Vector space Definition and Examples

Añadido: 2026-05-09

[00:00:02]dear students today I am going to explain you about Vector spaces before explaining about Vector spaces let me recall some basic definitions in 1844 Harman Grassman gave the introduction of vector space in 1888 piano gave the definition of vector space and linear Maps you know what is a vector vector has both magnitude and Direction it is a physical quantity which has both magnitude and Direction but in mathematics vectors denoted with line segments suppose here it is p x 1 comma x 2 is any point bar is equal to R Bar Delores the line segment we represent Vector as a line segment in mathematics scalars are often real numbers it may be complex numbers also usually the elements in the field F we denote with as scalars and you know the definition of the field a field is an it is a non empty set with two binary operations plus one dot it is said to be a field f plus is a billion group and with respect to multiplication it is closed it is associative ident One belongs to F and every non-zero element has multiplicative inverse has multiplication inverse and a b is equal to ba for all a comma B belongs to F and if it satisfies distributive loss we call it as a field these are the Field properties so you already know in the degree second year once again I am recollecting the definition of the field next one more definition we need is internal composition suppose V is a non empty set and F is anonymity set f is a mapping from V cross B to V then we call this mapping as internal composition a mapping F from V cross e f to V we call it as external composition so this is the meaning of internal composition and external composition in the vector space we denote internal composition as vector addition suppose V is a non empty set you take two vectors suppose U comma V belongs to capital V then U plus v belongs to again V so V cross V to V you take any two elements in V the again the wrapping should be an element in B so here U plus b is again belongs to V so addition of two vectors is again a vector and you take a scalar and to take a vector scalar into Vector a u it should be again belongs to V that is called scalar multiplication so the internal composition in Vector space we call it as vector addition and external composition we call it as scalar multiplication so usually in the external composition we take two different states here V is one set and F is another set in internal composition we take V so the elements in V Should again map out the elements in V okay remember that next a vector space so Vector space we usually denote the symbol with v so V is a non-empty set defined over the field F and the compositions here are one is vector addition and other one is scalar multiplication so anonymity set B together these two operations is said to be a vector space if it satisfies the following properties what is the first property V plus should be abelian group so you know with respect to addition it is a billion group means it has to satisfy closure property associative property identity property inverse property with respect to addition and commutative property so with respect to addition you have to satisfy these five properties see the first one the sum of U plus r plus v bar U Bar plus v bar addition of two vectors is again a vector so we call it as the elements in V we call it as vectors the elements in the field we call it as scalars so addition of two Vector is a vector and it is a commutative property U Bar plus v bar is equal to V bar plus U Bar and this is associative property U Bar plus v bar plus W bar is equal to U Bar plus v bar plus W bar there is a zero Vector Zero Bar in B such that U Bar plus G bar is equal to U Bar for All U Bar belongs to B next for each U Bar in V there is a vector minus U Bar in v such that U Bar plus minus U Bar is equal to 0 bar and the scalar multiplication is closed that means you take n vector you take a scalar and CU bar should belongs to B so this is called scalar multiplication is closed and the other property C into U Bar plus v bar is equal to c u bar plus c v Bar C plus d into U Bar is equal to c u bar plus d u Bar C into D U Bar is equal to C D into U Bar one into U Bar is equal to U Bar if any normative set satisfies all these 10 properties we call it as a vector space so to show that any set is a vector space to show that any set is a vector space the set has to satisfy all these five properties so V plus dot is said to be a Vector space V plus is an abelian group under scalar multiplication it is closed and it has to satisfy these four properties then we can say that V is a vector space okay next I'll give some example the set V together with this definitions of vector additional scalar multiplication satisfying the about 10 exams is called a vector space V or V is a vector space some examples for example I'm taking and I'm explaining you the examples of actresses for example you all know that set of all real numbers is appeal I am taking v as Q I am asking you is Q of R is a vector space this is my question you know that set of all rational numbers is a billion group under addition so no problem Q Plus is a billion group number one number two take an element in rational number some U belongs to Q and A belongs to R my question is a u belongs to q r naught for example I will take some half in q root 3 in R my question is is root 3 by 2 belongs to Q no root 3 is a rational number root 3 by 2 does not belongs to q that means scalar multiplication is not closed scalar multiplication is not closed therefore it is not a vector space same example if I take R of Q you know that R plus is a billion group an element in r in element in u the product again belongs to R and all the remaining four properties that is C into U plus v is equal to c u plus CV C plus d into U is equal to c u plus d u C into d u is equal to C D into U 1 into U is equal to U all these properties are true in this example R of Q so you take any product of real number into rational number you will get a real number in the number system all these properties are true therefore R of Q is a vector space but Q of R is not a vector space because the product of rational into real is again real not rational number so scalar multiplication is not closed so it is not a vector space take one more example R of C this is a field you know that real numbers and complex numbers their fields so R plus is a billion group very good you know that and the product of real into complex the product of a real number into complex number is again a complex number not real for example you take three belongs to R and you take one complex number one plus I belongs to c three into one plus I that is three plus three I it belongs to C but there does not belongs to R therefore the product of a real into complex they belongs to complex not real therefore scalar multiplication is not closed scalar multiplication is not closed is not closed therefore R of C is not a vector space same example you take C of r element in R again belongs to C on all the remaining properties are satisfied in C therefore C of R is a vector space so to check whether any set is a vector space or not what you have to do you have to check all the 10 properties that is V plus is a billion group scalar multiplication is closed in v and demining these properties C into U plus v is equal to c u plus CV C plus d into U is equal to c u plus d u C into d u is equal to C D into U 1 into U is equal to U all these proper 10 properties are satisfied we can say that it is a vector space so for any problem you have to do this if one property is not satisfied then it is not a vector space so one more example show that R cube of R is a vector space this is one more question now we will check it is a vector space or not the first thing is what is r Cube R Cube means ordered Triads X comma y com sorry x 1 x 2 x 3 each x i is a real number now my question is whether it is forms a vector space or not so for that what we have to do we have to prove that it is an abelian group under addition so let let us take a vector U is x 1 x 2 x 3 let us take one more vector v y one y two y three let us take one more Vector w z one Z two Z three now first question U plus v how will you define addition in the trades this is x 1 plus y one x two plus y two x three plus Y3 so each each x i is a real number y a is a real number x i plus y a is a real number x 1 plus y 1 is a real number x two plus y two is a real number so three real numbers so it belongs to R Cube so U plus v belongs to R Cube so additionally is closed now my question is U plus V plus W so what is U plus V plus W so U Plus V plus W is equal to x 1 plus y 1 plus Z1 comma x 2 plus Y 2 plus Z2 comma x 3 plus y 3 plus Z3 so this is the addition of the vectors okay so this can be written as because they are the each one is a real number each one is a real number x i plus y i plus Z I can be written as x i plus y i plus z a this is true in the number system so we can written as x 1 plus y one plus Z1 this can be written as x 2 plus y two plus Z 2 and this can be written as x 3 plus y three plus Z3 so this is same as x 1 plus y1 comma x 2 plus Y 2 comma X3 plus y 3 Plus Z1 comma Z 2 comma Z 3 so this is nothing but U plus V plus W so U plus b plus W is equal to same as U plus V plus W associative property is true next identity property U Bar is equal to x 1 x 2 x 3 and 0 bar is equal to zero zero zero U Bar plus 0 bar is equal to x 1 plus 0 y 1 plus 0 X sorry x 2 plus 0 x 3 plus 0 that is nothing but x 1 x 2 x 3 that is equal to U Bar Next U Bar plus minus U Bar that is x 1 U minus U Bar is nothing but minus U Bar is nothing but minus x 1 minus x 2 minus x 3 so U Bar plus minus U Bar is x 1 minus x 1 x 2 minus X 2 X 3 minus x 3 that is zero zero zero that is nothing but 0 bar so U Bar plus minus U Bar is equal to 0 bar inverse property is satisfied next U plus v that is equal to x 1 plus 1 x 2 next X 1 plus x 1 plus x 2 y 1 plus y two and Z1 plus Z sorry one minute let us take that U Bar is equal to x 1 x 2 x 3 and V bar is equal to Y one y two y three U Bar plus v bar is equal to x 1 plus y 1 comma x 2 plus Y 2 comma x 3 plus y 3 that is same as y 1 plus x 1 Y 2 plus x 2 y 3 plus x 3 because they are real numbers x 1 plus y 1 is equal to Y 1 plus x 1 that is V bar plus U Bar therefore V plus is a billion closely satisfied associate is satisfied identity inverse and commutative the next thing is take an element in v that is R Cube here your V is R Cube taken take a real number c this is a field here and multiply c u bar that is c x 1 c x 2 c x three C is a real number x i is a real number so c x i is again a real number that means c x 1 is a real number c x 2 is real number c x three is the real number so they belongs to R Cube so scalar multiplication is closed scalar multiplication is closed next what is the other properties you have uh the fourth four more properties what is that U Bar is equal to x 1 x 2 x 3 so let us C comma D belongs to r c plus d into U that is C plus d into x 1 x 2 x 3 that is C plus d into X1 C plus d into X2 C plus d into x 3 that is same as c x 1 plus dx1 c x 2 plus d x 2 c x 3 plus d x 3 that is same as c x 1 c x 2 c x three plus d x 1 d x 2 d x three that is same as C into x 1 x 2 x 3 plus d into x 1 x 2 x 3 that is same as c u plus d u so C plus d into u b c u plus d u what is one more property C into U plus v so that is C into U is x 1 x 2 x 3 V is y one y two y three both belongs to R Cube C into U plus b c into x 1 plus y1 x 2 plus y two x three plus y 3 that is c x 1 plus c y 1 c x 2 plus C by 2 c x 3 plus C by 3 this can be written as c x 1 c x 2 c x 3 plus C by 1 C by 2 C by 3 that witness C into x 1 x 2 x 3 plus C into y 1 Y 2 y 3 that can be written as c u plus CV so C into U plus v is equal to c u plus CV next one more thing one more property C into d u that is C into D into x 1 x 2 x 3 that is C into d x 1 d x 2 dx3 it can be written as c d x 1 c d x 2 c d x 3 c d into x 1 x 2 x 3 that is c d into U so C into d u is equal to C D into U next one into U that is 1 into x 1 x 2 x 3 that is X1 comma X2 comma X3 that is nothing but U so all the 10 properties are satisfied because R cube with respect to addition is a billion group scalar multiplication is closed and all these four properties are true therefore R cube of R is a vector space so any problem if you ask you to prove it is a vector space what we have to do we have to check all these properties if any property at least one property is not satisfied then we can say that it is not a vector space similarly you can show that R power n r also Vector space r power N means you have n tuples x 1 x 2 x n each one is a real number so in the same manner you can show that R power n of R is also Vector space okay one more problem show that the set of all points of R square of the form three s comma 2 plus 5S is not a vector space so by showing that it is not closed underscal R multiplication so what is your H the set of all points three s comma 2 plus 5 is for example uh let us take let us take an element for example you take let s is equal to one just if I am taking an example you will get 3 2 plus 5 7 so let us take three comma 7 belongs to H let us consider a real number let C is equal to sum to a mini value that's your wish let us take C that is equal to 3 R 2 whatever it may be belongs to uh r I am asking you to find out and take it as U let us take c u what is c u nine seven threes are twenty one so my prob my question is either CU should belongs to H because in scalar multiplication should be closed so c u should belongs to h means 9 comma 21 is it belongs to h r naught that is our question so if it is belongs to H it should be in this form that means three s should be 9 and 2 plus 5S should be 21 3s is equal to 9 means s is equal to 3. for 2 plus 5S is equal to 21 means 5S is equal to 21 minus 2 that is s is equal to nineteen by five so you are getting two different values one is s is equal to 19 by five and one one is s is equal to 3 that means the CU does not belongs to H so use the elements in h should be in this form that is 3 S comma 2 plus 5S so if this 9 21 should be there in h means there should be same s value for that this element should be there but I am getting two different values s is equal to 3 and S is equal to 19 by 5 so this is not possible so it this 921 does not belongs to H this means scalar multiplication is not closed is not closed therefore H is not a vector space Vector space means scalar multiplication should be closed okay if one property is not satisfied then it is not a vector space so it is H is not a vector space take one more example C the H value is the set of all elements such that x square plus y square is less than or equal to one let us take a vector U Bar that is 0 comma 1 here 0 square plus 1 square that is 1 less than or equal to 1 Mills definitely U belongs to H let us take one more Vector 1 comma 0 1 square plus 0 square that is one this is also belong stage I have taken two vectors U Bar V bar belongs to H now U Bar plus v bar that is 0 1 plus 1 0 so what is the answer you will get 0 plus 1 1 plus 0 1 but 1 square plus 1 square that is 1 plus 1 2 so here the condition is it should be less than or equal to 1 2 is greater than 1 so U Bar plus v bar does not belongs to H that means vector addition is not closed vector addition is not closed so if a tradition is not closely means it is not an abelian group so this property is not satisfied similarly you can take a vector for example you have taken U is equal to 0 comma 1 belongs to H for example let us take C is equal to 2 belongs to a real number c u that is two into zero zero zero comma two zero square plus 2 square 4 greater than 1 c u does not belongs to H so scalar multiplication is also not closed is also not closed therefore so I I should produce an example so if you give an example that vectorization is not closed it is enough or you can give an example that scalar multiplication is not closed so if any of these properties are not satisfied then we can say that it is not a vector space so these are the some examples of vector space so tomorrow I will explain you about the [Music] um properties of vector spaces okay