南航双语矩阵论 matrix theory第三章部分题解

1 Solution Key to Some Exercises in Chapter 3 5 Determine the kernel and range of each of the following linear transformations on 2 P a p xxp x b p xp xp x c 0 1 p xpxp Solution a Let p xaxb p xax if and only if if and only if Thus 0p x 0ax 0a ker b bR The range of is 2 P ax aR b Let p xaxb p xaxba if and only if if and only if and 0p x 0axba 0a 0b Thus ker 0 The range of is 2 P 2 Paxba a bR c Let p xaxb p xbxab if and only if if and only if and 0p x 0bxab 0a 0b Thus ker 0 The range of is 2 P 2 Pbxab a bR 备注备注 映射的核以及映射的像都是集合映射的核以及映射的像都是集合 应该以集合的记号来表达或者用文字来叙述应该以集合的记号来表达或者用文字来叙述 7 Let be the linear mapping that maps into defined by 2 P 2 R 1 0 0 p x dx p x p Find a matrix A such that xA Solution 1 1 1 1 2 0 x 11 21 1 2 1010 x Hence 1 1 2 10 A 10 Let be the transformation on defined by 3 P p xxp xpx a Find the matrix A representing with respect to 2 1 x x b Find the matrix B representing with respect to 2 1 1 xx c Find the matrix S such that 1 BSAS d If calculate 2 012 1 p xaa xax n p x Solution a 1 0 xx 2 22 22xx 002 010 002 A b 1 0 xx 22 1 2 1 xx 000 010 002 B c 2 1 1 xx 2 1 x x 101 010 001 The transition matrix from to is 2 1 x x 2 1 1 xx 101 010 001 S 1 BSAS d 22 01212 1 2 1 nn aa xaxa xax 11 Let A and B be matrices Show that if A is similar to B then there exist matrices nn nn S and T with S nonsingular such that and AST BTS Proof There exists a nonsingular matrix P such that Let Then 1 AP BP 1 SP TBP and AST BTS 12 Let be a linear transformation on the vector space V of dimension n If there exist a vector v such that and show that 1 v0 n v0 n a are linearly independent 1 vvv n b there exists a basis E for V such that the matrix representing with respect to the basis E is 0000 1000 0010 Proof 3 a Suppose that 1 011 vvv0 n n kkk Then 11 011 vvv0 nn n kkk That is 1221 0110 vvvv0 nnnn n kkkk Thus must be zero since 0 k 1 v0 n 211 111 vvv0 nnn n kkk This will imply that must be zero since 1 k 1 v0 n By repeating the process above we obtain that must be all zero This 011 n k kk proves that are linearly independent 1 vvv n b Since are n linearly independent they form a basis for V 1 vvv n Denote 1 12 v v v n n 12 23 1 nn 0 n 12 n 121 nn 0000 1000 0010 13 If A is a nonzero square matrix and for some positive integer k show that A can not k AO be similar to a diagonal matrix Proof Suppose that A is similar to a diagonal matrix Then for each there 12 diag n i exists a nonzero vector such that xixx iii A since xxx0 kk iiiii A k AO This will imply that for Thus matrix A is similar to the zero matrix Therefore 0 i 1 2 in since a matrix that is similar to the zero matrix must be the zero matrix which contradicts AO the assumption This contradiction shows that A can not be similar to a di