Math 489AB Exercises for Chapter 2 Fall 2008 Section 2.3 2.3.3. Let A M n (R). Then the eigenvalues of A are the roots of the characteristic polynomial p A (t). Since A is real, p A (t) is a polynomial with real coefficients. Thus, the roots of p A (t), and hence the eigenvalues of A, come in complex conjugate pairs. (See Appendix C.) 2.3.6. Suppose A, B M n are simultaneously similar to upper triangular matrices. That is, there is a nonsingular matrix S such that S 1 AS = 1 and S 1 BS = 2. Then AB BA = S 1 S 1 S 2 S 1 S 2 S 1 S 1 S 1 = S 1 2 S 1 S 2 1 S 1 = S ( 1 2 2 1 ) S 1 Therefore the eigenvalues of AB BA are the eigenvalues of 1 2 2 1. Now, let t 11,..., t nn be the diagonal elements of 1 and let s 11,..., s nn be the diagonal elements of 2. Then t 11 s 11 t 11 s 11 0 t 22 0 s 22 1 2 =...... = 0 t 22 s 22..., 0 0 t nn 0 0 s nn 0 0 t nn s nn where the s represent possibly nonzero elements. Thus the diagonal elements of 1 2 are t 11 s 11, t 22 s 22,..., t nn s nn. A similar calculation shows that the diagonal elements of 2 1 are the same. Thus, 1 2 2 1 is an upper triangular matrix with 0 s on the diagonal. Therefore, the eigenvalues of 1 2 2 1, and hence the eigenvalues of AB BA, are zero. Section 2.4 2.4.9. Let A M n, B M m and suppose A and B have no eigenvalues in common. Suppose X M n,m satisfies AX XB = 0, i.e. AX = XB. Multiply both sides of the above equation by A on the left to get Repeatedly performing this k times, we obtain A 2 X = AXB = (AX)B = (XB)B = XB 2. A k X = XB k, for any integer k. So for any polynomial p(t) = a k t k + a k 1 t k 1 + + a 1 t + a 0, [ ] p(a)x = a k A k + a k 1 A k 1 + + a 1 A + a 0 I X = a k A k X + a k 1 A k 1 X + + a 1 AX + a 0 X = a k XB k + a k 1 XB k 1 + + a 1 XB + a 0 X [ ] = X a k B k + a k 1 B k 1 + + a 1 B + a 0 I = Xp(B).
In particular this is true for the characteristic polynomial of A, which can be factored as p A (t) = (t λ 1 )(t λ 2 ) (t λ n ), where λ 1,..., λ n are the eigenvalues of A. By the Cayley-Hamilton Theorem, p A (A) = 0, so we have 0 = p A (A)X = Xp A (B). Notice that p A (B) = (B λ 1 I)(B λ 2 I) (B λ n I). Since λ i is not an eigenvalue of B, B λ i I is nonsingular for every i, thus p A (B) is nonsingular. (This is true because the eigenvalues of B are exactly those values λ for which B λi is singular.) Therefore Xp A (B) = 0 has only the zero solution X = 0. And, since AX = BX implies p A (A)X = Xp A (B), the equation AX XB = 0 has only the zero solution. Now consider the the function T (X) = AX XB. This is a function T : M m,n M n,m between vector spaces of the same dimension (mn). The above equation can be written as T (X) = 0. Moreover, for any scalars α, β, and matrices X, Y, we have T (αx + βy ) = A(αX + βy ) (αx + βy )B = αax + βay αxb βy B = α(ax XB) + β(ay Y B) = αt (X) + βt (Y ), so T is a linear transformation on M m,n. Since T (X) = 0 has only the zero solution, T is a nonsingular linear transformation, hence T (X) = C has a unique solution for every C M n,m (see section 0.5). In summary, we have shown that if A and B have no eigenvalues in common, then the equation has a unique solution for every C. AX XB = C Section 2.5 2.5.1. Proposition: A M n is normal if and only if Ax = A x for all x C n. Proof. : Suppose A is normal, i.e. A A = AA. Then for any x C n, Ax 2 = Ax, Ax = (Ax) Ax = x A Ax = x AA x = A x, A x = A x 2
Thus Ax = A x for all x C n. : Suppose Ax = A x for all x C n. Then for any x C n, x A Ax = Ax, Ax = Ax 2 = A x 2 = A x, A x = x AA x, i.e. x A Ax = x AA x. Thus it must be that A A = AA, so A is normal. (See problem 4.1.6.) 2.5.3. We know already that the eigenvalues of a Hermitian matrix are real. Thus we will show that if the eigenvalues of a normal matrix A are real, then A is Hermitian. So suppose A is normal. Then A is unitarily similar to a real diagonal matrix D. That is, there is a unitary matrix S such that S AS = D is real. Since D is real, D = D, so take the Hermitian adjoint of the above equation: D = S AS = S A S = D. Multiply on the left by S and on the right by S : so A is Hermitian. SS ASS = SS A SS A = A, 2.5.19. Proposition: Let A M n and a C. Then A is normal if and only if A + ai is normal. Proof. : Let A M n be normal and a C. Then (A + ai) (A + ai) = (A + ai)(a + ai) = A A + aa + aa + aai = AA + aa + aa + aai = (A + ai)(a + ai) = (A + ai)(a + ai), so A + ai is normal. : Suppose A + ai is normal. Then, as in the above calculation, (A + ai) (A + ai) = (A + ai)(a + ai) A A + aa + aa + aai = AA + aa + aa + aai Subtract aa + aa + aai from both sides to obtain A A = AA, thus A is normal. 2.5.20. Suppose A M n is normal and Ax = λx. Thus (A λi)x = 0, and by the previous problem, A λi is normal. Moreover, since (A λi)x = 0, problem 1 implies that (A λi) x = 0. Thus (A λi) x = (A λi)x = 0, so A x = λx.
Section 2.6 2.6.5. Let A M n and let A k be the kth iterate in the QR algorithm, A = A 0 = Q 0 R 0, and A k = Q k R k, A k+1 = R k Q k, where Q k is orthogonal and R k is upper triangular. Notice that Q k A k+1 Q k = Q k R k Q k Q k = Q k R k = A k, so A k+1 is unitarily similar to A k. By induction, each A k is unitarily similar to A. Thus, for each k there is a unitary matrix U k such that U k A k U k = A. Now suppose A k B. By the selection principle (Lemma 2.1.8), there is a subsequence U k1, U k2,... such that all the entries of U ki converge to a unitary matrix U 0 as i. For this subsequence, Take the limit of the above subsequence, Thus B is unitarily similar to A. U k i A ki U ki = A lim U k i i A ki U ki = U0 BU 0 = A.
Exercises for Chapter 3 3.1.2. For A = Section 3.1 [ ] 1 1, the characteristic polynomial is 1 1 p A (t) = det [ ] t 1 1 = (t 1) 2 1, 1 t 1 so A has eigenvalues λ 1 = 0 and λ 2 = 2. Since there are two distinct eigenvalues, A is diagonalizable, so the Jordan form of A is [ ] [ ] 0 0 2 0, or. 0 2 0 0 3 1 2 For A = 0 3 0, the matrix is triangular, so the eigenvalues are given by the diagonal 0 0 3 entries. Hence A has a single eigenvalue, λ = 3, of multiplicity 3. To determine the Jordan form we must find the dimension of the eigenspace of λ = 3. We first find the eigenvectors of A, i.e. the solutions of (A 3I)x = 0, or 0 1 2 x 1 0 0 0 x 2 = 0 x 2 + 2x 3 = 0. 0 0 0 x 3 There are therefore two LI eigenvectors associated with λ = 3, 1 0 v 1 = 0 and v 2 = 2. 0 1 There are therefore two Jordan blocks. Since n = 3 one of these is size 2 and the other is size 1. Thus the Jordan form of A is 3 1 0 0 3 0. 0 0 3 3.1.3. Let A M n be a matrix with complex entries, but with only real eigenvalues. Then A is similar to a matrix in Jordan form, i.e. A = SJS 1, where S is nonsingular and J is in Jordan form. The Jordan form J of A has the eigenvalues of A on its diagonal. The only other possible nonzero elements of J are 1 s in superdiagonal positions. Thus, since the eigenvalues of A are real, J contains only real elements. Thus, S must contain complex entries, otherwise the product SJS 1 would be real, which contradicts the assumption that A has complex entries. So the similarity matrix cannot be real.