x then a transpose will have a value of lambda only but igon vector of y which we cannot comment. There is no relation between y and x. Is that understandable to you or not?
A is having a igon value of lambda and igon vector of x then a transpose will have a igon value of lambda only. Why value of lambda? That is pretty understandable to you. If you take a transpose determinant will not change characteristic equation will not change the same value of lambda you will be getting but on vector you cannot comment. So only for a transpose you need to remember but for all of them on vector will be same. It is a very very important point. People focus on values as only but if I need to frame a good question in the examination I will frame it on igon vectors.
I will ask them what will be the igon vector corresponding to that particular value. That is say a is having a on value of two and three and they are they are saying that with respect to two you are having on vector v1 and here you are having vector v2. So now a² will be having a value of four and 9. So what will be the ig vector corresponding to value 9. So answer would be s v_sub2 only and here it will be v1. So vector will not change. Okay, corresponding igon vectors will not change but igon values will be changing and how they will be changing that also we know. So this question we have seen right now we need to study about the Kylie Hamilton theorem. Okay, Kylie Hamilton theorem. This theorem we will study but before that let's see some basic prerequisites very basic prerequisites that you already know about the equality equation. Here I will be talking about the roots, addition of roots, multiplication of roots. Here also I will talk about the roots, addition of roots and multiplication of roots. And similarly we will write down the characteristic equation for 2 +2 matrix by our conventional method. After that I will tell you a short trick and then we will study about Kylie Hamilton theorem.
Understandable. So till this point on top level we have seen these 10 properties. First we saw these 10 properties and this also you can treat as a separate property only. This is also a separate property that a is having value of lambda. Then a square is having value of lambda square. A to the power n will have the igon value of lambda to the power n. A inverse will have the IG value of lambda 1 by lambda.
Adjoint A will have the IG value of determinant of A divided divided by lambda. K will have the IG value of K into lambda. A plus CI will have the IG value of lambda plus C. All the vectors will be same. All these will be same.
Understandable and then we solve this particular question. Now we will talk about the basics of quadratic equation.
After a break, I will take a break and then I will come back and we will talk about the basics of a quadratic equation. I hope you are enjoying I hope you are enjoying the actual depth of value and igon vector. This is the actual depth otherwise I could have told on the surface level that this is how you find out the igon values on vectors independent vectors all those I would have skipped right but you are actually understanding the depth of igon value and igon vector I hope you understand let's study about basic quadratic equation after a break yeah welcome back so let's study about the quadratic equations basic quadratic equation we will see then basic cubic equation also we will see so this is your quadratic equation can I write down this equation in this manner that I'm taking a as common so if I take a as common what I will get x² + b by a + c by a equ= to 0.
Is it a quadratic equation? Yes sir. You have just taken a as common a goes to zero. This will become 0 by a that will be equal to zero only. So sir it is a quadratic equation. That means it will have two roots alpha and beta. So it will have s two roots.
What will be the addition of the roots?
So addition of the roots is given by simply this value with the negative sign and multiplication of the roots will be given by simply this value with the positive sign. I hope you know this thing in basic elementary classes you might have studied this right. This is not very tough to remember. Similarly, what I can do is that I can write down this equation as xq + b by a x² + c by a x + b by a equals to zero. Is it a cubic equation? Yes sir. How many roots it will have? Three roots. It will have alpha beta gamma.
Understandable? So what will be the addition of roots?
The addition of roots will be this value with the negative sign.
And what will be the multiplication of roots?
So here the multiplication of roots will be this is d by a right d by a but with the negative sign.
So basically when you write down the addition of roots always negative sign will come up. But in case of multiplication of the roots negative sign would come up if the maximum power is odd. And if the maximum power is even then here positive sign would come up.
Is that clear to you? We will write it down. But as of now, can you try remembering it that here negative for addition of roots always negative sign would come up. For multiplication of roots, negative sign would come up. If the maximum power is odd. If the maximum power is even then positive sign would come up. And always make sure that you will make the cofficient of maximum power 1. Then only you will write down the values. Let's just say equation is given 4xqub + 2x² + 3x + 1 =0 and you are supposed to find out how many roots it will have alpha beta gamma. So you are supposed to find out the multiplication of roots. What will be the answer? One. No the final value I need to check. But divide by 4 you doide by 4 you do divide by 4 you do. So answer would be 1x4. What is the maximum power here?
O so minus sign. So multiplication of the roots will be minus 1x4. Is that clear to you? What will be the addition of roots? Alpha plus beta plus gamma.
Addition of roots always this value but with the negative sign. - 2x4 is that clear to you understandable? So always make sure that first I will make the coefficient one here and then I will comment. So what does this represent here? C by A. So C by A represent alpha into beta plus beta into gamma plus gamma into alpha. That will be C by A.
Although it is not a matter of concern for you. You don't need to remember it.
Only thing you need to remember addition of roots and multiplication of roots.
Let us say I write down the equation in terms of power n. Okay. So next would be x ^ n minus 1. Next would be x ^ n minus 2 and it will keep on going and let's just say last constant value is d that is equals to zero. So this is the equation in power n. So what is the first thing that I will do? I will make this coefficient as 1. So I will have to divide by a.
I will have to divide by a.
I will have to divide by a. So this divide I have done. What will be the addition like how many roots it will have? It will have n root. It will have n roots right? Alpha 1, alpha 2 and total n roots it will have. So what will be the addition of roots? So alpha 1 plus alpha 2 dot dot dot dot alpha n.
Addition of roots will be this value ba with the negative sign always. Right?
And what will be the multiplication of roots? So multiplication of roots will be this value which is dy. So multiplication of roots will be s d by a but it can have minus d by a as well.
Multiplication of roots will be s last constant value. But it can have the value as minus d by a as well. When it will have minus d by a if n is odd and if n is even then it will be plus d by a. That's it. This is what we have written. So always at least remember for quadratic and cubic. If you remember for quadrant and cubic you will be able to make for power n as well. Very very very important. Okay. Always very very very important.
Only addition and multiplication is important. Why is that so? Because we are going to make a characteristic equation for the determinant for the matrix. We will make the characteristic equation the characteristic equation will be of this form only. So there I will be able to tell that this will be the trace of A and this will be the determinant of A. Right? Isn't it the trace of A and is isn't it the detent of A? So will you be able to remember whatever the equation is written write that equation down in the form such that here the maximum power is having coicient one. Then just next value that will represent the trace of A that will represent the trace. Okay simply BA with a negative sign and the last value will represent the determinant. You just need to check if the power is even or odd. If the power is even then simply the positive sign otherwise the negative sign. Understandable, straightforward.
Okay. Can you write down the characteristic equation of this 2 +2 matrix sir? If I need to write down the characteristic equation, I know that sir characteristic equation is given by this thing. So what I will do, sir, I will 2 - lambda 2 1 3 - lambda equals to 0.
Can I find out the determinant sir? Yes sir. 2 - lambda multiply with 3 - lambda - 2 = to0. If I take minus common from here from both of the end I'm taking minus common. So can I write this down as 2 - lambda into 2 lambda - 3 equals to 0 lambda minus 2 because I have taken negative sign from lambda - 3 also I have taken negative sign. So negative sign multiply with negative sign this will become positive sign right. So now just simply solve it. So lambda square - 5 lambda + 6 - 2= to0. So lambda square - 5 lambda + 4 = to 0.
Okay. Lambda square - 5 lambda + 4= to0.
This is the characteristic equation I'm getting.
Understandable. Another thing someone can say that sir I know that trace of I know that trace of a I know that trace of a I know that trace of A. What will be the trace of A? So trace of A is five. Sir I know that and I'm writing this down as trace of a sir I know the determinant of a as well determinant of a sir what is the determinant 6 - 6 - 2 that will be four so determinant of a also I know sir I know that if I'm writing down the correct equation it will be of the quadratic form why it will be of the quadratic form because this is the matrix of 2 +2 so for 2 +2 matrix you can write down your quadratic equation in this manner.
Lambda square minus trace of a multiply with lambda plus determinant of a equals to zero. So lambda square minus trace of a what is the trace of a five plus determinant of a that is 4 equals to zero. There you go. So this you need to remember. Okay.
So 4 2 +2 matrix quickly how will I write down that equation? So lambda square minus trace trace is 5 plus 3 into 2 6 6 - 2 4 equals to 0. That's it.
You will not do all this process.
Understandable? All good. And can you let me know what is this five is representing? I told you lambda 1 plus lambda 2. If this equation is having two roots, if this equation is having two roots, if this equation is having two roots which are lambda 1 and lambda 2. So addition of roots will be given by this value multiply with a negative sign. So minus 5 multiply with the another negative sign. It will become plus 5. And lambda 1 into lambda 2 what will be? That will be the last value. Now you need to check the power. Power is even that's why positive sign would come up. So lambda 1 plus lambda 2. Addition of lambda 1 and lambda 2 that means these are the ig values. No addition of values is represented by the trace only. So is it is it the trace? Yes sir.
And multiplication of the igon values is represent by the determin. So is it the det? Yes sir. This is also the determinant there.
Trace is also there. Determent is also there. So is there any issue? Absolutely not.
Say yes or no right so shortcut you have understood lambda square minus trace of a into lambda plus dment of a so can you write down now we will study about the kyl theorem we will study about it but let's just write down the characteristic equation of this matrix a can you write down the characteristic equation yes sir characteristic equation will be lambda square minus trace of a 3 + 2 5 plus determinant of a 4 equals to zero okay this was the same matrix yeah so can you write down the equation Yes. Now what Kylie Hamilton theorem will say that every square matrix satisfy its own characteristic equation. What does that mean? This is the characteristic equation is there. No. In place of lambda simply put a a² minus in place of lambda simply put a plus 4. This is a matrix. This is a matrix. So in the addition matrix should be there. No. So four alone four is a scalar value that cannot come. So 4 I equals to zero. So this will that is equals to zero matrix.
Basically lambda was a value uh lambda square was a scalar value. Lambda was a scalar value. In place of lambda I'm putting matrix A. This is your matrix A.
What I'm putting? I'm putting matrix A.
So basically you do you take the matrix A and do it square. What is the meaning of A square?
A square simply means matrix A multiply with matrix A. Then minus 5 multiply with matrix A + 4. And here I will write down matrix I. Matrix I that will be equals to zero matrix.
Okay.
So you try doing it on your own and you will get to know that yes this equation will be satisfied. What you do is that you find out a square. What will be a square? So 2213 multiply with 22 23. So what answer you will be getting? 22 2 4 6. In that manner you can do the calculation. You will get value of a square. Then you will get then you find out 5 a. Then you subtract from here and then you add 4 I from there. Okay. So you will get to know that it is coming out to be 0 0 0. So this is what it will say that whatever the characteristic equation if you are having the characteristic equation then you can simply write down this particular thing plus 4 I that will be equals to zero.
Understandable okay how it can be useful?
Can you let me know the value of a cube here? What I need to find out? I need to find out the value of a cube. Value of a cube. So what I'm doing? Let us say this is equation. On both of the side I'm multiplying with a. Premultiplication I'm doing. Premultiplication of a I'm doing. So this will become a square.
Sorry a cube. This will become a square.
This will be 4 into a into i. That will be 4 a. And a into zero will be zero only. Is that clear to you?
Premultiplication of a I have done in this equation. Premultiplication of a.
Premultiplication of a I have done.
Understandable? So this is my a cube.
What is the value of a square? From here what is the value of a square? a square is nothing but 5 a - 4 i. So a square is nothing but 5 a - 4 i + 4 a =0. So a cube - 25 a + 20 i + 4 a =0. So what is the value of a cube you are getting? a cube you are getting is 21 a minus 20 I. Is that clear to you?
This is the value of a cube you are getting. And this is the value of a square you are getting. A square was 5 a - 4 i. What is a cq? Multiply with a cube here. 5 a square 4 a. 5 a square will be 5 a - 4 i + 4 a. So 25 a + 20 I + 4 a. So this will become - 21 a + 20 I so 21 a. So this is the equation that you are getting.
Can you find out value of a to the power 4? A to the power 4. So premultiplication of a on both of the side. What I'm doing here? This is 20 I.
So again what I'm doing is that I'm doing the premultiplication of a on both of the side. So what it will become? It will become 21 A square - 20 I right 21 A square - 20 I into A. So this will become - 21 A square into A. Now what is the value of A square? So value of A square is 5 A - 4 I 5 A - 4 I - 20 A. So if you solve it what you are getting whatever that I'm getting that is not my matter of concern. I want to just prove something here.
This will be 84 minus 20 a. Okay. So this is coming out to be your a to ^ 4. So from here I can sense that a to the power 4 is coming out to be nothing but a 85 a minus 84 i. This is your a to the power 4.
So a to the power 4 is written in this manner. What is that? A I A I A A I. So we will make a conclusion but as of now you just need to see the results that this is how I can write down A to the power 4. This is how I can write down A to the power two. Understandable. This is how I can write down A to the power 2. A to the power 2 A to the power 3 A to the power 4. Okay. We will make a con conclusion. Have some patience. Let's do the same for 3 + 3 matrix. So for this 3 + 3 matrix, I got to know that this is my characteristic equation. So how did you find it out? By doing a minus lambda equals to0. So this character equation I got so that you can do on your own. So if this is the charactery equation I'm do I'm getting then I know that in place of lambda I can simply replace a this will also be satisfied. So can you write down the value of a cube here? Yes sir. Value of a cube is coming out to be value of a cube is coming out to be 3 a square minus 3 a + 3 i. This is the value of a cube you are getting. Can you find out the value of a ^ 4? Yes sir, I can find out the value of a to the^ 4.
So that would come out to be premultiplication of a I'm doing. So - 3 a square + 3 a it would be. So 3 into a cube will be 3 a² - 3 a + 3 i - 3 a² + 3 a. So 9 a² - 9 a + 3 i - 3 a² + 3 a. So 3 a square minus this will become 6 a square - 9 a + 3 a what you are getting - 6 a + 3 i this is your a to the power 4 you are getting this is the value of a to the power 4 you are getting 6 a square - 6 a + 3 i similarly what will be the value of a ^ 5 so for a ^ 5 you can multiply again 6 a cube 6 a² + 3 a here what I have done I have multiply the a right so this will become a 2 to the^ 5 this will become 6 a cube and so on so in place of a cube what you will write down 3 a² - 3 a + 3 i - 6 a² + 3 a so 18 a² - 18 a + 18 I - 6 a² + 3 a 12 a² square - 15 a + 18 I there might be some error in calculation forget about the calculation and all those things I want to prove some point here so here a to the power 3 3 a square something a the power 4 6 a square something a to the^ 5 12 a square something here a a to the power 2 was 5 a something a ^ 3 was 21 a something a 4 was 85 a something what you understood this was a 2 +2 matrix any higher higher power of a I could have written in terms of a and identity matrix. This is a 3 + 3 matrix. Any higher power of a I could have written in terms of a² a and i. Are you able to understand this point? For a 2 +2 matrix any higher power of a. Okay. Where k is greater than or equals to 2. Here I have written a a square. No a square. So that I could have written in terms of C1 A plus C2 into Y any higher power where K is greater than or equals to 3 3 + 3 matrix I could have written in terms of here it is three no so it will become A to the power 2 plus C2 into A plus C3 into I this is the like everyone would teach this Kylie Hamilton theorem this everyone would teach but I don't know why this topic is not taught.
So this is the second statement of kylton theorem. This is the first statement that lambda could be replaced with a. But one interesting observation we saw here that if I'm calculating a square or a cube or a ^ 4 one strong thing I'm noticing what I'm noticing that no matter whatever the power is that could be written in terms of a and i because it was a 2 +2 matrix. So let's just say if this was n then this was n minus one. Here the power was n minus 1 right. So similarly if this is 3 + 3 so here the power is n minus 1 as well which is 2. So for n cross n matrix what I will do a to the power k where k is greater than or equals to this value which is n. So a to the power k would be c1 a to the power this was n this was 3. So 3 would come as 2 a ^ n - 1 + c2 a ^ nus 2 and so on till cn i.
So in statement someday this particular thing could be written for kylie hamilton theorem for kylton theorem which of the following statements are true for a 2 +2 matrix a to the power k for k greater than equals to 2 it could be written as c1 a plus c2 i. What I have seen is that in a somewhere like forget about it. This is the concept.
Okay. Like the misconception is there that this particular statement is written for any kind of matrix. No, this particular statement could be written only for 2 +2 matrix only. This could be written only for 3 + 3 matrix and this could be written only for n + n matrix.
Are you able to understand this point?
Usually what has happened in the gate examination the question has always come on 2 +2 matrix. So people think that this particular statement would work for any n matrix. No.
Are you able to answer this point? Here I have got the value of a to the power 3 as 3 square - 3 a + 3 i. That's it. I cannot reduce it more. This is the maximum reduction that I can do. So here a to the power 2 would come. So if I am asking about 5 + 5. If let's just say there is 5 + 5 matrix then any higher power let's just say a ^ 9 could be written as c1 into this is 5 a ^ 4 plus c2 into a ^ 3. So let's write down this example only. So four 5 + 5 matrix 4 5 + 5 matrix a to the power 99 could be written as c1 this is five no. So here one lesser value will come plus C2 plus C3 plus C4 and plus C5. There you go. This is how we can write down.
But saying this particular thing that is wrong. So you cannot write down a ^ 5 as a to the^ 5. Sorry not a to the power 99 here a to the power 99 cannot be written in terms of c1 a plus c2 i. This is wrong. Okay this you cannot write down.
So the problem is that in all the time all the time everywhere the question has always been asked on 2 +2 matrix. So since the question has been asked on 2 +2 matrix so people have made this particular statement for any n + n matrix. But no for 3 + 3 this will be the scenario. For n cross n this will be the scenario. For 5 + 5 this will be the scenario for 2 +2 this will be the scenario whatever that n value is there here the power n minus 1 would be there okay so always remember this particular thing so in kylton theorem if you just able to write down the characteristic equation then your job is done if I'm able to write down the characteristic equation then this is how I can satisfy the condition and after that I go to know one more thing which is this particular slide that a to the power k could be written as c1 a + c2 i in terms of 2 +2 in case of 3 + 3 it could be written as c1 a² + c2 C2 A + C3 I. In case of five peroxide, it could be written as C1 A ^ 4 + C2 AQ + C3 A plus C2 AQ + C3 A² + C4 A + C5 into I and so on. Understandable? Let's solve gate question only. This is not a gate question. This is a question that is framed by me and then we will solve a gate question. So here a very rough answer you will get. You need to represent a to the power 10 in terms of a and i.
Okay. So I would go to know that it is a 2 +2 matrix. That means any higher power of a could be written as c1 a + c2 i.
Now I know one thing. What do I know that here it a I can replace with lambda? Basically if I know this equation can I write down the correct equation? Yes sir. A I can replace with lambda. What will be the value of a to the power 10? That will be lambda 2 the^ 10 c1 into lambda. Basically wherever you see a there you can replace lambda.
A to the power 10 lambda to the^ 10 a lambda C1 C2 into what is the ig value of 1 that is one only. Sorry what is the value of identity matrix that is one only. So I can write down the equation that lambda to the power 10 is nothing but c1 lambda to the power c1 lambda to the^ 1 + c2. This is something that I can write down.
Is that clear to you?
Say yes or no. Now what are the igon values of this particular matrix?
values. Can you find out the igon values? Yes sir, we can find out the values. So lambda square what what is the trace? Five lambda what is the detent? Four. If we solve what we will get four and 1, right? 4 16 21 1. Yes. So lambda value you will get 1 and four. So these are the on values.
Now if I put one here 1 would be equals to 1 + c2, right? And if I put four here so 4 ^ 10 would be equals to 4 C1 + C2.
Why I did so? I did so to find out the value of C1 and C2 because I need two equations. There are two variables. C1 and C2 are the variables. I need two equation. I got to know that my values are coming 1 and four. That means they will satisfy this equation because this is the characteristic equation. So they will satisfy here. So I put one here one then I put one here 1 equ= to C1 plus C2. Then here 4 to the^ 10 C1 plus A2.
If I subtract both of the equations what I will get 3 C1 would be equals to 4 ^ 10 minus 1.
So value of C1 is coming out to be 4 ^ 10 - 1 / 3.
Right? And what is the value of C2? You are getting C2 is 1 - C1.
1 - C1. So 1 - 4 ^ 10 - 1 / 3.
Right? So 3 - 4 ^ 10 + 1 / 3.
So what is the value of C2? You will get C1 + C2 is 1. 1 - C1. Value of C1 I have put 4 the^ 10 - 1 divid 3. So 3 I have taken here. So 3 goes here 3. And then when I open that bracket so it will become 4 the^ 10 - 1. So what is the value we are getting? 4 - 4 ^ 10 divided by 3. So this is the value of C2 we are getting. So C1 C2 we have got. Now using this equation using this particular equation I can write down the value. So A to the power 10 would be given as what is the value of C1 that is 4 ^ 10 4 ^ 10 - 1 / 3 * a + c2 is I can take 4 common from here. No. So it will become 4 into 1 - 4 ^ 9 / 3 right I have taken 4 common so this will become 1 - 4 ^ 9 / 3 * with i. So this is the value of a to the power 10 I can write out this is the answer. So this could be mentioned in one of the option that a to the power 10 could be expressed as this expression. So will you find out a to the power 10? No. What you will do? You will follow this particular procedure that's 2 +2 any higher power could be written in this format and then I can get the answer. Let's sol this. This is a gate question only. If you are able to solve it, then job done. First one is MSQ.
Next one is also MSQ or whatever you want to treat. Okay. Every question is MSQ. Every question is MSQ. Multiple answer could also be possible. Single answer could also be possible. So can you write down the correct equation? Yes sir. What will be the charact equation?
Lambda square minus trace of A. What is the trace? Minus 3. So minus minus will become plus what is the determinant? determinant will be 2 equals to zero. Did you understand this point?
Lambda square + 3 lambda + 2. Did you understand this? Absolutely sir.
Lambda square minus trace trace is - 3.
So lambda square + 3 lambda plus what is the det? Det is 2 - 3 into 0 0 - 2. So det is 2. Right?
So from here what I can get? A² + 3 a + 2 i. In place of lambda I can put a. In place of lambda I can put a. So a square + 3 a + 2 i.
Now what I can answer this I could be written as say yes or no. Right? I'm just doing the factorization nothing else. So a I can take common. So what I will get? A + 2 i. And here I can take common. So what I will get? A + 2 i. I hope you understand what I'm doing. X² + 3x + 2 what you will get?
Just solve from here. If you just solve from here, can you write down lambda + 1 into lambda + 2 equals to zero? Say right. So, similar thing I have done here. So from here what I will get a + 2 y is coming. So a + i multiply with a + 2 i is coming out to be zero. So if you write down this equation lambda + 1 into lambda + 2 equ= to 0. So in place of lambda you can put a in place of one you can put I in place of lambda you can put a in place of two you can put 2 I equals to zero the same answer you are getting right so either you do by this method which feels tougher for you so what you do is that do the factorization write down in this manner lambda + 1 into lambda + 2 then replace lambda with a so what option is matching this option is matching a plus i into a plus 2 i but other option may also match what option may also match a square + 3 a + 2 i= to0 is given. We need to find out the value of now a square + 2 a + 2 i will that be equals to z? No. This option is gone. What is a to the power a? E to the power a. What is e to the power a? Sir the power a is nothing but this we have already understood in basics. Right? So what is the expansion of a ^ x? 1 + x + x² by 2. So I + a + a square by 2 factorial plus a c by 3 factorial and so on like there is no meaning of this how it would be equals to zero no meaning of this this particular thing is equals to zero right so here nothing we can include so this option is also one let's find out here we need to find out the value of a inverse so can I find out the value of inverse what I need to do multiply a inverse on both of the side premultiplication I'm doing so this is a inverse I'm multiplying then a square is nothing but a into a + 3 a inverse into + 2 a inverse into i equals to zero. So a inverse into a will become i. I into a will become a plus a inverse into a will become 3 i plus a inverse into i will become 2 a inverse equals to zero. So a + 3 i + 2 a inverse a + 3 i + 2 a inverse equals to zero. Or in the examination they could have simply asked what is the value of a inverse. So that you will say sir it is nothing but a + 3 i divided by 2 and with the negative sign this will be the value of a inverse right you can find out a inverse from here a inverse goes here so negative sign would come up a plus b i divid by 2 so a and c are correct and how would you write down a to the power 9 it is a 2 +2 matrix so what you will do sir a to the power 9 could be written as c1 a + c2 i from here what I can sense Lambda to the power 9 could be written as C1 lambda plus C2. What is the value of lambda minus1 and - 2? So - 1 ^ 9 C1 - 1 + C2 C1 -1 is equals to - C1 + C2. Right? This is the first equation we are getting.
And then what is the value of lambda?
Lambda is minus2 as well. So if you put minus2 here in this equation if I'm putting minus2 so -2 to ^ 9 equals to c1 into -2 + c2 it is completely mathematics now there is no concept the simply mathematics you need to find out the value of c1 c2 so 2 ^ 9 is 512 minus sin - 2 c1 + c2 right if I subtract the equation 2 from 1 1 - 2 if I do so what I will get 511 1 and this will become C1. So C1 value is coming out to be 511.
Okay. And -1 - 511 + C2. That means C2 value will come out to be 510.
So C1 is 511. C2 value is 510. So a to the power 9 is coming out to be c1 is 51 a plus c2 is 510 i 510 I this is the value of a to the power 9 you are getting 511 into a 510 into i where is the question this is the answer Understandable, well and good, very straightforward.
There was another way as well like if on values of a are -1 and -2. So what will be the values of a to the^ 9? They will be -1 to the^ 9 and -2 the^ 9. So values will be -1 to the power 9 will be -2 and here this will become - 512. So a to the power 9 will have on values of minus1 and - 512. Right? These will be the ig values. These will be the ig values. Now you could have checked the option that which option is having -1 and - 512 as the value. First option what is the first option? First option is saying 5 11 a plus 510 I 510 I. So what will be its igon values?
Its igon values will be nothing but 511 lambda plus 510. This will be their ig values. Right? This will be their values. So here I need to put two values. I need to put lambda= to minus1 and lambda= to minus2 right lambda= to -1 and lambda equ= to -2 I need to put so if I put lambda= to minus1 what is the value? I'm getting lambda= to minus1. If I'm putting so on value I'm getting is minus1 and if I put lambda equ= to minus2 what value I'm getting that I'm getting is - 512 because 51 into 2 that is 2 to 10 subtracted from 510 you will be getting 512 only but with the negative sign that means the first option is getting these values -1 and - 512 and we are supposed to get these values only minus 1 and - 512 that means option A would be correct understandable here if you put you will not get the answer here also you will not get the answer here also So you will not get the answer.
Understandable?
Did you understand this question or not?
Right. A to the power n you need to find out. So it's simple by this particular method also you could have find out C1 A into C2 I in that manner you can find out or if you just want to see by the options. So in the options what you can see directly you can sense that sir if I put up the values. This is the first option.
What will be the igon values of matrix B which is equals to 511 A + 510 I. So what will be the IG value? So on value will be 511 into value of A plus 510 into value of I. I value of I is nothing but 1. So 511 lambda + 510. So put the value of lambda which is minus1 and minus2 and get the IG values which are same as the value of a to the power 9. That means a to the power 9 and this will be equivalent.
That's it.
Okay. and check rest of the options. If all three you are able to cut that means a will be the answer. Understandable?
Well and good. Let's proceed in the next question. So what this question is saying consider a non singular 22 matrix. Now there is question and question only. Okay. For very long time we are going to solve problems only now only it's all about questions. Okay. So consider non-s singular 2 +2 matrix.
Trace of A is given as four. Trace of A square is given as five. Determinant of a matrix A. You need to find out. Let's just say matrix A is there. It will have two igon values lambda 1 and lambda 2.
So what will be the igon values of A square. So that will be lambda 1 square and lambda 2 square. What is given?
Trace of A is given. What does that mean? Sir lambda 1 + lambda 2 equ= to 4 is given. Trace of a square is given. So what is the ig values? Lambda 1 square and lambda 2 square. What does that mean? Lambda 1 square + lambda 2 square= to 5 is given. If I do the square on both of the side, what I will get? a square + B square + 2 A B that will be equals to 16. What I have done? I have done the square on both of the side. So lambda 1 + lambda 2 whole square that will be lambda 1 square + lambda 2 square + 2 lambda 1 into lambda 2 that would be equals to 16. Now lambda 1 square + lambda 2 square is given as 5. So 2 lambda 1 into lambda 2. Here it is five right that is equals to 16. If you solve from here then lambda 1 into lambda 2 is coming out to be 11 by2 which is 5.5.
5.5 and what is this 5.5? 5.5 is nothing but the determinant of a. Say yes or no.
So determinant of a is nothing but 5.5.
That is the answer. Very simple question. You just need to find out the value of lambda 1 into lambda 2. 2. What is the determinant of a? So in my mind it will click that s I need the value of lambda 1 in lambda 2. If these two equations are given I will think yes I can find out the value of lambda 1 and lambda 2. How will I do it? I will just simply do the square lambda 1 square plus lambda 2 square plus 2 lambda 1 lambda 2. So this will be five here 11 11 11id 2 5.5 will be the answer there you go understandable now matrix A is given matrix B is also given I is the identity matrix you need to find out the determinant of B what is matrix B is given matrix B is given as AQ - A² - 4 A + 5 I this is your matrix B Okay.
And I is the identity matrix. You need to find out that determinant of P.
So what I will do?
Lambda B is given as lambda A square.
What I'm saying is that what I'm saying is that matrix A is having the igon values of lambda A. So and I'm saying that matrix B is having value of lambda B.
Understandable? So I'm finding the igen values of matrix B. What that would be equivalent to? So that would be equivalent to lambda AQ minus lambda A² - 4 lambda - 4 lambda A + 5. Fair enough. Now I need to find out the igon values of matrix A. I need to find out the igen values of matrix A. So how will I find out the values of matrix A? Why I need to find out the igon values? Why I need to find out the igon values?
Because if I able to find out the igen value of matrix A, if you know the igen values, if you know the igen values of matrix A, then you know the igen values of matrix B as well. And if you know the igen values of matrix B, then you know the determinant as well because simply if I do the multiplication of all the igon values, then my job would be done.
This pi sign means multiplication. So if I do the multiplication of all the igon values of matrix B, then my job would be done. So matrix A is there find out their igon values using those igon values find out the values of matrix B.
Whatever values that you are getting do the multiplication you will get the determinant of B.
Is that understandable to you? Okay. So this is what we need to do. We need to find out the igon value of matrix A. So what those values would come out to be?
Simply I can sense that on values will be 1 2 and minus 2. I values of A. We need to find out A. We need to find out. So a minus lambda i would be equated to zero.
So through the diagonal element I need to subtract lambda. So 1 - lambda 2 - lambda and - 2 - lambda - 2 - lambda 0 - 1 0 - 1 0 0 okay so what you do you do the expansion with respect to the last row right so with respect to last row 0 0 will be gone so here I can see -2 minus lambda 2 - lambda multiply with 1 - lambda.
Right? So values on values are coming out to be on value of a are coming out to be -2 2 and 1 - 2 2 and 1 -2 2 and 1.
These are the igen values of a. Now we are supposed to find out the igen values of b. So on values of b we are finding out what those ig values will be. So lambda a cube minus lambda a square.
So do the cube then negative sign four times five times. So first do the cube then do the square. First do the cube then do the square then minus 4 into -2 + 5. This is the scenario you need to do. Okay. So just take care of the calculation here. Anything that can go wrong is calculation only. So what will be the answer? - 8 - this will become - 2 square that will be 4 right + 8 + 5. So 1.
Yes. Other value would be 2 ^ 3 - 2 ^ 2 - 4 into 2 + 5. So 8 - 4 - 8 + 5 again 1. Next would be 1 ^ 3 - 1² - 4 into 1 + 5. This is 1 - 1 - 4 + 5 again 1. So on values of lambda b is coming out to be 1 1 1 only.
This is the igon values you are getting.
These are the values you are getting.
What does that mean? What will be the determinant? Retrintment will be 1 into 1 into 1.
Now, can anyone tell what will be B? Can anyone tell what will be B? B will be nothing but a identity matrix.
Why is that? So, because it is a 3 + 3 matrix. It is a 3 + 3 matrix. So, B is also a 3 + 3 matrix. Right? B is also a 3 + 3 matrix. And this is also a 3 + 3 matrix. And all the values are coming out to be one. So for what kind of matrix we get values to be one? Identity matrix. No identity matrix because identity matrix is a diagonal matrix and for the diagonal matrix diagonal elements are the on values right diagonal elements are the values. That's why it would be a identity matrix only. Is it necessary?
Is it necessary that B will be identity matrix only?
What if B is this matrix? What is B?
What if B is this matrix?
What are the values here? Is it a lower triangular matrix? It is a lower triangular matrix. Right? So what are the values? Sorry, it is one one and one only.
Right? And what is the determinant?
Determinant is also one.
So for this particular matrix as well, you are having the same values. So can you conclude that B is I3? Absolutely not. Okay. So conclusion we cannot make that B will be equals to I3. It could be anything else as well. No, it could be this matrix as well because this matrix is also having 1 as as the value and determinant is also one. So we cannot conclude what will be B. We just needed to find out the determinant of B which is coming out to be one. So how did you find out? So relation between B and A was given. Okay sir. I can I get the ig values of A? Yes sir. If I know the IG values of A then sir I will be able to determine the ig values of B as well. If I know the igen values of B then simply their multiplication would be the determinant of B. There you go.
Understandable very straightforward question. Let's go to the next one. Let A be a 7 + 7 matrix with characteristic equation. This is the characteristic equation is given. Which of the following statements are true? You need to find out that determinant and all those things determinant and trace 2 lambda to the power 7 3 lambda to the power 6 + 4 lambda to the power 4 + 7 lambda square = 16 = 0 can I conclude something? No.
First target is take this out take this two out. So if I take 2 common so lambda ^ 7 - 3x2 lambda the^ 6 + 2 lambda power 4 + 7x2 lambda ^ 6 + 16x2 8. So now I can conclude something right from here I will be able to conclude what I will be able to conclude.
Just one second mobile.
Yeah. So what I was saying what will be the trace?
Trace will be summation of roots or trace will be the summation of igon values. Summation of values and what would be what that would be equal to that would be equal to negative of this vector negative of this vector that would come out to be 1.5. And what will be the determinant? Determinant will be simply the multiplication of all the roots. Basically this will have seven roots. No lambda 1, lambda 2 and lambda 7. This will have seven roots. So what will be the multiplication?
Multiplication will simply be this value. So this value but seven is there which is old. So answer will be minus 8.
So is minus 8, trace is 1.5. Very important question and a lot of people will mess it up. They will forget this negative positive sign. So always in the summation that means trace will be 3x2 multiply with the negative sign. Just one second. Yeah sorry for the interruption actually some call came in between. Yeah. So we were solving this question. We found out the value of trace of a and we find out the value of determinant of a. How did we find out?
So sir this was our characteristic equation. We need to make sure that lambda to the power 7 the maximum power should have cofficient one. I made the cofficient one. Then I check that the very next power lambda 2 the^ 6 that is having a cofficient of minus 3x2. So negative of this coefficient will be my trace right and the very last value the very last constant value will represent my determinant but I need to see this power lambda to the^ 7 is there lambda to the^ 7 what does that mean that means it is odd power so minus sign I will put so determinant will be minus 8 and trace will be 1.5 that's the answer okay let's solve this question it is a good question actually basically we are going to make some conclusion from here so a matrix A is there it is having four values Matrix A is there. It is having four on values which are lambda 1, lambda 2, lambda 3 and lambda 4. You need to find out the characteristic polomial and characteristic equation.
What is the characteristic equation? You know that sir characteristic equation is nothing but this is equals to zero. This is the characteristic equation.
Sir, when I solve when I solve this equation, I get four roots, right? That means when I solve this particular equation, these are the four roots I will be getting. So can I write down lambda 1 - lambda 2 lambda - lambda 3 into lambda minus lambda 4.
This will be my character equation. What will be the roots of this equation? What will be the roots of this equation? So roots of this equation would be nothing but a roots here will be lambda 1, lambda 2, lambda 3 and lambda 4. So these are this is my characteristics equation. So if I have the roots or if I have the igon values can I write down the characteristic equation? Yes sir.
Lambda minus lambda 1 multiply by lambda minus lambda 2. So this will be my characteristic equation.
Can you write down the characteristics polomial sir? characteristics polomial characteristics polomial s we we used to show it by P lambda right sir we used to show it by P lambda and P p lambda used to be nothing but s this a minus lambda only right sir and sir can I write down in this manner that I'm writing down lambda minus lambda 1 basically I will remove this zero in place of zero I will write down p lambda so this will be my characteristics polomial L but my question would be is that correct? Is this answer correct?
This is your characteristic polinomial.
So this is what you have to determinant of a minus lambda I determinant of a minus lambda I this was known to be characteristic polomial and when I when you equated to zero this is your characteristic equation. This was your polomial and this was your characteristic equation. This is what you have told sir. So yeah this is what I have written when you equate it to zero let's just say my P lambda was 2 lambda + 3 this was your p lambda when I equate it to zero this is my characteristic equation this is my characteristics polomial when I equate it to zero let's take it on the next page what I'm saying is that my characteristics polomial let my characteristics polomial was 3 into lambda + 3 into lambda + 2. What will be my this was my characteristics polomial.
What will be from here? What will be my characteristics equation? So that will be nothing but p lambda equ= to0 or what you can say 3 into lambda + 3 into lambda + 2= to0. So from here what you can get lambda + 3 into lambda + 2 equals to 0. So this is your characteristic equation. This is your characteristics equation.
And what is your characteristics polomial? This is your characteristics polomial.
So is this P lambda?
Just compare.
Is this value P lambda? No. P lambda is this much.
What does that mean? It may have some constant factor as well. No. So this is something that I want to tell that here it may have some constant factor as well which is K.
So if you know the on values then you can write down the characteristic equation very easily. But if someone says write down the characteristics polomial write down the characteristics polomial then what you will see let's just say value as 1 2 3 4. So what will be the characteristics equation? The characteristics equation would have been lambda minus1 multiply with lambda - 2 multiply with lambda - 3 multiply with lambda - 4 equals to 0. What will be your character six polomial?
Lambda minus 1 multiply with lambda - 2 multiply with lambda - 3 multiply with lambda minus 4. But that value multiply with the constant factor k. Is that understandable to you or not? If you know the equation that doesn't mean you know the characteristic polomial as well because it may have some constant factor that you don't even know. So this was our characteristic polinomial because three can go outside.
But if sorry this was our characteristic equation but if I need to write down the characteristics polomial it may have some constant factor that constant factor three is there this constant factor three is there which did not appear here a very important point very very important point that using the characteristics equation you can never write down the characteristics polomial but using the characteristics polomial you can write down the characteristic equation for sure just put p lambda equals to zero you will get the characteristic equation but if I am having the characteristics equation you are writing down the characteristic polinomial but there will be a constant factor for Okay. So this is a concept building question just to solve next particular problem. This question you need to solve. A is a 4 + 4 matrix whose characteristics polinomial is given. B is defined as this. Which of the following statements are true? You need to find out the trace of A. You need to find out the determinant of A. Then you need to find out the value of A inverse.
Then you need to find out the determinant of B. This is the question that is being asked.
Is that clear to you?
Okay. So shall we get started?
characteristics polomial is given. What is given? Characteristics polomial is given. So can I get the characteristic equation? Yes sir, I can get the characteristic equation by simply putting P lambda equals to zero. So sir if I put it put it put this equals to zero what I will be getting? So sir from here I need to make sure one thing that cofficient of lambda^ 4 should be 1. So lambda to the^ 4 + 1x2 lambda square + 1x2 equals to zero. So this is my equation. From here sir how this equation could be written lambda to ^ 4 plus 0 lambda power 3 +.5 lambda square + 0 lambda +.5 = to0. This is how the equation would be written. So what will be the trace trace of a sir after lambda to the power 4 whatever the next cofficient is there whatever the next power is there that is lambda lambda lambda to the power 3. Now what is the cofficient of lambda^ 3?
That is 0. That will come with a negative sign. So minus 0 is nothing but 0 only. So what is the trace? That is zero. And what is the determinant sir?
Whatever the last constant value is there 0.5.
Then sir I need to see that then I need to see the maximum power that is even.
So positive sign would come up. So determinant of a is.5. This is also correct. Can I find out the value of a square? Yes sir. This is the characteristic polinomial that is given.
Right. So since this is given what what does that mean sir? 2 lambda 2 the^ 4 lambda square + 1= to0 is given. So what I can do sir in place of lambda I can put a right now what I can do I can multiply with a inverse on both of the sides. So if I multiply with a inverse then what it will become a inverse into a will become i. So this will become 2 a to ^ 3. This will become a only and this will become a inverse. So what is the value of a inverse? You are getting a inverse we are getting is minus 2 aq minus a. So this is the value of a inverse that we are getting. This is also correct. Find out the determinant of B. This is the question. First of all, just simple things. Find out that determinant of B.
Do it on your own. Try it on your own.
So what is B? B is nothing but 2 to the 2 2 A to the power 8 then A ^ 6 then 2 A to the power 2 A to ^ 5 plus A to ^ 4 + A ^ 3 plus 5 4 A + 3 I 4 A + 3 I this is your B right one thing I already know that this particular thing is equals to zero.
What do I know? That 2 a ^ 4 2 a ^ 4 + a to the power 2 + i is equal to zero. So can I use it somewhere? Basically I want to make this form. If I'm able to make this form here then I will be able to reduce the expression. No this expression is looking very much big. So I want to reduce it. So what I thought here is that let's take a the^ 4 common.
So what I'm getting 2 a ^ 4 plus then what I'm getting a² then I need I what do I need I can I get I from here yes sir you can get I from here plus what it is 2 a ^ 8 plus a ^ 6 plus a ^ 4 so this term I have taken this term also I have taken this term also I have taken now there is 2 a the^ 5 so I thought that if I take a ^ 4 common if I take a common why I'm taking a common because I know that inside I will get to 2 2 a ^ 4 this factor I'm considering if I take a common then I will get 2 a ^ 4 plus this will become a² + 4 i + 3 i that is coming out to be b is that clear to you or not 2 a the^ 8 a^ 6 a the^ 5 a power 4 a^ 3 4 a + 3 i so a power 4 I I have taken common. So 2 a ^ 4 + a² + i from here I have taken then 2 a ^ 5. So this is a then 2 a ^ 4 + a² + 4 i + 3. Now I know that this particular thing is zero. What I can do here?
What I mean to say is that I know that 2 a ^ 4 + a² + i is zero. What I can do about this particular thing?
4 I could be written as I + 3 I. 4 I could be written as I + 3 I. Say yes or no. So 2 A ^ 4 + A² + I. This will also be equals to zero. This will also be equals to zero. So what I can write? A ^ 4 * with 0 plus a * 0 + 3 I + 3 I. This is my B. So what I'm getting b= to a ^ 4 into 0 will be zero. This will become 3 A.
3 I plus A will become 3 A and this will be 3 I. This is the value of B you are getting. This is your matrix B. Matrix B.
Is that clear to you? This is my matrix B. So I just simply used it. I just simply used it. And this is how I reduced it. Here I thought of taking A to the^ 4A. This is the factor I'm getting. Just I used this A to the power 4 here. Then A to the A I have taken common. then 2 a to ^ 4 + a² + 4 i I was getting but I + 3 I okay so this will become 3 a + 3 i so 3 a + 3 i this was the scenario I was getting so 3 a + 3 i that is I'm getting what I need to find out I need to find out the determinant of b so determinant of b would be nothing but this value right so can I take out three if I take out three power four would come how did I write down this step I hope Hope you are able to understand how this tab came. How this step came into some matrix C determinant would be K to the power N determinant of C. This is the identity that I I'm using. So what is A here? A A A is 4 + 4. I will also be 4 + 4. Right? I is identity matrix of order 4 + 4 only. So it will also be 4 + 4. This will also be 4 + 4.
So if you are taking three outside so this a plus i will be a matrix of 4 + 4 only. So if you are taking three outside so 3 to the^ 4 will come up. So what is your target now?
Target is now what is my target? My target is finding the determinant of a plus i. This is what I need to find out. What I need to find out determinant of a + i.
So I am I am assuming that there is a new matrix c which is a plus i. Now I know one thing that a is having the on values of a is having the igon values of lambda 1 lambda 2 lambda 3 and lambda 4. Why is that so? Because it can have four values. It is 4 + 4. So it is having values of lambda 1, lambda 2, lambda 3, lambda 4. So what will be the igon values of C?
So that will be lambda + 1, lambda 2 + 1, lambda 3 + 1 and lambda 4 + 1. And what I'm supposed to find out? I am supposed to find out the determinant of C. So that will simply come out to be this value lambda 2 + 1, lambda 3 + 1 and lambda 4 + 1. So sir, what I can do is that I can solve this equation. I can solve this particular equation from there. I can get the value of lambda 1, lambda 2, lambda 3 and lambda 4. Once I get this equation, then what I will once I get these values then I will add plus one there and I will get the value of determinant of C. Then I will put it here. I will get the answer.
Try finding these values. So lambda 1, lambda 2, lambda 3, lambda 4. Solve this equation and try finding these values.
Lambda 1, lambda 2, lambda 3, lambda 4.
They will be complex.
So first you need to get the roots. Sir, if I know the roots of A, that means I I would know the roots of if I know the IG values of A, then I know the values of C as well because C is nothing but A plus I. So their values will be nothing but lambda + 1, lambda + 2, lambda + 3, sorry, lambda 1 + 1, lambda 2 + 1, lambda 3 + 1, lambda 4 + 1. So I know these values as well. I need to find out the determinant of C because what is my target? My target is determinant of C would be nothing but determinant of A plus I. So that means this is the value I need to find out. So as of now my matter of concern is that I am supposed to find out this value.
This is what I am supposed to find out.
This is my target. If I'm able to get this value then your job is done because this value is nothing but determinant of a + i. This is something I need to find out.
Now we have the characteristics equation of a right. We have characteristics polomial. We have the characteristics polomial of matrix A. What is that characteristics polinomial of matrix A?
So that is given as P lambda. Answer that is given as 2 lambda 2 the^ 4 + lambda square 2 lambda to the power 4 + lambda square + 1.
Can I write down the characteristic equation of A? Can you write down the characteristics equation of A?
Yes sir. Characteristic equation of A will be sir lambda - lambda 1 lambda - lambda 2 lambda - lambda 3 and lambda minus lambda 4 equals to zero. Say yes.
This will be the characteristic equation. Why is that? So because I'm having four roots from here can I find out from characteristic equation can I find out the characteristics polomial from characteristic equation from characteristics equation can I find out the characteristics polinomial? Someone would say that yes sir I you can find out just make sure one thing in characteristics polomial some constant value k can also come up. So sir simply this will be your answer.
Why you are writing down again again and again sir? Simply this will be your answer. This will be your characteristics of polomial.
So this will be your characteristics polomial.
My matter of concern is what will be the value of K here if I only consider this particular thing lambda multiply with lambda multiply with lambda multiply with lambda. So if I solve this equation, if I solve this equation, what I will get? If I solve this much part, I will get lambda to the power 4 then plus something lambda to the^ 3 or whatever it is, I don't know something I will get. But I am pretty sure that this will be lambda to the power 4 only because lambda is having maximum power 1 here, lambda is having maximum power 1 here, lambda, maximum power 1, maximum power 1. So if lambda multiply with lambda multiply with lambda multiply with lambda, what you will get? lambda to the power 4 and all are having cofficient of 1 1 1 1 multiply with 1 multiply with 1 multiply with 1. So lambda 2 the^ 4 you will get that means what you will get k lambda to the power 4 plus something here lambda to the power 3 something lambda to the^ 2 whatever it is I don't care this can be compared with this so from here value of k has to be two so your characteristics polomial of matrix a could be written as 2 into lambda minus lambda 1 lambda minus lambda 2 lambda minus lambda and lambda minus lambda 4. Say yes or no.
And P lambda value you already know.
This value you already know. Is that clear to you or not? The very important part is this two.
I know the roots are roots are lambda 1, lambda 2, lambda 3, lambda 4. I know the roots. Okay. If I know the roots, I know the characteristic equation as well.
This is the characteristic equation. If you know the characteristic equation, can do you know the characteristic polinomial? Yes sir. Characteristic polomial also I know. But there will be a constant factor K. sir what will be the value of that constant factor? The value of that constant factor I don't know as of now. Then what I saw that sir characteristic polomial this is how I can write down and characteristics polinomial is given in the question as well. From there I got to know that value of K will come out to be two. Why it is coming coming out to be two?
Because whatever the equation that I wrote down when I compared with the equation that is given lambda to the power 4 is having a cofficient of two.
and I am having a cofficient of K that means K is equals to 2. So this is my characteristics polomial. In this characteristics polomial if I put lambda= to minus1 if lambda= to minus1 what I will get if I if I put lambda equ= to - 1 so this will come p of minus1 that will be equated to two in place of lambda you will put minus1 minus lambda 1 -1 - lambda 2 -1 - lambda 3 -1 - lambda 4 do you understand why did I put minus one here because now I will what I will do I will take minus sign outward from here as well I will take minus sign outward from here as well minus sign outward from here as well minus sign outward. So min -1 multiply with minus1 multiply with minus1 multiply with minus one what it will become + one and if you are taking minus sign out what you are going to get from here what you are going to get we don't have any page in the end as well we don't have any page so what do we do now we are lacking the page guys so I will have to push it a bit downward okay because we don't have the page in this slide I can have maximum 50 ways only. So what is the value of P of minus sign? P of minus 1. I'm getting P of minus one. So this is two only.
Then minus sign I'm taking outward. So what I'm getting lambda 1 + 1, lambda 2 + 1, lambda 3 + 1 and lambda 4 + 1.
So that is equivalent to P of minus 1.
Can you find out the value of P of minus 1? What is P of lambda? P of lambda is nothing but 2 lambda to ^ 4 plus lambda square + 1. What will be the value of p of minus 1? sir p of minus 1.
If I put minus 1 here, this was the value only. No. Yes. So what will be the value of p of minus 1? If I put minus1 here then if I put minus1 here that is coming out to be four. So p of minus 1 is nothing but four only. So from here from this equation can you let me know the value of lambda 1 + 1 multiply with lambda 2 + 1 multiply with lambda 3 + 1 multiply with lambda 4 + 1 that will be equated to 4 divided by 2 that will be equated to two only right so this is the value that we are getting and what is this lambda + 1 and all those things let's call this some equation forget about it so lambda + 1 into lambda lambda lambda plus 1 lambda 1 + 1 plus lambda multiply with lambda 2 + 1. What is this? This is nothing but determinant of c. So this is nothing but determinant of c which is equals to determinant of a + i. So what will be the determinant of b that you are getting? So determinant of b that we are getting is s nothing but determinant of b. Determinant of b is 3 to the power 4 3 ^ 4 multiply with determinant of a + i multiply with 2. Okay. So this will be 81 81 multiply with 162. Absolutely brilliant problem. If this kind of questions are coming then hardly one out of thousand student would be able to solve it in the examination. So answer is 162. This is wrong.
Why I put 324 here? Because I know that those guys who have done very well in game mathematics, they would be able to think but they will forget this factor too. I'm pretty sure they would have forgot this. They would have forgot this factor too. What they would have done correctly polinomial they would have done lambda minus lambda 1 multiply with lambda minus lambda 2 multiply with lambda minus lambda 3 multiply with lambda minus lambda 4 then they would know that sir I need the value of lambda 1 + 1 that means in place of lambda I need to put minus 1 then they would put it and they would get the answer p of minus 1 value is 4 from here they would get p of - 1 is 4 then they will multiply 3 to the^ 4 with 4 so 81 into 4 or that would be 324 yes sir this is the answer 324 is the answer so this is what a lot of people would have done right but always always remember that's why I took this question that if I know the roots if I know the roots which are lambda 1 lambda 2 lambda 3 lambda 4 then I know the characteristics equation for sure that would be lambda minus lambda 1 multiply with lambda - lambda 2 multiply with lambda - lambda 3 multiply with lambda - lambda 4 I know the characteristics equation for sure but characteristics polomial will have a constant factor as well that constant factor could be there why is that so using this example Right? Using this example, I can sense that yes sir, here characteristics polinomial was given which was having a constant factor of three. But when I wrote down the characteristic equation then I was not getting this constant factor three. That means from this equation if I write down the characteristic polinomial from here if I write down the characteristics polomial then characteristic polomial for B would be lambda + 3 lambda + 2. But here some constant factor would come up. So how will I find out the value of this constant factor? From here I will find out that sir here what was the value given three that mean k was three. So similarly here as well in this problem as well what we did this much analysis was easy for us then we got to know that a is having four values then I can write down the characteristic equation of a the way I wrote down the characteristic equation of a from that characteristic equation can I get the characteristic polinomial yes you can get but k would be there now can I get the value of k if I compare with the actual expression actual expression is saying that lambda to the^ 4 is having a coefficient of two that means k value has to be two so this is the value of k I'm getting now my target is getting this value lambda 1 + 1 multiply with lambda 2 + 1 multiply with lambda 3 + 1 multiply with lambda 4 + 1. What does that mean? That means if I am having this equation in this equation if I put lambda equ= to minus 1 then I can take minus sign outward. Then 2 into 1 + lambda 1 1 + lambda 2 1 + lambda 1 1 + lambda 4 then this is something I can get right. This is equated to four only. Why it is equated to four? Because it is equivalent to p of minus 1. So from here this is the value that we are getting that is coming out to be 2. So 32 to the^ 4 mult* 2 is 162. That is the answer. Say yes for now.
Understandable very straightforward.
Okay. So we are just solving the problems. More questions we are going to solve. But before that let's understand a very simple concept. Very very straightforward concept that you already know. So can you write down the correct equation of this? Yes sir. Lambda square minus what is the trace? Two. What is the determinant? Two. Right? 1 - 1 determinant is two. So this will be the correct equation. Can you solve it?
Yes sir, I can solve it. What is the value of lambda? You will get - b + - root b² - 4 a c upon 2 a. Right? So 2 + - 2 j / 2. So value of lambda you are getting 1 + - j that means it is having two on values 1 + j 1 - j. Now what is the meaning of s real matrix? Real matrix means all the elements are real. All elements are real. Okay. All elements are real. This is the meaning of real matrix. Real matrix means all the elements of this matrix A are real. From there we are getting on values which are complex.
values which are complex.
These values are complex. Right? We have seen a lot of examples where values were real.
on value as a real. So if there is a real matrix having real elements then on values could be complex as well. They could be real as well. But if they are complex then they will be complex conjugate.
They will be complex conjugate. Can I say that lambda 2 is simply lambda 1 conjugate only. If you take the conjugate of this J will be replaced with minus J will be replaced with plus J. So for a real matrix for a real matrix you can get complex values surely you can get complex values but whatever the complex values that you are getting they will be in the conjugate form. So if I say that I am having a matrix A which is having three values one value is 2 + J other value is three. Can I find out the third value?
Yes sir. It is a real matrix that means third value will be simply the conjugate of this which will be 2 minus J. So these three will be the values. Is that clear to you or not? So these kind of question could be asked in the examination that only one complex value they will mention one some other value they will mention and then other thing you need to comment because if this I know this complex value which is 2 + j then other one would be 2 minus j only understandable. Now what about the complex matrix sir? Complex matrix what is the meaning of complex matrix?
Complex matrix means one of the elements should be complex. One of the elements should be complex. Here real elements are also there. Here complex elements are also there. But only one element you need complex. One of the elements should be complex. One of the element should be complex.
Only one complex element is there. No, that means it is a complex matrix. Okay.
So when I solve this the these are the igon values I'm getting. So what allig values I'm getting? What allig values I'm getting? 2 + i, 3 - i and 4. So are they in the conjugate? No conjugate. No conjugate.
So basically whatever the concept that we have studied here that they will be in the conjugate, they will be in the complex conjugate. This works only for real matrices. So what is the conclusion sir? What you are saying? Conclusion is that matrices could be real as well. It could be complex as well. If we find out the igon values, if you find out the igon values. So sir if we find out the igon values in both of the cases the s then sir values could come out to be real as well or complex as well. It could be real as well or it could be complex as well. We have seen multiple examples right here as well sir it could be real as well it could be complex as well because value came out to be real as well and complex as well.
real or complex.
But sir, here there is a special case.
Here there is a special thing that sir if it is coming out to be complex in case of real matrices if it is coming out to be complex then they will be in the conjugate.
But here it is not necessary to be conjugate. They may be conjugate. They may not be conjugate. Not necessary conjugate. Not necessary conjugate.
So here they are surely conjugate. Here they may be conjugate, may not be conjugate. Is that understandable to you or not?
Okay. So we will solve the equation related to it. Again what is the conclusion? Matrices could be real as well. Matrices could be complex as well.
Both of the matrices can have both kind of on values. It either it could be real as well. It could be complex as well.
But in case of real matrices if you are getting one complex on value then its conjugate will also be project then its conjugate will also be project. Is that clear to you? So using this question we will using this property we will solve a question but let's write down the conclusion. What is the conclusion? For real matrix if complex number a plus iota b is an igon value then its complex conjugate a minus iota b will also be igon value. For complex matrix there is no such restriction restriction on value need not appear in conjugate pairs. Very simple thing very straightforward. Say yes or no. Very straightforward. It is real matrices. If you are getting complex on values then if you are getting a plus iota b then a minus iota b will also be a value. But for complex matrix there is nothing like that.
Understandable? What is given? A is a real 3 + 3 matrix. A is a real 3 + 3 matrix. One of the igon value is 4 + 3 i. One of the igon value is 4 + 3 i. So if a is a real 3 + 3 matrix here I'm doing the solution.
Okay. If a is a real 3 + 3 matrix and how many igon values it will have s it will have lambda 1, lambda 2 and lambda 3. Three igon values it will have. So one of the igon value is given that is 4 + 3 i. Since it is a real matrix since it is a real matrix I know that other on value will be nothing but 4 - 3 I can write down i as well. I can write down j as well. Okay. Sometimes it is written as J. Sometimes it is written as I.
Don't get confused. Here I'm writing down I. Okay. So which of the following statements are true? Lambda 1 I know.
Lambda 2 I know. Do I know the lambda 3?
I don't know lambda 3 but I'm pretty sure that it will be real 101% it will be real. Why is that?
If it was complex, let's let us say if lambda 3 was 2 + 3 J, then another on value has to be there which will be 2 - 3 J. Because if it is a real matrix and you say that A plus IOTA B is present, then A minus IOTA B will also be present.
But it is not possible. Why is that so?
Because it can have only three on values. It can maximum have only three igon values. So if it can have only three values, one value I know which is 4 + 3 I then second has to be 4 - 3 I and third has to be real because if it was complex then its conjugate should also be present which is not possible.
You can write down here okay in your rough space you can write down if you feel like that I will be forgetting it but here I'm just writing down that it is real for sure 101% will be real. Why is that so? Because if it was complex then its conjugate also has to be present. But since it is a 3 + 3 matrix only three igon values are possible. So fourth igon value will not even be there. Is that understandable to you? So lambda 3 has to be real.
Okay. So first option A must have one real igon value and two complex values.
Yes. A will surely have three linearly independent vectors. Are they different?
Are all of them different? If they were same then I need to think. But if they are different lambda 1, lambda 2, lambda 3, all these are different.
Say yes or no lambda 1, lambda 2 and lambda 3 all three are different values or not 4 + 3 i is not equals to 4 - 3 I right they are different no someone would say that s they are conjugate but lambda 1 is not equals to lambda 2 they are different igon values okay so they are different and lambda 3 is also different so with respect to different igon values you will get linearly independent different igon vectors so it will surely have three linearly independent igon vectors say yes or no it will surely have three linearly independent ig vectors Understandable?
A may have repeating values. Absolutely not. And now what they have given determinant of A equals to 21.
Right? So what will be the real value?
So determinant of A is 21. Then this will be 4 + 3 I. This will be 4 - 3 I.
And the real value is lambda 3 that is 21. What is the multiplication of both of them? I hope you know. It is a square + B square. Okay. A square + B square. A square + B square. So 4 square is 16. 3 square is 9. So it will be 25. So 25 into lambda 3 equals to 21. So lambda 3 will come out to be 21 by 25.
Okay. So this will be the value.
So this option is also wrong. Is that clear to you?
So this kind of question could be asked.
3 + 3 matrix is there. One of the values they have mentioned. Other values you already know which is 4 - 3 I. And third one you know that it will be real. All three are different. This is already real. So it will be different than the complex value. Lambda 1 is not equals to lambda 2 is not equals to lambda 3 and surely it is not equals to lambda 3 right? Understandable all three are different. If all three are different then what will happen?
If all three are different then they will have three independent vectors. If they were same then I would have then I would have to think that what will be the geometric multiplicity and all. But here different values will have different vectors. I don't need to think. So they will have three linearly independent igon vectors. Can they have repeating igon values? No. And if determinant of a is given that means multiplication of lambda 1 lambda 2 lambda 3 is given. 4 + 3 i 4 - 3 i multiply by lambda 3 will be equal to 21. So value of lambda 3 is coming out to be 21 by 25. Understandable? Let's go to the next question. These two questions are there. Try solving it.
Let a be a real 5 + 5 matrix. Two of the igon values are given 4 + 3 i and 2 - 4 i. It is given that rank of a is four.
Then find the trace of a.
Try doing it on your own.
Rank of a is given four which is less than n or you can say which is less than five. From here you can understand that this will be zero. Okay. Now two of the igon values are mentioned. A can have five values. No. Two of the igon values are mentioned. 4 + 3 I then other value I already know it will be 4 - 3 I then 2 - 4 I is mentioned then other value I already know it will be 2 + 4 I now can you let me know what will be this IG value so it will be a real value if determinant is coming out to be zero what does that mean that means lambda 1 lambda 2 lambda 3 lambda 4 and some lambda 5 will make the determinant zero if the determinant is zero then you know that zero has to be one of the value no because nullity of A if a rank is four then what will be the nullity of A? Nity of A is 1 and nity of A represents the number of independent igon vectors corresponding to lambda equals to0. So if nullity is non zero if nity is greater than or equals to 1 that means lambda equals to0 is present. Lambda equ= to 0 is present. So third value is lambda equals to 0. Sorry fifth value is lambda equ= to 0. So if you know all these values what you are supposed to find out you are supposed to find out the trace of a. What is the meaning of trace of a? You need to add all of them.
2 - 4 I + 2 + 4 I. So if I add all of them what I will get 12. So answer is 12.
Okay. This is the answer. These kind of questions are asked in gate examination actually. Okay. Don't think that sir these kind of questions will not be asked. These are important with respect to gate examination. Okay. Now next question.
Let A be a 7 + 7 matrix 1 1 1 1 2 2 3 4.
What is the minimum possible number of linearly independent vectors? And what is the maximum possible number of linearly independent vectors? I think you should be able to tell what will be the answer. So 1 1 1 2 2 3 4 with respect to these same on value at least one I will have at least one linearly independent vector I will have or maximum three I will have right at least one or maximum three at least one or maximum three and here at least one or maximum three right and here one independent vector we have and here one independent vector we will have so to minimum will be four and maximum will be 3 + 3 sorry maximum will be two no why three because this value is repeated twice here arithmetic arithmetic mult Multiplicity of 1 is three. So what could be the geometric multiplicity of one? That could be that has to be less than or equals to 3.
Right? Because a m is greater than gm.
Arithmetic multiplicity of 2 is 2. So what will be the geometric multiplicity of two? That has to be less than equal to two. So minimum I can have one, maximum I can have three. So here four and here seven.
So this is repeated slice with respect to these values. Minimum I will have one independent vector. Minimum I will have one independent vector here. Surely I will have one independent vector. Here surely I will have one independent vector. So total four vectors I will have that is a minimum number and maximum linearly independent vector that I can have is seven. That's it. Is that clear to you? Okay. So this lecture is done. Next lecture we will see we will start solving the problems only. So this is the first part of values and igon vector. I hope you are enjoying the depth of the lecture. Okay. Otherwise this this particular topic could have been taught in some 2 hours. Very simple thing you could have been told. A magnitude of a sorry det of a minus lambda equals to zero. Find out the value of lambda. Find out the vector.
Chapter is done. But we are seeing the actual concept actual depth that no matter how tough problems they are framing we will be able to solve it. If they are framing this kind of problem as well then also we will be able to solve.
Okay. So let's meet in the next lecture.
There we will solve some problems and we will see some more concepts. Okay.
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