Math 3, Fall Jerry L. Kazdan Problem Set 9 Due In class Tuesday, Nov. 7 Late papers will be accepted until on Thursday (at the beginning of class).. Suppose that is an eigenvalue of an n n matrix A and let E be the set of all eigenvectors with the same eigenvalue. Show that E is a linear subspace of R n. Say that both ~v and ~w are in E. We need to show that both c~v and ~v + ~w are in E cor any scalar c. Now A(c~v) ca~v c~v (c~v so c~v E. Also, A(~v + ~w) A~v + A ~w ~v + ~w (~v + ~w). Thus ~v + ~w E.. Let A be a real matrix whose eigenvalues are not real. a) Suppose one of the eigenvalues has absolute value. Explain why the other must as well. Let + i be a complex eigenvalue with 6. Since A is real, then its complex conjugate, i is also an eigenvalue. But jj jj. b) Explain why A must be diagonalizable. Since the eigenvalues of A are distinct, it is diagonalizable. 3. This asks you to come up with four examples. In each case, nd a matrix (perhaps ) that is a) Both invertible and diagonalizable. The identity matrix, I ; the matrix b) Not invertible, but diagonalizable. The zero matrix, ; the matrix c) Not diagonalizable but is invertible.. d) Neither invertible nor diagonalizable... 3. 4. If the matrices A and B are similar and if A 3, must B 3? Proof or counterexample. True. Since B S AS for some S, then B 3 S A 3 S.
5. In a large city, a car rental company has three locations the Airport, the City, and the Suburbs. One has data on which location the cars are returned daily Rented at Airport % are returned to the City and 5% to the Suburbs. The rest are returned to the Airport. Rented in City Rented in Suburbs % returned to Airport, % returned to Suburbs. 5% are returned to the Airport and % to the city. If initially there are 35 cars at the Airport, 5 in the city, and 35 in the suburbs, what is the long-term distribution of the cars? Lat A k, C k, and S k be the number of cars on day k at the Airport, City, and Suburbs, respectively. Then A k+ 73A k + C k + 5S k C k+ A k + 8C k + S k S k+ 5A k + C k + 73S k 73 5 Thus the transition matrix T is T @ 8 A 5 73 To nd the he eigenvector P with eigenvalue one needs to solve This gives A S 5C. 7A + C + 5S A C + S 5A + C 7S In addition, since P is supposed to be a probability vector, A + C + S. Thus C so A S 5. Using the initial state, there are 35 + 5 + 35 cars in all. Thus in the long run city / cars airport cars suburbs card. 6. Let R be a (real) 3 3 orthogonal matrix. a) Show that the eigenvalues,, which may be complex, all have absolute value. Since R is an orthogonal matrix, then kr~vk k~vk for every vector ~v. In particular, if R~v ~v, then k~vk k~vk jjk~vk so jj.
b) If det R show that is one of the eigenvalues of R and that if R 6 I, no other eigenvalue can be. none are. Since R is a real matrix, either two of its eigenvalues are complex or Case Two complex eigenvalues, say and, so and then. Since det R 3, then 3. Case All three eigenvalues are real. Since the real eigenvalues can only have values, the only possibilities are ; ; ; ; ; ; ; ; ; ; ; Because det R and R 6 I, the only possibility is ; ;. For the remainder of this problem assume det R and R 6 I. c) Let N be an eigenvector corresponding to and. let Q be the plane of all vectors orthogonal to N. Show that R maps Q to Q. An orthogonal matrix preserves the inner product hr~x; R~yi h~x; ~yi for all vectors ~x, ~y. Thus, since R N ~ N ~, if ~x Q, so h~x; Ni ~, then that is, R~x Q. hr~x; ~ Ni hr~x; R ~ Ni h~x; ~ Ni ; d) Why does this show that R is a rotation of the plane Q with N as the axis of rotation? Q is a two dimensional plane and R is an orthogonal transformation from Q to itself. Thus R acts on Q as either a rotation or a reection. Since det R, it is a rotation. 7. [Bretscher, Sec. 7. #36] Find a matrix A such that eigenvectors with corresponding eigenvalues 5 and. 3 and There are several approaches. Here is one. Since A can be diagonalized (it 3 has a basis of eigenvectors) the matrix S whose columns are the eigenvectors of A has the property that S AS D, where D is the diagonal matrix whose elements are 5 and. Thus 3 5 4 3 A SDS 5 3 are 3
8. [Bretscher, Sec. 7. #38] We are told that @ A is an eigenvector of the matrix 4 M @ 5 3A. What is the associated eigenvalue? @ so the eigenvalue is. 4 5 3A @ A @ A @ A 9. [Bretscher, Sec. 7. #] Find all of the eigenvalues of M B C @ 3 4A 3 and determine their algebraic multiplicity. det(m I) det ( )( ) 3 4 det 3 The eigenvalues are thus,,,. The eigenvalue has algebraic multiplicity, the others have algebraic multiplicity. For this example the geometric multiplicities agree with the algebraic multiplicities so the matrix can be diagonalized. B C. [Bretscher, Sec. 7. #4] Consider a 4 4 matrix A, where B, C, D and D are matrices. What is the relationship between the eigenvalues of A, B, C, and D? R S First a preliminary result. In the above notation, if M, then T det M (det R)(det T ). Case If R is not invertible then its columns { and hence the corresponding columns of M { are linearly dependent so both det R and det M. Case R is invertible. This is an exercise in block multiplication of matrices. Let X be an unknown matrix. Then R S I X R RX + S T I T 4
Since R is invertible we can pick X so that RX + S. The result should now be clear. Applying this to M A I we conclude that the eigenvalues of A, including their algebraic multiplicities, are exactly those determined by B and D. Note, however, that even if both B and D are diagonalizable, A may not be. The simplest example is B D and C.. The characteristic polynomial p A () of A is p A () det(a I) ( ) n + c n ( ) n + + c ; () Caution many books dene the characteristic polynomial as det(i A), which changes some signs.) In class we showed that similar matrices have the same characteristic polynomial. Recall that the trace of a matrix A (a ij ) is the sum of the diagonal elements trace(a) a + a + + a nn. In this problem you will see that the trace and determinant of A are two of the coecients in the characteristic polynomial. a) Show that c det(a). Let in equation (). b) Since also the eigenvalues of A are the roots of the characteristic polynomial, show that the trace of A is the sum of its eigenvalues trace(a) + + n These are the coecient c n of ( ) n. [Although this is true for all n., only do this for n 3. The procedure in the general case is identical.] In the case of n 3, the characteristic polynomial is a cubic. We compute the coecient of for both sides of equation (). First we expand det(i A) by minors using the rst column a a a 3 p A () det @ a a a 3 A a 3 a 3 a 33 a a (a ) det 3 a a a a 3 a 33 det 3 a a + a a 3 a 33 3 det 3 a 3 a 33 Note that the second and third terms on the last line do not contribute any quadratic terms in so we can ignore them. We indicate such terms by writing +. Continuing from above p A () (a )(a )(a 33 ) + 3 + (a + a + a 33 ) + () 5
Next we factor p A () using its roots, the eigenvalues j. similar to that above. The computation is p A () ( )( )( 3 ) 3 + ( + + 3 ) + (3) Comparing the coecient of in equations () and (3) we conclude that the trace of a matrix is the sum of its eigenvalues.. Let A be the transition matrix of a Markov Chain. If ~v (; ; ) T show that A ~v ~v. Why does this imply that is an eigenvalue of A? Thus if A is the transition matrix of any Markov Chain, is always an eigenvalue. Since the sum of the elements in each column of A is, the sum of the elements in each row of A is. This is exactly the same as A ~v ~v. Because for any matrix A the eigenvalues of A and A are the same, we conclude that is an eigenvalue of A. 3. Say a sequence x fx ; x ; x ; x 3 ; g has the properties x, x, and, recursively, x k+ x k+ + x k for k ; ; ; 3. For instance, x, x 3, etc. uk Let u k x k, v k x k+ and write W k. Note W. a) Show that v k vk W k+ v k + u v W k Since v k+ x k+ x k + x k+ u k + v k, then W k+ uk+ v k+ vk u k + v k W k AW k ; where A is the evident matrix. b) Let A denote the matrix above. Show that W k A k W. W AW A W, W 3 AW A 3 W, etc. c) Diagonalize A and use this to compute A k and thus W k explicitly. The characteristic polynomial of A gives, so ( p 5. The corresponding eigenvectors are ~v + and ~v. Then S AS D, where D + +, and the change of coordinates matrix 6
is S whose columns are the corresponding eigenvectors of A. Thus + A SDS. Therefore, using +, A k SD k S k + p + k 5 + p k + k k + k 5 k + k k+ + k+ Therefore W k A k W p 5 d) Use this to get an explicit formula for x k. k + k k+ + k+ Since x k u k, then x k is the rst component of W k, x k p k + k p 4 + p! k p! 3 k 5 5 5 5 5 It is interesting that although by the recursive formula x k+ x k + x k+ are clearly integers, the explicit formula we just found involves p 5. the x k k 4. [Bretscher, Sec. 7.3 #8] Let B B k C @ k A where k is an arbitrary constant. Find the eigenvalue(s) of B and determine both their algebraic and geometric k multiplicities. [Note First try the analogous case.] Since B is upper triangular its eigenvalues are 3 4 k (so the algebraic multiplicity is 4). To determine the geometric multiplicity we want the dimension of the kernel of B ki, that is, the solutions of v B C Bv C @ A @ v 3 A B C @ A v 4 Clearly v can be anything, while v v 3 v 4. Thus the kernel is just multiples of ~v (; ; ; ) and is one dimensional. The geometric dimension of this eigenvalue is. a 5. [Bretscher, Sec. 7.4 #-] Let A and B choices of the constants a and b are these diagonalizable? a. For which b 7
The matrix A has distinct eigenvalues and so it can be diagonalized for any choice of a. Similarly for B, if b 6, it can be diagonalized for any choice of a. However, if b, then has algebraic multiplicity but if a 6 it has geometric multiplicity. It can be diagonalized if and only if a. Remark I found the problems Bretscher, Sec. 7.3 #43-44 and many other applications at the end of Section 7.3 interesting. You might too. These are not assigned. Bonus Problem [Please give this directly to Professor Kazdan] -B Let S be the space of smooth real-valued functions u(x) that periodic with period, so u(x + ) u(x) for all x R with the inner product hf; gi R f(x)g(x) dx. Let D S! s be the derivative operator, Du dudx and let L D. a) Find the adjoints of both D and L. Note that by denition, the adjoint A of a linear map A is the map that satises the identity hv; Awi ha v; wi for all v and w. [Hint Integrate by parts]. b) Find the eigenfunctions u k (x) and corresponding eigenvalues, k of L, so u(x) is periodic and Lu k k u k. c) Use the result of the previous part to conclude that if k 6 ` are integers, then sin kx is orthogonal to sin `x and cos `x. [Last revised December, ] 8