Math 215 HW #11 Solutions

Transcription

1 Math 215 HW #11 Solutions 1 Problem 556 Find the lengths and the inner product of 2 x and y [ 2 + ] Answer: First, x 2 x H x [2 + ] 2 (4 + 16) , so x 6 Likewise, so y 6 Finally, x, y x H y [2 + ] y 2 y H y [2 ] 2 + (4 + 16) + 16, 2 + (2 + ) 2 () 2 ( i) i 2 Problem 5516 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 On the other hand, Therefore, A x, A x λ x, λ x λ 2 x, x λ 2 x 2 A x, A x (A x) T A x x T A T A x x T x x, x x 2 meaning that λ 2 1, so λ 1 λ 2 x 2 x 2, (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

2 (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 must be an eigenvalue of A 3 Problem 5522 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 (ie Hermitian part) of Z is half of Z + Z H Find a similar formula for the imaginary part (ie skew-hermitian part) K, and split these matrices into A + K: i i i Z and Z 5 i i Answer: Notice that so indeed 1 2 (Z + ZH ) is Hermitian Likewise, (Z + Z H ) H Z H + (Z H ) H Z H + Z, (Z Z H ) H Z H (Z H ) H Z H Z (Z Z H ), is skew-hermitian, so K 1 2 (Z ZH ) is the skew-hermitian part of Z Hence, when we have A 1 2 (Z+ZH ) 1 2 and K 1 2 (Z ZH ) 1 2 On the other hand, when we have Z ( i + 5 ( i 5 A 1 2 (Z + ZH ) 1 ( i i 2 i i i, 5 ) i 5 2 ) 3 1 [ 4 2i 5 2 i i Z i i ) i i + 1 i i i 4 2i i 4 2i 5 ] [ ] 8i 4 + 2i 4 + 2i 4 + 2i 4 + 2i 2i 2i [ ] i i and K 1 2 (Z ZH ) 1 ( i i 2 i i ) i i 1 i i 2 2i 2i i i 2

3 4 Problem 5528 If A z, then A H A z If A H A z, multiply by z H to prove that A z The nullspaces of A and A H are A H A is an invertible Hermitian matrix when the nullspace of A contains only z Answer: Suppose A H A z Then, multiplying both sides by z H yields meaning that A z z H A H A z (A z) H (A z) A z, A z A z 2, Therefore, we see that if A z, then A H A z and if A H A z, then A z, so the nullspaces of A and A H are equal A H A is an invertible matrix only if its nullspace is { }, so we see that A H A is an invertible matrix when the nullspace of A contains only z 5 Problem 5548 Prove that the inverse of a Hermitian matrix is again a Hermitian matrix Proof If A is Hermitian, then A UΛU H, where U is unitary and Λ is a real diagonal matrix Therefore, A 1 (UΛU H ) 1 (U H ) 1 Λ 1 U 1 UΛ 1 U H since U 1 U H Note that Λ 1 is just the diagonal matrix with entries 1/λ i (where the λ i are the entries in Λ) Hence, (A 1 ) H (UΛ 1 U H ) H U(Λ 1 ) H U H UΛ 1 U H A 1 since Λ 1 is a real matrix, so we see that A 1 is Hermitian 6 Problem 568 What matrix M changes the basis V 1 (1, 1), V2 (1, 4) to the basis v 1 (2, 5), v 2 (1, 4)? The columns of M come from expressing V 1 and V 2 as combinations mij v i of the v s Answer: Since and we see that V V 2 M v 4 1 v 2 1 v 4 2, Problem 5612 The identity transformation takes every vector to itself: T x x Find the corresponding matrix, if the first basis is v 1 (1, 2), v 2 (3, 4) and the second basis is w 1 (1, ), w 2 (, 1) (It is not the identity matrix!) Answer: Despite the slightly confusing way this question is worded, it is just asking for the matrix M which converts the v basis into the w basis Clearly, 1 1 v w w 2 3

4 and so the desired matrix is v M 3 w w 2, Problem 5638 These Jordan matrices have eigenvalues,,, They have two eigenvectors (find them) But the block sizes don t match and J is not similar to K 1 1 J 1 and K 1 For any matrix M, compare JM and MK If they are equal, show that M is not invertible Then M 1 JM K is impossible Answer: First, we find the eigenvectors of J and K Since all eigenvalues of both are, we re just looking for vectors in the nullspace of J and K First, for J, we note that J is already in reduced echelon form and that J v implies that v is a linear combination of 1 Hence, these are the eigenvectors of J, Likewise, K is already in reduced echelon form and K v implies that v is a linear combination of 1, 1 Hence, these are the eigenvectors of K Now, suppose 1 m 11 m 12 m 13 m 14 M m 21 m 22 m 23 m 24 m 41 m 42 m 43 m 44 such that JM MK Then 1 m 11 m 12 m 13 m 14 m 21 m 22 m 23 m 24 JM m 21 m 22 m 23 m 24 1 m 41 m 42 m 43 m 44 m 41 m 42 m 43 m 44 4

5 and m 11 m 12 m 13 m 14 1 m 11 m 12 MK m 21 m 22 m 23 m 24 1 m 21 m 22 m 31 m 32 m 41 m 42 m 43 m 44 m 41 m 42 Therefore JM MK means that m 21 m 22 m 23 m 24 m 11 m 12 m 41 m 42 m 43 m 44 m 21 m 22 m 31 m 32 m 41 m 42 and so we have that m 21 m 24 m 22 m 41 m 44 m 42 Plugging these back into M, we see that m 11 m 12 m 13 m 14 M m 23 m 43 Clearly, the second and fourth rows are multiples of each other, so M cannot possibly have rank 4 However, M not having rank 4 means that M cannot be invertible Therefore, M 1 JM K is impossible, so it cannot be the case that J and K are similar 9 Problem 564 Which pairs are similar? Choose a, b, c, d to prove that the other pairs aren t: a b b a c d d c c d d c a b b a Answer: The second and third are clearly similar, since and [ 1 1 ] [ b a d c ] [ 1 1 ] [ ] a b c d c d a b Likewise, the first and fourth are similar, since 1 d c 1 1 c d a b 1 b a 1 1 a b c d There are no other similarities, as we can see by choosing a 1, b c d 5

6 Then the matrices are, in order Each of these is already a diagonal matrix, and clearly the first and fourth have 1 as an eigenvalue, whereas the second and third have only as an eigenvalue Since similar matrices have the same eigenvalues, we see that neither the first nor the fourth can be similar to either the second or the third 1 (Bonus Problem) Problem 5614 Show that every number is an eigenvalue for T f(x) df/dx, but the transformation T f(x) x f(t)dt has no eigenvalues (here < x < ) Proof For the first T, note that, if f(x) e ax for any real number a, then T f(x) df dx aeax af(x) Hence, any real number a is an eigenvalue of T Turning to the second T, suppose we had that T f(x) af(x) for some number a and some function f Then, by the definition of T, x f(t)dt af(x) Now, use the fundamental theorem of calculus to differentiate both sides: Solving for f, we see that so Therefore, exponentiating both sides, f(x) af (x) f (x)dx f(x) 1 a dx, ln f(x) x a + C f(x) e x/a+c e C e x/a We can get rid of the absolute value signs by substituting A for e C (allowing A to possibly be negative): f(x) Ae x/a Therefore, we know that T f(x) x f(t)dt x Ae t/a dt aae t/a] x aaex/a aa a(ae x/a A) a(f(x) A) On the other hand, our initial assumption was that T f(x) af(x), so it must be the case that af(x) a(f(x) A) af(x) aa Hence, either a or A However, either implies that f(x), so T has no eigenvalues 6