Math 2040: Matrix Theory and Linear Algebra II Solutions to Assignment 3

Transcription

1 Math 24: Matrix Theory and Linear Algebra II Solutions to Assignment Section 2 The Characteristic Equation 22 Problem Restatement: Find the characteristic polynomial and the eigenvalues of A = Final Answer: The characteristic polynomial of A is λ 2 λ + 6 The eigenvalues of A are λ = 2 and λ 2 = 8 λ Work: det(a λi) = det = ( λ) λ 2 9 = 6 λ + λ 2 = (λ 2)(λ 8) 28 Problem Restatement: Find the characteristic polynomial and the eigenvalues of 7 2 A = 2 Final Answer: The characteristic polynomial of A is λ 2 λ + 2 The eigenvalues of A are λ = and λ 2 = 7 λ 2 Work: det(a λi) = det = (7 λ)( λ) + 4 = λ 2 λ 2 λ + 2 = (λ ) 2 24 Problem Restatement: Find the characteristic polynomial and the eigenvalues of A = Final Answer: The characteristic polynomial of A is ( λ)(λ 7)(λ + 4) (Also, although not requested, the eigenvalues of A are λ =, λ 2 = 7, and λ = ) Work: det(a λi) = det λ 2 λ Cofactor expansion along the second λ λ row gives det(a λi) = ( λ)det = ( λ)( ( λ)(2 + λ) 8) = 6 (2 + λ) ( λ)(λ 2 λ 28) = ( λ)(λ 7)(λ + 4) 22 Problem Restatement: Use the properties of determinants to show that A and A T have the same characteristic polynomial Final Answer: If is any square matrix then det() = det( T ) Therefore, we have det(a λi) = det((a λi) T ) = det(a T λi T ) = det(a T λi)

2 224 Problem Restatement: Show that if A and are similar, then det(a) = det() Final Answer: Suppose A and are similar Then there is an invertible matrix P such that A = P P Therefore, det(a) = det(p P ) = det(p )det()det(p ) = det(), with the final equality justified by det(p ) = /det(p ) for an invertible matrix P Section Diagonalization 6 Problem Restatement: Find the eigenvalues of A = = = P DP Final Answer: A has the eigenvalues λ = of multiplicity 2 and λ 2 = 4 of multiplicity A basis of the eigenspace of λ = is given by the first two columns of P and a basis of the eigenspace of λ 2 = 4 is given by the last column of P 2 Problem Restatement: Diagonalize A = 4 Final Answer: A = P DP where D = and P = λ Work: det(a λi) = det = (2 λ)( λ) 2 = λ 4 λ 2 λ = (λ )(λ+2) Therefore, the eigenvalues ( of A are ) λ =, ( and λ 2 = ) 2 Nul(A I) = Nul = Nul = Span{ } 4 4 /4 /4 Nul(A ( 2)I) = Nul = Nul = Span{ } = Span{ } 4 4 2

3 6 Problem Restatement: Diagonalize A = λ = 2, as given on page 2 Final Answer: A = P DP where D = and P = 2 2, assuming the eigenvalues are 2 2 Work: Nul(A 2I) = Nul = Nul = Span{ 2, 6 Nul(A I) = Nul = Nul = Span{ 2 } Problem Restatement: With A =, find an invertible matrix P 2 = P such that A = P 2 2 DP2 with D = Final Answer: Any other eigen-basis will do Thus, we can use any nonzero multiples of the columns of P to construct P 2 This works by the Diagonalization Theorem on page 2 P 2 = 2P will suffice 2 Problem Restatement: Construct a non-diagonal 2 2 matrix that is diagonalizable but not invertible Final Answer: Many examples are possible One is A =, since A = P DP where D = and P = a b Work: If A = is not invertible then it must have linearly dependent columns Therefore, the non-invertibility of A implies A = for some scalar α We require one c d a αa c αc of αa or c to be non zero, in order for A to be non-diagonal The advantage of allowing c a αa to be zero is that A will be, and this triangularity implies the eigenvalues are λ = a and λ 2 = Thus, setting a = and λ =, say, gives us A = }, non-diagonal

4 and not invertible Still, we need to see this A is diagonalizable A basis of the eigenspace corresponding to λ = is { }, and a basis of the eigenspace corresponding to λ 2 = is { } Therefore, A is diagonalizable as required Section 4 Eigenvectors and Linear Transformations 42 Problem Restatement: Let D = {d, d 2 } and = {b, b 2 } be bases for vector spaces V and W, respectively Let T : V W be a linear transformation such that T (d ) = 2b b 2 and T (d 2 ) = b + b 2 Find the matrix of T relative to D and Final Answer: 2 Work: Let a a 2 be the 2 2 matrix of T relative to D and We have P D : R 2 V, and P : R 2 W a = P T P De = P T (d ) = P (2b 2 b 2 ) = Similarly, a 2 = 44 Problem Restatement: Let = {b, b 2, b } be a basis for a vector space V and T : V R 2 be a linear transformation with the property that T (x b + x 2 b + x 2 b ) = 2x 4x 2 + x Find the matrix for T relative to and the standard basis of R x 2 + x 2 2 Final Answer: Work: If S is the standard basis of R 2 then P S = I 2 The first column of the required matrix 2 is P S T P e = I 2 T (b ) = T (b ) = T (b + b 2 + b ) = Similarly, the second column is T (b 2 ) = and the third column is T (b ) = 48 Problem Restatement: Let = {b, b 2, b } be a basis for a vector space V Find T (b 4b 2 ) when T is a linear transformation from V to V whose matrix relative to is Final Answer: T (b 4b 2 ) = 24b 2b 2 + b 4

5 Work: T (b 4b 2 ) = P T P (b 4b 2 ) = P T b 4b 2 + b = P T = P T = P 24 2 = 24b 2b 2 + b 42 Problem Restatement: Find the -matrix for the transformation x Ax when 4 A = and = {b 2, b 2 } = {, } 2 2 Final Answer: 2 Work: A = P AP = = Problem Restatement: Define T : R 2 R 2 by T (x) = Ax where A = Find a basis of R 2 such that T is diagonal Final Answer: = {, } 7 7 Work: Essentially, we are just required to diagonalize A After examining the eigenvalues and 8 the corresponding eigenspaces of A, we get P AP = D, where D = and 2 P = We have A = T 7 S = P S T P S where S = {e, e 2 } Notice P S = I 2, in which case we get T = P T P = P I 2 T I 2 P = P P S T P SP = P AP = D