SOLUTIONS TO HOMEWORK #7, MATH 54 SECTION 001, SPRING 2012

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1 SOLUTIONS TO HOMEWORK #7, MATH 54 SECTION, SPRING JASON FERGUSON Beware of typos These may not be the only ways to solve these problems In fact, these may not even be the best ways to solve these problems Congratulations to you if you find better solutions! Ex ] 56: Find the characteristic polynomial and the eigenvalues of the matrices in Exercises A λi λ λ ( λ)(8 λ) ( 4)4 λ λ + 4 From the quadratic formula, the eigenvalues of A are: ± 6 ± 9i There are no real eigenvalues is also an acceptable answer Ex 5: Exercises 9-4 require techniques from Section Find the characteristic polynomial of each matrix, using either a cofactor expansion or the special formula for determinants described in Exercises 5-8 in Section Note: Finding the characteristic polynomial of a matrix is not easy to do with just row operations, because the variable λ is involved] 4 λ A λi 4 λ (4 λ) λ λ λ (4 λ)( λ)( λ) Ex 57: For the matrices in Exercises 5-7, list the eigenvalues, repeated according to their multiplicities Solution The eigenvalues of a triangular matrix with multiplicities are the entries along its main diagonal, with multiplicities The answer is,,,, 4 Ex 58: It can be shown that the algebraic multiplicity of an eigenvalue λ is always greater than or equal to the dimension of the eigenspace corresponding to λ Find h in the matrix A below such that the 5 6 eigenspace for λ 5 is two-dimensional: A h 5 4

2 Solution Let B equal: 6 A 5I h 4, 4 and let b,, b 4 be the columns of B Then the eigenspace for 5 is Nul B, so we want to find all h for which dim Nul B From the rank-nullity theorem, this is the same as needing dim Col B 4 Because b and b 4 are nonzero and not constant multiples of each other, {b, b 4 } is linearly independent Then dim Col B will be if and only if {b, b 4 } is a basis of Col B, if and only if {b, b 4 } spans Col B Since b is already in Span{b, b 4 }, we only need to find those h for which b is in Span Col B, ie when there are constants x, x for which: x b + x b 4 b Rewriting this in matrix form, we have: h 4 h h 6 h Therefore, there are constants x, x for which x b + x b 4 b if and only if h 6, and tracing back through gives that this is the final answer 5 Ex 5: In Exercises and, A and B are n n matrices Mark each statement True or False Justify each answer a If A is, with columns a, a, a, then det A equals the volume of the parallelepiped determined by a, a, a b det A T ( ) det A c The multiplicity of a root r of the characteristic equation of A is called the algebraic multiplicity of r as an eigenvalue of A d A row replacement operation on A does not change the eigenvalues Solution a False Let A Then det A <, and so cannot be the volume of any shape in R at all (The correct statement is that det A is the volume of the parallelepiped determined by a, a, a ) b False Let A I Then det A T det A, but ( ) det A ( ) (The correct rule is that det A T det A) c True This the definition of algebraic multiplicity as found on p8, just before Example 4 of Section 5 d False Let A, and B I Then since A is triangular, its eigenvalues are the entries on the main diagonal, ie and Likewise, is the only eigenvalue of B Therefore, A and B have different eigenvalues, even though B can be obtained from A by a single row replacement operation 6 Ex 5: In Exercises and 4, use the factorization A P DP to compute A k, where k represents an arbitrary positive integer a a (a b) b b Solution A k P D k P ] a k b k ] ] a k a k b k ] a k ] (a k b k ) b k

3 7 Ex 56: In Exercises 5 and 6, the matrix A is factored in the form P DP Use the Diagonalization Theorem to find the eigenvalues of A and a basis for each eigenspace Solution By the diagonalization theorem, the eigenvalues of A are 5 and 4 Bases for the eigenspaces of 5 and 4 are, and, respectively 8 Ex 58: Diagonalize the matricesin Exercises 7-, if possible For Exercise 8, one eigenvalue is λ 5 and one eigenvector is (,, ) 6 6 Solution Let A be the matrix in the problem statement We are given that (,, ) is an eigenvector of A We compute: ( ) Therefore, is the eigenvalue of A corresponding to (,, ) We are also given that 5 is an eigenvalue of A Since we need to find the eigenspaces of 5 and anyway, we will start by computing them: 6 4 A 5I Therefore, if (A 5I)x, then x and x are free and x 4 x + x Thus: x x 4 + x x x Therefore, λ 5 has two associated linearly independent eigenvectors, for example (4,, ) and (,, ) We also know that λ has (,, ) as an associated eigenvector This gives three linearly independent eigenvectors, so A is diagonalizable, and by the Diagonal Matrix theorem: Ex 5: In Exercises and, A, B, P, and D are n n matrices Mark each statement True or False Justify each answer (Study Theorems 5 and 6 and the examples in this section carefully before you try these exercises) a A is diagonalizable if A has n eigenvectors b If A is diagonalizable, then A has n distinct eigenvalues c If AP P D, with D diagonal, then the nonzero columns of P must be eigenvectors of A d If A is invertible, then A is diagonalizable Solution a False As was shown in Example 4, A is not diagonalizable, but it does have λ as an eigenvalue, with associated eigenvector (,, ) But then (,, ) and (,, ) are

4 also eigenvectors of A, so A has three different eigenvectors (One possible corrected statement is A is diagonalizable if A has n linearly independent ] eigenvectors ) b False Let A I Then because A is diagonal, A is diagonalizable (since A I AI) However, since A is triangular, its eigenvalues are the entries on the main diagonal, so A has only one eigenvalue (One possible corrected statement is If A has n distinct eigenvalues, then A is diagonalizable ) c True (Warning: The problem did not say that we could assume P is invertible! My solution will not assume it) Let p,, p n be the n columns of P, and let d,, d n be the diagonal entries of D Then the ith column of AP is Ap i, while the ith column of P D is d i p i Since AP P D, we have Ap i d i p i Therefore, if p i, then p i is an eigenvector of A with associated eigenvalue d i d False As was shown in Example 4, A is not diagonalizable However, its determinant is: , a and so is invertible (The matrix for any a is the simplest counterexample I could think of) a Ex 544: Let B {b, b, b } be a basis for a vector space V and T : V R be a linear transformation with the property that x 4x T (x b + x b + x b ) + 5x x + x 5 Find the matrix for T relative to B and the standard basis for R Solution For any vector v in R, the coordinate representation of v relative to the standard basis of R is v itself Therefore, the formula for the matrix of T relative to B and the standard basis for R simplifies to: T (b ) T (b ) T (b ) ] 4 5 Ex 546: Let T : P P 4 be the transformation that maps a polynomial p(t) into the polynomial p(t) + t p(t) a Find the image of p(t) t + t b Show that T is a linear transformation c Find the matrix for T relative to the bases {, t, t } and {, t, t, t, t 4 } Solution a T ( t + t ) ( t + t ) + t ( t + t ) t + t t + t 4 b Let p and q be any polynomials in P, and let c be any real scalar Then: T (p + q) (p + q) + t (p + q) (p + t p) + (q + t q) T (p) + T (q) T (cp) (cp) + t (cp) c(p + t p) ct (p), which is what it means for T to be linear c Let C be the basis {, t, t, t, t 4 } of P 4 Then the matrix of T relative to {, t, t } and C is: T ()]C T (t)] C T (t ] )] C + t ] C t + t ] C t + t 4] ] C Ex 54: Verify the statements in Exercises 9-4 The matrices are square If B is similar to A and C is similar to A, then B is similar to C 4

5 Solution Since B is similar to A, there is an invertible matrix P with P AP B Multiplying on the left by P and on the right by P gives A P BP Since C is similar to A, there is an invertible matrix Q with Q AQ C, and plugging in A P BP gives: Q P BP Q C Then P Q is invertible, and its inverse is Q (P ) Q P Therefore, (P Q) B(P Q) C, so B is similar to C Ex 54: If A is diagonalizable and B is similar to A, then B is also diagonalizable Solution Since A is diagonalizable, there is a diagonal matrix D for which D is similar to A Then by the previous exercise, D is similar to B, so B is diagonalizable 4 Ex 554: Let each ] matrix in Exercises -6 act on C Find the eigenvalues and a basis for each 5 eigenspace in C det A λi 5 λ λ (5 λ)( λ) ( ) λ 8λ + 7 Its roots, ie the eigenvalues of A are: 8 ± ± i Because 4 + i is an eigenvalue of A, A (4 + i)i must have row rank, so: i i A (4 + i)i i Therefore, {} + i {} i is a (complex) basis of the eigenspace of A corresponding to 4 + i Since A is real, is a (complex) basis of the eigenspace of A corresponding to 4 i 5 Ex 556: 4 ] 4 det A λi 4 λ 4 λ (4 λ) (λ 4) Its roots, ie the eigenvalues of A are therefore 4 ± i Therefore, { i]} { i]} i i A (4 + i)i i i i is a (complex) basis of the eigenspace of A corresponding to 4 + i Since A is real, is a (complex) basis of the eigenspace of A corresponding to 4 i 5

6 a b 6 Ex 554: In Exercises -, find an invertible matrix P and a matrix C of the form such b a that the] given matrix has the form A P CP For Exercises -6, use information from Exercises det A λi 5 λ 5 λ (5 λ)( λ) ( 5) λ 6λ + Its roots, ie the eigenvalues of A are: 6 ± 6 4 ± i Because i is an eigenvalue of A, A ( i)i must have row rank, so: A ( i)i + i 5 + i 5 + i 5 Therefore, v is a complex eigenvalue of A corresponding to λ i + i By Theorem 9, if: P Re v Im v ] 5, C, then P is invertible and A P CP, as needed 7 Ex 556: 5 Solution By my solution to Ex 554, 4 i is an eigenvalue of this matrix, with corresponding eigenvector 4 Then by Theorem 9, if P and C then P is invertible and A P CP + i 4 6