南航双语矩阵论matrix theory第4章部分习题参考答案

of
AT
are the nonzero vectors in
N ( AT ) {k (0,1)T | k R} . The intersection
N ( AT ) N ( A) {(0,0)T } . Hence, Aand AT do not have eigenvectors in common.
T T T T T c1 (u1u1 ) c2 (u2uT 2 ) cn (u n u n )
T T c1u1u1 c2u2uT 2 cn u n u n A
Hence, A is a symmetric matrix.
T T Aui (c1u1u1 c2 u 2 uT 2 cn u n u n )u i T T c1u1 (u1 ui ) c2u 2 (uT 2 u i ) cn u n (u n u i )
Exercise 11
Let [u1 , u 2 ,, u n ] be an orthonormal basis for R n and let A be a linear combination of rank
T T 1 matrices u1u1 , u2uT 2 ,…, u n u n . If T T A c1u1u1 c2u2uT 2 cn u n u n
Show that A is a symmetric matrix with eigenvalues c1, c2 ,, cn and that u i is an eigenvector belonging to ci for each i. Proof
T T AT (c1u1u1 c2u2uT 2 cn u n u n )
of
Q
cFra Baidu bibliotekrresponding
to
. Then
Qx x .
x x Qx x
hence, 1 (b) QH Q I , det(Q H ) det(Q) 1 , det(Q) det(Q) 1 . Hence, |det(Q)|=1.
Exercise 9
Exercise 6
Let Q be a unitary or orthogonal matrix. (a) Show that if is an eigenvalue of Q, then 1 (b) Show that |det(Q)|=1. Proof (a) Let x be an eigenvector
(The matrix C is called the companion matrix of p( x) .) (a) Show that if i is a root of p( ) 0 then i is an eigenvalue of C T with eigenvector x (1, i ,, in 2 , in 1 )T . (b) Use part (a) to show that if p( ) has n distinct roots then p( ) is the characteristic polynomial of C. (The result is true even if all the eigenvalues of p( ) are not distinct.) Hint: C and C T have the same characteristic polynomial. Proof (a)
characteristic polynomials. Hence, Aand AT have the same eigenvalues. Aand AT do not necessarily have the same eigenvectors. For example,
2 1 2 0 T Let A , then A . 0 2 1 2 The eigenvectors of A are the nonzero vectors in N ( A) {k (1,0)T | k R} . The eigenvectors
k 1 for k 1, 2,3 .If the eigenvalue of Q are all real, then
k 1 or 1 k 1, 2,3 .
Since det(Q) 1 , 12 3 1 . Hence, the possible triples of eigenvalues (1 , 2 , 3 ) with
Exercise 16
Let be an orthogonal transformation on a Euclidean space V (an inner product space over the real number field). If W is a -invariant subspace of V, show that the orthogonal complement of W is also -invariant. Proof Let V W W , where W is -invariant. Let {u1 , u2 ,, uk } be an orthonormal basis for
ci ui (uT i u i ) ci u i
Hence, u i is an eigenvector belonging to ci for each i.
Exercise 13
Let A and B be n n matrices. Show that (a) If is a nonzero eigenvalue of AB, then it is also an eigenvalue of BA. (b) If 0 is an eigenvalue of AB, then 0 is also an eigenvalue of BA. Proof (a) If is a nonzero eigenvalue of AB, then there is a nonzero vector x such that ABx x . From ABx x , we see that Bx 0 . Since BA( Bx) Bx , we obtain that Bx 0 is an eigenvector of BA corresponding to the eigenvalue .Hence, is also an eigenvalue of BA. (b) If 0 is an eigenvalue of AB,then det( AB) 0 . Hence, det( BA) det( AB) 0 . Thus, 0 is an eigenvalue of BA.
Exercise 15
Let p( x) x n a1 x n 1 an be a polynomial of degree n 1 , and let
0 1 C 0 0 an 0 0 0 an 1 1 0 0 an 2 0 0 1 a1 0 0 0
)
If i is a root of p( ) 0 , then p(i ) 0 . We obtain that eigenvalue of C T with eigenvector x (1, i ,, in 2 , in 1 )T .
Let Q be a 3 3 orthogonal matrix whose determinant is equal to 1. (a) If the eigenvalue of Q are all real and if they are ordered so that 1 2 3 , determine the values of all possible triples of eigenvalues (1 , 2 , 3 ) (b) In the case that the eigenvalues 2 and 3 are complex, what are the possible values for 1 ? Explain. Solution (a) By Exercise #6,
Exercise 14
Prove that there do not exist n n matrices A and B such that AB BA I . Proof Let A (aij ) and B (bij ) . tr ( AB ) aij b ji , tr ( BA) bij a ji .
1
1 2 3 are
(1,-1,-1),(1,1,1). (b) If the eigenvalues 2 and 3 are complex, then 2 and 3 must be the conjugate of each
other. 2 3 =1 . Hence, 1 =1 since 12 3 1 .
i 1 j 1 i 1 j 1 n n n n
Hence, tr ( AB) tr ( BA) .
2
tr ( AB BA) tr ( AB) tr ( BA) 0 , tr ( I ) n
It is impossible to have matrices A and B such that AB BA I .
C T x i x . Then i is an
(b) If p( ) has n distinct roots, then all roots of p( ) are eigenvalues of C T . We obtain that the characteristic polynomial of C T and p( ) have the same n distinct roots. And also they have the same degree and the same leading coefficient. Hence, the characteristic polynomial of C T is the same as p( ) . Since C and C T have the same characteristic polynomial, we know that p( ) is the characteristic polynomial of C.
0 1 T C x 0 0 0 0 1 0 0 0 0 0 0 an 0 an 1 0 an 2 1 a1
T
i i 1 2 2 i i i n2 n 1 n 1 i i i n 1 n n 1 a a a p ( i n i n 1 i 1 i i
第四章部分习题参考答案 Exercise 4
Show that Aand AT have the same eigenvalues. Do they necessarily have the same eigenvectors? Explain. SolutionSince det( I A) det(( I A)T ) det( I AT ) , Aand AT have the same