# MATH 511 ADVANCED LINEAR ALGEBRA SPRING 2006

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1 MATH 511 ADVANCED LINEAR ALGEBRA SPRING 26 Sherod Eubanks HOMEWORK : 3, : : 4, 6, 12, 13, : 1, 5, 8 Section 1.1: The Eigenvalue-Eigenvector Equation Problem 3 Let A M n (R). If λ is a real eigenvalue of A with Ax λx, x C n, let x ξ + iη, where ξ, η R n are the entry-wise real and imaginary parts of x. Show that Aξ λξ and Aη λη; conclude that there is a real eigenvector of A associated with λ. Must both ξ and η be eigenvectors of A? Can there be a real eigenvector associated with a complex non-real eigenvalue? Solution. First, if A M n (R), and λ is a real eigenvalue of A with Ax λx, x C n, then as x ξ + iη where ξ, η R n are the entry-wise real and imaginary parts of x, it follows that and Ax A(ξ + iη) Aξ + iaη C n λx λ(ξ + iη) λξ + iλη C n hence equating real and imaginary parts yields Aξ λξ and Aη λη, and as x, one of ξ and η is nonzero, and so there is a real eigenvector of A associated with λ. Secondly, observe that x ξ + iη implies that at least one of ξ and η is nonzero. If ξ, then clearly ξ cannot be an eigenvector of A by definition, while the assumption that x is an eigenvector remains intact (the same result follows if we assume η ). As such, at least one of ξ and η is an eigenvector of A, but this need not hold for both. Finally, to answer the third question, suppose λ is purely imaginary, and suppose η in the above, leaving ξ to be determined. Then λ iζ for some ζ R, hence as ξ R n, and Ax Aξ R n λx iζξ C n thus the only way for Ax λx, that is, Aξ iζξ, since ζ, is for ξ. Therefore, x, and so there cannot be a real eigenvector associated with a purely imaginary eigenvalue. Problem 5 A M n is called idempotent if A 2 A. Show that each eigenvalue of an idempotent matrix is either or 1. Solution. Let λ be an eigenvalue of A M n corresponding to the eigenvector x, and put p(t) t 2. Then by Theorem 1.1.6, p(λ) λ 2 is an eigenvalue of p(a) A 2 corresponding to the eigenvector 1

2 x, but as A 2 A, both λ and λ 2 correspond to the same eigenvector x, hence λ λ 2. As such, λ 2 λ λ(λ 1), for which λ or λ 1. Therefore, each eigenvalue of an idempotent matrix is either or 1. Section 1.2: The Characteristic Polynomial Problem 4 Let A M n, and let A i A({i} ) M n 1, the principal submatrix of A resulting from deleting row and column i, i 1,..., n. Show that d dt p A(t) p Ai (t). Solution. First, by Theorem , if A M n, then p A (t) t n ( 1) k+1 E k (A)t n k, so d dt p n 1 A(t) nt n 1 ( 1) k+1 (n k)e k (A)t n k 1 n 1 Furthermore, as A i M n 1, by the same theorem we have p Ai (t) t n 1 ( 1) k+1 E k (A i )t n k 1 and thus n 1 p Ai (t) nt n 1 ( 1) k+1 t n k 1 which implies that, to show the desired equality, we must show E k (A i ), (n k)e k (A) E k (A i ), k 1,..., n. Let B be any principal k k submatrix of A; that is B A({i 1,..., i k }) for some set of indices i 1,..., i k. Then it is clear from the definition of E k that detb appears only once in each E k (A i ) where i i j for j 1,..., k. That is, detb appears (n k) times in E k (A i ); hence as B is arbitrary, and since this holds for each k 1,..., n, it follows that each term of E k (A) appears (n k) times in E k (A i ), thus (n k)e k (A) E k (A i ). Therefore, it follows that d dt p A(t) p Ai (t), as desired. Section 1.3: Similarity Problem 4 Give an example of two commuting matrices that are not simultaneously diagonalizable. Does this contradict Theorem ? 2

3 Solution. Let A M 2 be the 2 2 zero matrix, and 1 B Clearly AB BA, and so A and B commute, but, observing that if x x 1, x 2 T, then x2 1 Bx x 2, and thus as σ(b) {}, where is an eigenvalue of multiplicity 2, there is only one eigenvector corresponding to the eigenvalue (up to scalar multiples), and hence by Theorem 1.3.7, B is not diagonalizable. Notice also that A is diagonal, hence A and B cannot be simultaneously diagonalizable. This does not contradict Theorem , however, since the hypothesis of the theorem requires both matrices to be diagnalizable, and hence it does not apply.. Problem 6 If A M n is diagonalizable, consider the characteristic polynomial p A (t) and show that p A (A) is the zero matrix. Solution. Let A M n be diagonalizable, and let σ(a) {λ 1,..., λ n }. Then there is a nonsingular matrix S M n (whose columns, according to Theorem 1.3.7, are the n linearly independent eigenvectors of A corresponding to λ 1,..., λ n, respectively) such that A SΛS 1 where Λ diag{λ 1,..., λ n }. Now, let p(t) a i t i be any polynomial, and observe that i p(a) p(sλs 1 ) a i (SΛS 1 ) i i i a i i products {}}{ SΛS 1 SΛS 1 SΛS 1 a i SΛ i S 1 Sp(Λ)S 1. i Note that as Λ diag{λ 1,..., λ n }, we have Λ i diag{λ i 1,..., λ i n} for i,..., n, and moreover, p(diag{λ 1,..., λ n }) diag{p(λ 1 ),..., p(λ n )}. Now, put p(t) p A (t). Then, as p A (λ k ) for k 1,..., n, it follows that p A (A) Sp A (Λ)S 1 Sp A (diag {λ 1,..., λ n })S 1 Sdiag {p A (λ 1 ),..., p A (λ n )} S 1 SS 1, as desired. Problem 12 If Λ diag{λ 1,..., λ n } M n has n distinct diagonal entries, use the ideas from the proof of Theorem to show that ΛB BΛ for some B M n if and only if B is itself diagonal (but not necessarily with distinct diagonal entries). Solution. Let Λ diag{λ 1,..., λ n } M n have n distinct diagonal entries. Note first that the if part of the implication is trivial, since diagonal matrices commute. So, let B b ij and suppose ΛB BΛ. Then it follows that λ i b ij b ij λ j, from which it follows that (λ i λ j )b ij, hence b ij whenever 3

4 λ i λ j. But, by distinctness, λ i λ j if i j, hence b ij if i j. That is, entries not on the diagonal of B are zero, hence B is itself diagonal. Observe that this says nothing regarding the values of b ii, i 1,..., n, so the diagonal entries of B may or may not be distinct. Problem 13 Suppose A M n has n distinct eigenvalues. If AB BA for some B M n, show that B is diagonalizable and that A and B are simultaneously diagonalizable. Solution. If A M n has n distinct eigenvalues, then A is diagonalizable, and so A SΛS 1 for some nonsingular S M n and diagonal matrix Λ M n, hence if AB BA for some B M n, then SΛS 1 B BSΛS 1 implies SΛS 1 BS BSΛ and so ΛS 1 BS S 1 BSΛ. Hence S 1 BS and Λ commute, thus by Problem 12, it follows that S 1 BS is diagonal. This implies that B is diagonalizable, hence as S 1 AS and S 1 BS are diagonal, it follows therefore that A and B are simultaneously diagonlizable, as desired. Problem 16 Let A, B M n, and suppose that either A or B is nonsingular. If AB is diagonalizable, show that BA is also diagonalizable. Consider A and B to show that this need not be true if both A and B are singular. Solution. Let A, B M n, and suppose first that A is nonsingular. Then if AB is diagonalizable, there is a nonsingular matrix S such that AB SΛS 1 where Λ is a diagonal matrix. Then as A is nonsingular, B A 1 SΛS 1 and so BA A 1 SΛS 1 A A 1 SΛ(A 1 S) 1. Thus as A 1 S is nonsingular, it follows that BA is also diagonalizable. The same result follows if we replace A with B in the above as nonsingular, since with the same assumptions, AB SΛS 1 implies A SΛS 1 B 1 and so BA BSΛS 1 B 1 BSΛ(BS) 1, hence BA is diagonalizable. However, if A and B are the singular matrices A 1 and B 1 1 then AB and BA A. While is a diagonal matrix, we saw in Problem 4 that A has only one (linearly independent) eigenvector, hence BA is not diagonalizable. Therefore, the result in this problem need not hold if both A and B are singular. Section 1.4: Eigenvectors Problem 1 Show that A M n has rank 1 if and only if there exist two nonzero vectors x, y C n such that A xy 4

5 Show that (a) such an A has at most one nonzero eigenvalue (of algebraic multiplicity 1); (b) this eigenvalue is y x; and (c) x is a right and y is a left eigenvector corresponding to this eigenvalue. What is the geometric multiplicity of the eigenvalue? Solution. Let A M n. Suppose first that A xy. Then if we let x x 1,..., x n T C n and y y 1,..., y n T C n be nonzero, then A y 1 x y n x; that is, the i th column of A is the vector y i x. Since each column is a scalar multiple of x, it follows that α 1 y 1 x + + α n y n x (α 1 y α n y n )x has the solution α 1 y n, α n y 1, α 2 α n 1. Since y is nonzero by assumption, we may also assume α 1 and α n (otherwise we may select different α is as we did above). Thus, αx if and only if α, hence x is a linearly independent set. Therefore, as only one column of A forms a linearly independent set, it follows that ranka 1. Conversely, suppose ranka 1. Then there is one and only one column of A that forms a linearly independent set. As such, all of the other columns of A are scalar multiples of this one column. That is, if we let the linearly independent vector be denoted by x, then assuming that x lies in the first column A, without loss of generality, A x α 2 x α n x for some scalars α i C, i 2,..., n. Then A xα T where α 1, α 2,, α n T. Since α C n, observe that we can find a y C n such that y 1, α 2,..., α n T and y 1, α 2,..., α n α T which yields A xα T xy, as desired (observe that selecting x to lie in a column other than the first yields a 1 in the corresponding row of y). For (a), observe that as ranka 1, there is exactly one vector b such that Av b has a solution. If b, then each eigenvalue of A is, for Av λv for nonzero v would imply that λv, hence λ. Now, suppose b. Then if Av λv and Aw µw for v, w nonzero, we must therefore have λ, µ nonzero, and so as Av Aw b we also have λv µw. But this implies that v µ w, and so λ Av µ Aw Aw. Thus, µ 1, and hence λ µ. As a result, if A has a nonzero eigenvalue, it is unique. λ λ Also, assuming that λ is a nonzero eigenvalue of A, observe that dimn(a) n 1 since ranka 1, hence the geometric multiplicity of the eigenvalue is n 1, and so as λ, the algebraic multiplicity of the eigenvalue is n 1 as well. Hence, the algebraic multiplicity of λ is 1 if λ is nonzero. For (b), observe that Ax (xy )x x(y x) (y x)x, hence as y x C and x, it follows that y x is an eigenvalue of A corresponding to the (right) eigenvector x. Note that y x may be zero or nonzero depending on the choice of x and y. For (c), note that y A y (xy ) (y x)y, thus y is a left eigenvector of A corresponding to y x and x is a right eigenvector corresponding to y x (as in (b)). Finally, as ranka 1, we know that 1+dimN(A) n, hence the geometric multiplicity of the eigenvalue is n 1 (as stated above). Problem 5 Consider the block triangular matrix A11 A A 12 A 22, A ii M ni, i 1, 2. Show that the eigenvalues of A are those of A 11 together with those of A 22, counting multiplicities. If x C n 1 is a right eigenvector of A 11 corresponding to λ σ(a 11 ), and if y C n 2 is a left eigenvector of x A 22 corresponding to µ σ(a 22 ), show that C n 1+n 2 is a right eigenvector and C y n 1+n 2 is a left eigenvector of A corresponding to λ and µ, respectively. What can you say about left and right eigenvectors of A corresponding to λ and µ, respectively? Can you generalize these observations to block triangular matrices with arbitrarily many diagonal blocks? 5

6 x1 Solution. First, if Ax λx for x, then if we let x where x x 1 C n 1 and x 2 C n 2 then 2 A11 A Ax 12 x1 A11 x 1 + A 12 x 2 λx1 A 22 x 2 A 22 x 2 λx 2 implies first of all that A 22 x 2 λx 2, hence λ is an eigenvalue of A 22 if x 2. If x 2, then from the above we have A 11 x 1 λx 1, and so although λ is not an eigenvalue of A 22 if x 2, for x we must have x 1, and so λ is an eigenvalue of A 11. Therefore, if λ is an eigenvalue of A then it is an eigenvalue of A 22 or A 11 (or both), hence σ(a) σ(a 11 ) σ(a 22 ). Note that the above holds counting multiplicities since the sum of the number of columns of A 11 and A 22 is equal to the number of columns of A. Now, if x C n 1 is a right eigenvector of A 11 corresponding to λ σ(a 11 ), and if y C n 2 is a left eigenvector of A 22 corresponding to µ σ(a 22 ), then x and y, hence ˆx x, T and ŷ, y T in C n 1+n 2 satisfy the following: A11 A Aˆx 12 A 22 x ŷ A y A 11 A 12 A 22 A11 x λx λˆx, y A 22 µy µŷ. Hence ˆx is a right eigenvector of A corresponding to λ and ŷ is a left eigenvector of A corresponding to µ. Note also that ŷ ˆx, even if λ and µ are not distinct. The above results can easily be extended to block triangular matrices of any size, that is, with arbitrarily many diagonal blocks. In this setting, say, if A has size n n m, it would follow that the eigenvalues of such a matrix are the same as the eigenvalues, counting multiplicities, of the matrices along the diagonal, and an eigenvalue corresponding to A ii M ni has an eigenvector, from which an eigenvector of A whose only nonzero entries lie between row n i 1 and row n i 1 + n i, can be constructed as in the above. If this eigenvector is a right eigenvector, then it is orthogonal to every left eigenvector corresponding to every other matrix A jj, j i, and vice versa. Problem 8 Further eigenvalues (and eigenvectors) can be calculated using the power method via a bridge, called deflation, which delivers a square matrix, of size 1 smaller, whose eigenvalues are the remaining eigenvalues of A M n. Let λ n and y (n) be an eigenvalue and eigenvector of A (calculated using the power method or otherwise), and let S M n be nonsingular with first column y (n). Show that S 1 λn AS A 1 and that the eigenvalues of A 1 M n 1 are λ 1,..., λ n 1 in the notation of Problem 7. Another eigenvalue may be calculated from A 1 and the deflation repeated - and so on. Solution. Let A M n and λ n, y (n) be an eigenvalue and eigenvector pair of A, and let S M n be nonsingular with first column y (n). Observe that we can write S as follows: S y (n) s 2 s n 6

7 where s j C n denotes the j th column of S for each j 2,..., n. Since Ay (n) λ n y (n), observe that AS Ay (n) As 2 As n λn y (n) As 2 As n. So, as S 1 S I n, and the first column of I n is e (n) 1 C n (the vector with a 1 in its first row and zeros elsewhere), it follows that the first column of S 1 AS is λ n e (n) 1, hence, we have found that S 1 λn AS A 1 for some matrix A 1 M n 1. Now, as we have that A is similar a block triangular matrix, and since similar matrices have the same spectra, by Problem 5 (above), it follows that σ(a) {λ n } σ(a 1 ); that is, the other eigenvalues of A are the eigenvalues of A 1. 7

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