hw11solutions

# hw11solutions - Math 215 HW #11 Solutions 1. Problem 5.5.6....

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Unformatted text preview: Math 215 HW #11 Solutions 1. Problem 5.5.6. Find the lengths and the inner product of ~x = 2- 4 i 4 i and ~ y 2 + 4 i 4 i . Answer: First, k ~x k 2 = ~x H ~x = [2 + 4 i- 4 i ] 2- 4 i 4 i = (4 + 16) + 16 = 36 , so k ~x k = 6. Likewise, k ~ y k 2 = ~ y H ~ y = [2- 4 i- 4 i ] 2 + 4 i 4 i = (4 + 16) + 16 , so k ~ y k = 6. Finally, h ~x,~ y i = ~x H ~ y = [2 + 4 i- 4 i ] 2 + 4 i 4 i = (2 + 4 i ) 2- (4 i ) 2 = (4- 16 + 16 i ) + 16 = 4 + 16 i. 2. Problem 5.5.16. Write one significant fact about the eigenvalues of each of the following: (a) A real symmetric matrix. Answer: As we saw in class, the eigenvalues of a real symmetric matrix are all real numbers. (b) A stable matrix: all solutions to du/dt = Au approach zero. Answer: By the definition of stability, this means that the reals parts of the eigenvalues of A are non-positive. (c) An orthogonal matrix. Answer: If A~x = λ~x , then h A~x,A~x i = h λ~x,λ~x i = λ 2 h ~x,~x i = λ 2 k ~x k 2 . On the other hand, h A~x,A~x i = ( A~x ) T A~x = ~x T A T A~x = ~x T ~x = h ~x,~x i = k ~x k 2 . Therefore, λ 2 k ~x k 2- k ~x k 2 , meaning that λ 2 = 1, so | λ | = 1. (d) A Markov matrix. Answer: We saw in class that λ 1 = 1 is an eigenvalue of every Markov matrix, and that all eigenvalues λ i of a Markov matrix satisfy | λ i | ≤ 1. 1 (e) A defective matrix (nondiagonalizable). Answer: If A is n × n and is not diagonalizable, then A must have fewer than n eigenvalues (if A had n distinct eigenvalues and since eigenvectors corresponding to different eigenvalues are linear independent, then A would have n linearly independent eigenvectors, which would imply that A is diagonalizable). (f) A singular matrix. Answer: If A is singular, then A has a non-trivial nullspace, which means that 0 must be an eigenvalue of A . 3. Problem 5.5.22. Every matrix Z can be split into a Hermitian and a skew-Hermitian part, Z = A + K , just as a complex number z is split into a + ib . The real part of z is half of z + z , and the “real part” (i.e. Hermitian part) of Z is half of Z + Z H . Find a similar formula for the “imaginary part” (i.e. skew-Hermitian part) K , and split these matrices into A + K : Z = 3 + 4 i 4 + 2 i 5 and Z = i i- i i...
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## This document was uploaded on 08/20/2011.

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hw11solutions - Math 215 HW #11 Solutions 1. Problem 5.5.6....

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