Math 489AB Exercises for Chapter 1 Fall Section 1.0

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1 Math 489AB Exercises for Chapter 1 Fall 2008 Section We want to maximize x T Ax subject to the condition x T x = 1. We use the method of Lagrange multipliers. Let f(x) = x T Ax and g(x) = x T x. Then we want to maximize f(x) subject to g(x) = 1. Thus, the critical points are found by Lagrange multipliers: Note that f(x) λ g(x) = 0. so f(x) = x T Ax = x i a ij x j = a ii x 2 i + a ij x i x j i=1 j=1 i=1 i j f(x) = 2a kk x k + a kj x j + a jk x j x k j k j k = 2a kk x k + 2 j k a kj x j since A is symmetric = 2 a kj x j = (Ax) j, j=1 ( f(x) Thus f(x) =, f(x),..., f(x) ) x 1 x 2 x n = 2Ax and g(x) = x T x = x 2 i i=1 = (2x 1, 2x 2,..., 2x n ) = 2x = 2 ((Ax) 1, (Ax) 2,..., (Ax) n ) Thus the critical points are at solutions of 2(Ax λx), i.e. where Ax = λx. Moreover, at one of these critical points where x T x = 1 we have f(x) = x T Ax = x T λx = λx T x = λ, which is maximal when λ is the largest eigenvalue of A. Section If A is nonsingular and Ax = λx, then multiply both sides by A 1 : A 1 Ax = A 1 λx x = λa 1 x A 1 x = λ 1 x

2 1.1.3 Suppose A is real and λ is a real eigenvalue corresponding to a complex eigenvector, i.e. A(ξ + iη) = λ(ξ + iη). We obtain two equations by identifying the real and imaginary parts of the equation: Aξ + iaη = λξ + iλη (Aξ λξ) + i(aη λη) = 0. Since A, ξ, η and λ are all real, Aξ λξ and Aη λη are real. The real and imaginary parts of the above equation must be zero, so Aξ λξ = 0, and Aη λη = 0 Aξ = λξ, and Aη = λη. Now, since x = ξ + iη is an eigenvector it is non-zero. So one or the other of ξ and η (or both) must be non-zero. Thus either ξ or η (or both) are real eigenvectors associated with λ. Moreover, if ξ and η are any two real eigenvectors (or the same eigenvector) associated with λ, then ξ + iη is an eigenvector. Now suppose λ is a complex eigenvalue of A M n (R) (meaning that λ has a non-zero imaginary part). Then, if x is a real eigenvector of A, we have Ax = λx. But a real number times a real number is a real number and a real number times a complex number is a complex number. Thus the left hand side of the above equation is real, while the right hand side is complex, which is a contradiction. Thus x must be complex. We conclude that it is impossible for a real eigenvector to be associated with a complex eigenvalue of a real matrix Suppose that A 2 = A, and that λ is an eigenvalue of A with associated eigenvector x. That is, Ax = λx. Multiply both sides of this equation by A: A 2 x = λax = λ(λx) = λ 2 x, and, A 2 x = Ax = λx. λ 2 x = λx Thus λ must satisfy λ 2 = λ, which has the two solutions λ = 0 and λ = 1. Since λ was an arbitrary eigenvalue of A, all of the eigenvalues of A are either 0 or If A q = 0 and Ax = λx for x 0, then A q x = λ q x = 0. Thus λ must be zero for all eigenvalues of A. An example of a nonzero nilpotent matrix is A = Suppose A = A and Ax = λx. Multiply both sides of this equation by x on the left: x Ax = λx x = λ x 2.

3 Take the complex conjugate transpose of the left side. Since this is a number this is the same as the complex conjugate. Thus, using the reverse order rule for the conjugate transpose, x Ax = (x Ax) = x A x = x Ax. Any complex number which equals its conjugate is a real number, so x Ax is real, and thus must be real. λ = x Ax x 2 Section A is singular iff det(a) = 0. But det(a) = λ 1 λ 2 λ n, where λ 1,..., λ n are the eigenvalues of A. Thus det(a) = 0 iff one of the λ i s is zero, i.e. iff A has a zero eigenvalue The diagonal elements of AB are (AB) ii = n k=1 a ik b ki, so tr(ab) = m a ik b ki. i=1 k=1 Likewise, tr(ba) = m b ik a ki i=1 k=1 = m a ki b ik i=1 i=1 = tr(ab) We can use this now to switch the order of S 1 and AS in the trace of S 1 AS to show: ( ) tr S 1 AS = tr (ASS 1) = tr(ai) = tr(a) If A is nilpotent then the sum of the eigenvalues of A is zero. So by Theorem , tr(a) = 0. The characteristic polynomial of a nilpotent matrix is p A (t) = (t 0)(t 0) (t 0) = t n. ( Since σ A k) = i.e. { } λ k 1,..., λk n, use the fact that the trace of A k is the sum of its eigenvalues, ( tr A k) = λ k i. i=1

4 Section Suppose A and B are diagonalizable and commute. Then, by Theorem they are simultaneously diagonalizable. So let S be the similarity matrix that diagonalizes A and B. Then S 1 AS = D and S 1 BS = E are diagonal. Thus A + B = SDS 1 + SES 1 = S (D + E) S 1, so A+B is similar to D +E. But D +E is diagonal with entries λ i +µ i, so these are the eigenvalues of A + B Let B M n be diagonalizable. So B = SDS 1, where D is diagonal. Let D be the diagonal matrix of square roots of the diagonal entries of D. (For complex entries z = re iφ, we can take ( ) 2 re iφ/2, for instance.) Then D = D. Let A = S DS 1. Then so A is a square root of B. A 2 = S DS 1 S DS 1 = S ( D ) 2 S 1 = SDS 1 = B, Let A, B M n and suppose that B has n distinct eigenvalues. We have proven already that if A and B are simultaneously diagonalizable, then they commute. So, we must show that if A and B commute, then they are simultaneously diagonalizable. Suppose A and B commute. Let s 1,..., s n be the eigenvectors of B associated with the eigenvalues λ 1,..., λ n. I.e. Bs i = λ i s i for i = 1,..., n. Since A and B commute, we have B (As i ) = A (Bs i ) = Aλ i s i = λ i As i. Therefore, As i is an eigenvector of B associated with the eigenvalue λ i. By Lemma 1.3.8, the eigenvectors of B are linearly independent, so there cannot be two linearly independent eigenvectors associated with any of the eigenvalues λ i. This means that, for each i, As i = µ i s i for some constant µ i. Thus {s 1,..., s n } is a set of linearly independent eigenvectors for A. So let S = [s 1,..., s n ] be the matrix with columns s i. Then S 1 BS = diag (λ 1,..., λ n ), and S 1 AS = diag (µ 1,..., µ n ) so A and B are simultaneously diagonalizable Let A, B M n and suppose that AB is diagonalizable. On one hand, suppose that A is nonsingular. Let S be such that AB = S 1 DS

5 where D is diagonal. Multiply the above equation by A 1 on the left and A on the right. Then BA = A 1 S 1 DSA = (SA) 1 D(SA), so BA is diagonalizable. On the other hand, suppose that B is nonsingular. Then multiply by B 1 on the right and B on the left. Then BA = BS 1 DSB 1 = (SB 1) 1 ( D SB 1), so, again, BA is diagonalizable. To show that the assumption that either A or B is nonsingular cannot be dropped, let A =, and B = 1 1. Then AB =, and BA =. In this case AB is diagonalizable (it is diagonal already), but BA is not diagonalizable.

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