CSIR NET MATHEMATICS JUNE 2019 | Linear Algebra | Complete Solutions of Section C

Hello friends, in this video we are going to discuss the problem of linear algebra from CSIR NET June 2019. Example number 68 from booklet C marks 4.75. Let L of Rn be the space of real linear maps from Rn to Rn. If kernel of T or null space of T then which of the following is true? option there exists T belongs to L of R5 except 0 such that range of T equal to kernel of T that does not exist T belongs to L of R5 except 0 such that range of T equal to kernel of T. Third option there exists T belongs to L of R6 except 0 such that range of T is equal to kernel of T and the fourth option that does not exist T belongs to L of R6 except 0 such that range of T equal to kernel of T so it's a problem of range and the kernel of linear transformation so first we recall range and rank of T so range of T is equal to TX X belongs to Rn say all images of T for X belongs to Rn and the dimension of range of T is called the rank of T. The kernel and nullity kernel of T is equal to X belongs to Rn such that Tx is equal to 0. So the set of all X whose image is 0 and nullity is the dimension of kernel of T and most important theorem rank nullity theorem rank of T plus nullity of T is equal to dimension of domain of transformation T and in our case domain is Rn so we have rank of T plus nullity of T equal to n. Now we have given two domain one is R5 second one is R6 so we begin with the R5 so rank of T plus nullity of T equal to n and n is equal to 5 so we have rank of T plus nullity of T equal to 5 for R5. Since in first and second option we have given range of T is equal to kernel of T so range of T is equal to kernel of T so rank of T equal to nullity of T so rank of T plus nullity of T equal to 5 so rank of T plus rank of T because rank T equal to nullity of T equal to 5 so rank of T is equal to 5 by 2 which is a rational number so which is not possible therefore we cannot find range of T and kernel of T for T belongs to L of R5 which is a option number 2 so it is correct and option number one is wrong now for R6 rank of T plus nullity of T equal to n and rank of T plus nullity of T equal to 6 they are equal therefore we have rank of T plus rank of T is equal to 6 and hence rank of T equal to 3 so nullity of T equal to 3 so dimension of range and T is 3 so if we are able to find T such that then third option is correct if we are not able to find T then fourth option is correct so there is a possibility of range of T and kernel of T equal to 3 so example we are going to define one transformation R6 to R6 such that T of x1 x2 x3 x4 x5 x6 its image is 0 0 0 x1 x2 x3 now range of T is equal to span of images of basis vectors of R6 which is 1 0 0 0 0 0 these are the standard basis vector and its images so its images for image of first standard vector 1 0 0 0 0 0 it's 0 0 0 1 0 0 0 similarly for others and for the last three basis vectors 0 0 0 1 0 0 0 0 1 0 and 0 0 0 0 0 1 image is always 0 so it's a span of 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 and 0 0 0 0 0 1 Now we are going to find the kernel x such that T of x is equal to 0. So you can easily find if first three coordinate x1 x2 x3 0 then for any other values of x4 x5 x6 we will have 0 values Therefore in generalized 0 0 0 a b c its image is always 0 and you can rewrite in another form a of 0 0 1 0 0 0 plus b of 0 0 0 1 0 0 plus c of 0 0 0 0 1 0 in other words kernel is a span of 0 0 0 1 0 0 0 0 1 0 and 0 0 0 0 0 1 therefore kernel of T and range of T are the same for this transformation and both of the dimension 3 so we are able to find T such that range of T is equal to kernel of T from L of R6 so option number 3 is correct and option number 4 is wrong example number 69 from booklet C marks 4.75 let V be a finite dimensional vector space over R and T V to V be a linear map can you always write T equal to T2 o T1 composite map from some linear map T1 V to W and T2 W to V where W is some finite dimensional vector space and such that first option both T1 and T2 are onto, second option both T1 and T2 are one to one, third option T1 onto T2 one to one and the fourth option T1 one to one T2 one to one now it's a problem of decomposition of linear map T T2 o T1 and there is a different four option for T1 and T2 note that V is a finite dimensional vector space now option 1 suppose option 1 is correct that T can be decomposed into T2 o T1 where T1 and T2 both are onto map and this is true for all T we have result for finite dimensional vector spaces composition of onto linear map is onto therefore T1 and T2 are onto map therefore its composition T2 o T1 equal to T is all onto and it is always true because we assume that T can be decomposed into T2 or T1 for all T but we are able to find some linear transformation which are not onto just for example orthogonal projection T defined from R2 to R2 T(x,y) equal to (x,0) now every element of V in co-domain has at least one u in domain such that Tu equal to v then the map is called onto map this is a definition of onto map vector (0, 1) from R2 co-domain space we cannot associate any vector u such that u equal to v because the second component is always 0 in the image part and therefore the definition says that for every v there should be u in domain space here which is not possible therefore orthogonal projection is not an onto map and therefore the first option that both T1 and T2 are onto is wrong now option number 2 suppose option 2 is correct and which is both T1 and T2 are 1 to 1 for finite dimensional vector spaces composition of 1 to 1 linear map is 1 to 1 therefore T1 is 1 to 1 T2 is 1 to 1 therefore T is also 1 to 1 map and it is true for all T therefore always T is one-to-one map counter example the same example orthogonal projection T(x,y) equal to (x,0) from R2 to R2 it is not an one-to-one map so one-to-one map means what Tu equal to Tv then u equal to v that means image of two vectors is same then vector must be same here we have two vectors u equal to 1 1 and v equal to 1 2 then their images are equal that means vector is not equal but that images are equal or in other words images are equal but the vectors may be different therefore this map is not in one-to-one map so the second option this says that T is always one-to-one which is also wrong option number three where we required T1 is on to map and T2 is one-to-one map so we restrict W equal to TV TV is a range space of V. Now we define T1 from V to W and it is T of V equal to TV. That means image of T1 is nothing but the image of. So we can easily prove that this is a on to map. Now the T2 we require one to one map and the simplest one to one map is identity map and T2 V equal to V. Now if we take the composition of T2 T1 T2 or T1 equal to T2 T1 V and image of T1 V is nothing but the TV and T2 is identity map so we will get the TV therefore T equal to T2 o T1 so we can decompose T into the composite map T2 o T1 when T1 is on to and T2 is 1 to 1 and it is true for all the linear transformation therefore option number 3 is correct now once you are able to prove the option number 3 we can easily prove option number 4 if you remember this result, result is for finite dimensional vector space V and W, the linear map T: V to W is 1 to 1 map if and only if it is onto, so it is 1 1 then it is onto map so the same map is defined for the option 3, we will work here W is equal to T: V, T1: V to W is equal to T1: V equal to T: V, it is 1 to 1 map because it is onto from this result and T: W to V equal to T2: V equal to V, it is onto map because it is a 1 1 from this result and the same map will work and we'll get the same result, therefore T equal to T2 o T1 and therefore option number 4 is correct. Example number 70, booklet C, marks 4.74, let A equal to a_ij be a 3 cross 3 complex matrix. Identify the correct statement from the following. First option determinant minus 1 power i plus j a i j is equal to determinant a or it has value minus determinant a. Third option square root minus 1 power i plus j a_ij has value determinant A, or in the fourth option it has value minus determinant A. We are given A equal to a_ij is a 3 cross 3 complex matrix. So, we can write down in the form a_11, a_12, a_13 similarly a_31, a_32, a_33 now corresponding determinant value, now we required in the first option minus 1 power i plus j a_ij determinant so every term is multiplied by minus 1 power i plus j just for example a_11 is multiplied by minus 1 power 1 plus 1 which is 2 a 1 2 is multiplied by minus 1 power 1 plus 2 is 3 a 1 3 multiplied by minus 1 power 1 plus 3 is 4 a 2 1 is multiplied by minus 1 power 2 plus 1 is 3 and similarly a 3 3 multiplied by minus 1 power 6 now we can see in the first row minus 1 power 2 is a common factor in the second row minus 1 power 3 is a common factor and in the third row minus 1 power 4 is a common factor so using the property of determinant we can write down minus 1 power 2 minus 1 power 3 minus 1 power 4 into the remaining part. Now in the remaining part second column has common factor minus 1 and the third column has common factor minus 1 power 2 so which is also take out from the determinant then we have minus 1 power 2 minus 1 power 3 minus 1 power 4 minus 1 and minus 1 power 2 and the remaining part is a 1 1 a 1 2 a 1 3 a 2 1 a 2 2 a 2 3 a 3 1 a 3 2 a 3 3 which is nothing but the determinant a now the total minus 1 power is 2 plus 3 plus 4 plus 1 plus 2 which is 12 and into determinant so minus 1 power 12 is 1 so it has value determinant a so first option is correct second option say it is minus determinant a which is wrong now for the third option we have to find the determinant value square root minus 1 power i plus j a i j but the square root minus 1 is represented by the complex number iota i power i plus j with the property i square equal to minus 1 so the corresponding determinant every value a i j multiply by i power i plus j so a 1 1 is multiplied by i power 1 plus 1 to a 1 2 multiplied by i power 1 plus 2 a 1 3 multiplied by i power 1 plus 3 which is 4 and similarly a33 multiplied by i power 3 plus 3 6. Now we can see first row has common factor i square second row has common factor i cube and third row has common factor i power 4. So now we can see in the second column i is a common factor and in the third column i square is common factor so take out from the determinant then we have i square i cube i power 4 i into i square and the remaining part is determinant a 1 1 a 1 2 a 1 3 a 2 1 a 2 2 a 2 3 a 3 1 a 3 2 a 3 3 which is nothing but the determinant a so it is i power 2 plus 3 plus 4 plus 1 plus 2 which is i power 12 into determinant a i power 4 has value 1 so 1 power 3 is 1 so remaining part is determinant a so it has value determinant a so third option says that has value determinant a which is correct and the fourth option says is determinant minus 1 which is wrong example number 71 from booklet c marks 4.75 let px is equal to a 0 plus a 1 x plus a 2 x square plus a n x power n be a non-constant polynomial of degree n greater than equal to 1. Consider the polynomials qx is equal to integration 0 to x pt dt rx is equal to d by dx px. Let we denote the real vector space of all polynomials in x then which of the following are true. First option q and r are linearly independent in v. Second option q and r are linearly dependent in v. Third option x power n belongs to linear span of q and r and the fourth option x power n plus 1 belongs to the linear span of q and r now we have given px is equal to a 0 plus a 1 x plus a 2 x square plus a n x power n is a non-constant polynomial of degree n greater than equal to 1 so a n must be non-zero qx is equal to integration 0 to x pt dt so integration 0 to x polynomial pt a 0 plus a 1 t plus a 2 t square plus a n t power n after the integration we have qx is equal to a0x plus a1x square by 2 plus a n x power n plus 1 divided by n plus 1 rx is a differentiation of p with respect to x and after the differentiation we have a1 plus 2a2x plus plus n a n x power n minus 1 now linearly dependent vectors two polynomials q and r are linearly dependent in V if Q is a multiple of R that means for some scalar alpha we can write Q equal to alpha R or in others we can write down R equal to 1 upon alpha times Q. Since Qx has degree n plus 1 and Rx has degree n minus 1 therefore Qx cannot be written in the form of alpha r or r cannot be written in the form of 1 upon alpha times q. Therefore q and r are linearly independent vectors. So the first option is correct and the second option is wrong which says that it is dependent. Now option 3 which says that x power n belongs to the linear span of q and r. So we begin with the assumption suppose x power n belongs to the linear span of q and r. so x power n can be written as a linear combination of q and r that means x power n is equal to alpha times q plus beta times r by substituting the value of q and r after comparing the coefficient of x power 0 and x power n plus 1 which is a n alpha divided by n plus 1 on the right hand side on the left hand side the coefficient of x power n plus 1 is absent so therefore it is a n alpha divided by n plus 1 equal to 0 and the coefficient of x power 0 which is constant so on the right hand side it is beta a1 on the left hand side it is absent so beta a1 equal to 0. Therefore alpha equal to 0 beta equal to 0 because a n and a 1 are not always be 0. Therefore x n power n cannot be written as a linear combination of q and r and hence x n power n does not belong to linear span of q and r. So option number 3 is wrong. Now for the option number 4 we begin with the assumption xn power n plus 1 belongs to linear span of q and r. So xn power n plus 1 can be written as a linear combination of q and r that means alpha q plus beta r. Now we compare the coefficient of x power n and x power 0. Then we have an minus 1 alpha divided by n equal to 0 because on the left hand side the coefficient of x power n is 0 on the right hand side we have a n alpha divided by n plus 1 and the coefficient of x power 0 is 0 on the left hand side on the right hand side it is beta a1 so beta a1 equal to 0 and a n minus 1 and a1 is not always be 0 therefore beta equal to 0 and alpha equal to 0 so x n power n plus 1 cannot be written as a linear combination of q and r and hence x n power n plus 1 does not belong to linear span of q and r therefore option number 4 is wrong so we have only one correct answer q and r are linearly independent Option 1. Example number 72 from booklet C marks 4.75. Let MNR be the ring of n cross n matrices over R. Which of the following are true for every n greater than equal to 2? First option, there exist matrices AB belongs to MNR such that AB minus BA is equal to identity matrix of nth order IN. Second option, AB belongs to MNR and AB equal to BA then A is diagonalizable over r if and only if b is diagonalizable over r third option a b and b have the same minimal polynomial and the fourth option a b and B have the same eigenvalues now we'll discuss one by one option first option A B minus BA is equal to identity matrix now we are taking trace on the both side trace means sum of diagonal now property of trace trace of M minus N is equal to trace of M minus trace of N so therefore trace of AB minus BA is equal to trace of AB minus trace of BA equal to n means identity matrix trace which is a sum of diagonal so sum of diagonal means 1 plus 1 plus 1 n times so trace of I equal to n now trace of AB is equal to trace of BA therefore trace of AB minus trace of BA equal to 0 and 0 is equal to n option number 1 is wrong for n greater than equal to 2 now option 2 we consider two matrices AB equal to 0 0 0 and B equal to 0 1 0 second option is about AB equal to BA then A is diagonalizable over R if and only if B is diagonalizable AB is equal to BA equal to 0 matrix so it satisfy this condition now A is 0 matrix so it is diagonalizable matrix but B is not diagonalizable matrix we can easily prove it and hence the second option is wrong which says that A is diagonalizable if and only if B is diagonalizable so we are able to find A is diagonalizable B is non-diagonalizable with the condition AB equal to 0 option number three is about the minimal polynomial so first we recall the definition of minimal polynomial minimal polynomial ma of a is the smallest monic polynomial of a such that m of a equal to zero that means any other polynomial qa equal to 0 then minimal polynomial must divide qa now we know the Cayley-Hamilton theorem which shows that characteristic polynomial pa equal to 0 therefore ma must divide pa we'll use this result now consider two matrices a equal to 0 1 0 0 and b equal to 0 0 0 1 now ab equal to 0 1 0 0 and its characteristic polynomial is determinant ab minus lambda equal to lambda square in form of ab so p of ab equal to ab square lambda replaced by ab square now by the Cayley-Hamilton theorem p of ab equal to 0 so ab square equal to 0 now we want minimal polynomials and minimal polynomial is divide characteristic polynomial so there is a two possibility for the minimal polynomial which divide characteristic polynomial ab square ab and ab square so there is a two possibility for minimal polynomial ab and ab square suppose ab then ab is non-zero because minimal polynomial will give the zero value but ab is non-zero so first option is wrong therefore minimal polynomial for ab must be ab square such that ab square equal to zero now minimal polynomial for ba so ba will be zero matrix its characteristic polynomial is P of lambda equal to lambda square and in form of BA it is BA square and by the Cayley-Hamilton theorem P of BA equal to 0 and BA square equal to 0. Now similarly minimal polynomial BA two possibility with divide the characteristic polynomial BA and BA square. Here BA equal to 0 therefore we have minimal polynomial BA is BA itself because this BA divide the characteristic polynomial as well as it is a smallest polynomial which gives zero value and hence minimal polynomial of BA is BA itself and minimal polynomial of AB is AB square so for this example the minimal polynomial of AB and BA are the different hence option number three is wrong now option number four is about the eigenvalues so we begin with the characteristic polynomial because eigenvalues are the root of the characteristic polynomial the characteristic polynomial if ab is equal to determinant ab minus lambda i if we rewrite ab into a inverse a inverse with the identity matrix and identity into ab is ab itself now lambda into identity equal to a inverse lambda is a eigenvalue is a scalar therefore we can write down a lambda a inverse now rewrite the terms a b a minus lambda i into a inverse using the property of determinant determinant of a into determinant of b a minus lambda i into determinant of a inverse determinant of a inverse is equal to 1 upon determinant determinant a cancel the determinant a and we have determinant of b a minus lambda i which is the characteristic polynomial of b so characteristic polynomial of a b is equal to characteristic polynomial of b now polynomial are equal therefore characteristic equations are equal and the eigenvalues are the root of the characteristic equation therefore we have the same root for the b and a b and hence therefore they have the same eigenvalues and option number four is correct example number 74 booklet c marks 4.75 let rn cross rn to r denote the standard inner product on rn function tw defined from rn to rn by tw v equal to v minus two times inner product v with w divided by w w into w for v belongs to rn which of the following are true first option determinant tw equal to one second option inner product tw tw v2 is equal to inner product v1 with v2 for all v1 v2 belongs to rn third option tw is equal to tw inverse and the fourth option t2w is equal to two times tw first we recall standard inner product on rn two vectors on rn u equal to u1 u2 un and v equal to v1 v2 vn so standard inner product on rn is nothing but the dot product so u1 v1 plus u2 v2 plus un vn now for n equal to 2 we are going to define function tw r2 to r2 which is tw of v equal to v minus 2 times inner product v with w divided by w with w into w here we don't have again any basis on r2 or rn so we consider the standard basis on rn so for r2 the standard basis are e1 equal to 1 0 e2 equal to 0 1 now we consider any non-zero vector 1 1 on r2 we are going to find the image of basis with respect to tw where w is equal to 1 1 so by the definition tw even equal to so here v equal to e1 so e1 minus 2 times e1 inner product w divided by w inner product w into w subtracting the value of w 1 1 and e1 equal to 1 0 we have 1 0 minus 2 times inner product 1 0 with 1 1 divided by 1 1 with 1 1 into 1 1 so inner product 1 0 with 1 1 1 and inner product 1 1 with 1 1 is 2 so 2 2 cancel into 1 1 so we have 1 0 minus 1 1 equal to 0 minus 1 so image of first basis vector e1 with respect to tw is 0 minus now similarly image of e2 with respect to tw where e2 is equal to 0 1 and w is equal to 1 1 so if we substitute the value then after the simple inner product we have image minus 1 0 for e2 with respect to tw now corresponding matrix for this transformation first column is a image of e1 with respect to tw and the second column is a second basis vector e2 is image with respect to tw so e1 image is 0 minus 1 and image of e2 is minus 1 0 and its determinant value is minus 1 so first option says that determinant value is always 1 but we are able to find the determinant value minus 1 so first option is wrong second option which is the inner product between twv1 twv2 so twv1 equal to v1 minus 2 times v1 w divided by inner product w with w into w and image of v2 with respect to tw is v2 minus 2 times v2 w divided by w inner product w into w. Now using simple properties of inner product so v1 inner product with the second component plus minus 2 times inner product v1 w divided by w w inner product into w with second component now inner product v1 with v2 minus inner product v1 with the second part where the this inner product ratio is a constant so take out from the inner product we have v1 inner product w now in the second term we are taking the inner product of first component with the v2 plus first component with the second term of the second component which is minus 2 times inner product v2 with w divided by inner product w with w into w now in the last term we have 2 inner product v1 with w divided by w with w is a constant and the same similarly 2 times inner product v2 with w divided by w is a constant so if we take the multiplication and take out from the inner product we have 4 times inner product v1 with w v2 with w divided by square of inner product w with w into inner product w with w now for the second and third term we have inner product two times v2 with w v1 with w divided by w with w so both the terms are the same always remember inner product w with v2 is equal to v2 with w so we have two equal terms coefficient are minus two minus two so it will convert into the four times inner product v2 with w v1 with w divided by w with w in the last term fourth term denominator has square of inner product w with w and the numerator has inner product w with w so one power cancelled and we have four times inner product v1 with w v2 with w divided by w with w so we can see the second and third term will give minus four times v1 inner product w v2 inner product w divided by w inner product w and the last term is a four times the same term so these terms are cancelled and we have inner product v1 with v2 and options is that inner product between tw v1 tw v2 equal to inner product between v1 and v2 is correct now we are going to discuss inverse part suppose image of any vector v with respect to tw is u so for the inverse inverse of tw of u is equal to v now twv equal to u so by the definition is v minus two times inner product v with w divided by w with w w equal to u now we take inner product both the side with w so inner product v with w and we have second term now we simplify then we have inner product V with W in the second term we have 2 times inner product V with W divided by W with W and W with W. So inner product W with W cancel with the denominator inner product W with W and we have simplified form inner product V with W minus 2 times inner product V with W is equal to U with W and we have relation inner product minus times V with W is equal to u with w and v is equal to from this relation u plus two times inner product v with w divided by w with w into w now for the inverse tw inverse u is equal to v but v equal to u plus two times inner product v with w divided by w with w into w but inner product v with w is equal to minus times in a product u with w substituting then we have t inverse u equal to u minus 2 times u with w divided by w with w into w which is the same definition of t w therefore t w is equal to t w inverse the third option is correct now we are going to check the fourth option t 2 w is equal to 2 times t w so t 2 w v equal to we replace w by 2 w then we have v minus 2 times v in a product with 2 w divided by 2 w in a product 2 w into 2 w then we have 2 into 2 into 2 so we have 8 times in a product v with w divided by 2 w with 2 w is equal to 4 times in a product w with w into w so we have v minus two times v inner product with w w inner product w into w now two times tw v equal to 2v minus four times inner product v with w divided by w with w which is not equal to t2w v so fourth option is wrong example number 75 from booklet c marks 4.75. Consider the matrix A equal to 0, 1, 1, 0 over the field Q of rational numbers. Which of the following matrices are of the form P transpose AP for the suitable 2 cross 2 invertible P over Q? Here P transpose denote the transpose of P. First option 2, 0, 0 minus 2. second option 2 0 0 2, third option 1 0 0 minus 1 and the fourth option 3 4 4 5. Now matrix A is given 0 1 1 0. We want to find matrix P which is invertible such that any one of the option can be written in the form of P transpose AP. Suppose P equal to ABCD is invertible 2 cross 2 matrix then for invertibility determinant ad minus bc should not be 0. Now p transpose ap is equal to ac bd 0 1 1 0 abcd. So after the multiplication we have 2ac ad plus bc ad plus bc and 2bd. Now we are going to check one by one option. So now p transpose ap is equal to 2 0 0 minus 2 option number one now p transpose ap equal to 2ac ad plus bc ad plus bc 2bd is equal to 2 0 0 minus 2 by comparing the components of matrix a we have 2ac equal to 2 ad plus bc equal to 0 2bd equal to minus 2 therefore ac equal to 1 ad plus bc equal to 0 bd equal to minus 1 for the invertibility of B determinant ad minus bc should not be 0 so we are going to find the value of ad minus bc so ad minus bc square equal to ad plus bc whole square minus 4 ad bc so we have the value of ad plus bc ac and bd so after the substituting the values we have ad minus bc square is equal to 4 which is possible for some p therefore p is exist and therefore the option number 1 is correct now we repeat the process for the option 2 p transpose ap equal to 2 0 0 2 then 2 ac ad plus bc ad plus bc 2 bd equal to 2 0 0 2 by comparing the elements of matrix a we have ac equal to 1 ad plus bc equal to 0 and bd equal to 1 now find the value of ad minus bc whole square so ad plus bc whole square minus 4 ad bc so we have ad plus bc equal to 0 ac equal to 1 bd equal to 1 so 4 into 1 1 therefore ad minus bc whole square equal to minus 4 so square of any number should not be negative for the real values so which is not possible therefore p does not exist and hence the option number 2 is wrong now for the option number 3 p transpose ap equal to 1 0 0 minus 1 then repeating the process we have ac equal to 1 by 2 ad plus bc equal to 0 and bd equal to minus 1 by 2 and value of ad minus bc whole square equal to 1 and therefore p can be exist for some values of a b c d and hence the option number 3 is also correct now option number 4 where p transpose a p equal to 3 4 4 5 so 2 a c a d plus b c a d plus b c 2 b d equal to 3 4 4 5 by comparing the component 2 a c equal to 3 a d plus b c equal to 4 and 2 b d equal to 5 therefore a d minus b c whole square is equal to 4 square minus 4 into 3 by 2 into 5 by 2 and we have ad minus bc whole square equal to 1 so p can be exist therefore option number 4 is correct hence this question has three correct answer 1 3 and 4 thanks for watching for more videos like share subscribe press the bell icon of Next Gen Maths