南航双语矩阵论第四章习题答案2016年版.pdf

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 .
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 ,
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.
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, A and AT do not have eigenvectors in common.
each other. 2 3 =1 . Hence, 1 =1 since 123 1 .
Exercise 11
Let [u1 , u2 , , un ] be an orthonormal basis for R n and let A be a linear combination of rank
characteristic polynomials. Hence, A and AT have the same eigenvalues. A and 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
. Th来自百度文库n
Qx x .
x x Qx x
hence, 1 (b) Q H Q I , det(QH ) det(Q) 1 , det(Q) det(Q) 1 . Hence, |det(Q)|=1.
Exercise 9
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, k 1 for k 1,2,3 . If the eigenvalue of Q are all real, then
CT 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.
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 .
k 1 or 1 k 1,2,3 .
Since det(Q) 1 , 123 1 . Hence, the possible triples of eigenvalues (1 , 2 , 3 ) with
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
(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)
T AT (c1u1u1 c2u2uT 2
, cn and that u i is an
cnunuT n)
T T c ( u nu ) n n
T T T T c1 (u 1u (u 2 u ) 1) c 2 2
T c1u1u1 c2u2uT 2
cnunuT n A
T T 1 matrices u1u1 , u 2uT 2 ,…, u n u n . If T A c1u1u1 c2u2uT 2
cnunuT n
Show that A is a symmetric matrix with eigenvalues c1 , c2 , eigenvector belonging to ci for each i. Proof
第四章部分习题参考答案 Exercise 4
Show that A and AT have the same eigenvalues. Do they necessarily have the same eigenvectors? Explain. Solution Since det( I A) det(( I A)T ) det( I AT ) , A and AT have the same
0 1 T C x 0 0 0 0 0 0 1 0 0 0 an 0 an 1 0 an 2 1 a1 0
T
i 1 2 i i n2 in 1 i n 1 a a i n n 1 i
n 1 n a1i i
in 1 p (i )
i i2
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 .
Exercise 15
Let p( x) xn a1 xn 1
0 1 C 0 0
an be a polynomial of degree n 1 , and let
0 0 0 0 1 0 0 0
an 0 an 1 0 an 2 1 a1 0
Hence, A is a symmetric matrix.
T Aui (c1u1u1 c2 u 2uT 2 T c1u1 (u1 ui ) c2 u 2 (uT 2 ui )
cn u n uT n )u i cn u n (uT n ui )
ci ui (uT i u i ) ci ui
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 of Q corresponding to