南航双语矩阵论matrixtheory第三章部分题解126.pdf

《南航双语矩阵论matrixtheory第三章部分题解126.pdf》由会员分享，可在线阅读，更多相关《南航双语矩阵论matrixtheory第三章部分题解126.pdf（5页珍藏版）》请在淘文阁 - 分享文档赚钱的网站上搜索。

1、 南 航 双 语 矩 阵 论 m a t r i x t h e o r y第 三 章 部 分 题 解(总 4 页)-本页仅作为文档封面，使用时请直接删除即可-内页可以根据需求调整合适字体及大小-2 Solution Key to Some Exercises in Chapter 3#5.Determine the kernel and range of each of the following linear transformations on 2P(a)()()p xxp x(b)()()()p xp xp x (c)()(0)(1)p xpxp Solution(a)Let()p x

2、axb.()p xax.()0p x if and only if 0ax if and only if 0a.Thus,ker()|b bR The range of is 2()P|ax aR (b)Let()p xaxb.()p xaxba.()0p x if and only if 0axba if and only if 0a and 0b.Thus,ker()0 The range of is 2()P2|,Paxba a bR (c)Let()p xaxb.()p xbxab.()0p x if and only if 0bxab if and only if 0a and 0b

3、.Thus,ker()0 The range of is 2()P2|,Pbxab a bR 备注: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.#7.Let be the linear mapping that maps 2P into 2R defined by 10()()(0)p x dxp xp Find a matrix A such that ()xA.Solution 3 1(1)1 1/2()0 x 11/21 1/2()1010 x Hence,1 1/210A#10.Let be the transformation on 3P defined b

4、y ()()()p xxp xpx a)Find the matrix A representing with respect to 21,x x b)Find the matrix B representing with respect to 21,1xx c)Find the matrix S such that 1BSAS d)If 2012()(1)p xaa xax,calculate()np x.Solution(a)(1)0 ()xx 22()22xx 002010002A (b)(1)0 ()xx 22(1)2(1)xx 000010002B (c)21,1xx21,x x10

5、1010001 The transition matrix from 21,x x to 21,1xx is 101010001S,1BSAS 4 (d)2201212(1)2(1)nnaa xaxa xax#11.Let A and B be n n matrices.Show that if A is similar to B then there exist n n matrices S and T,with S nonsingular,such that ASTand BTS.Proof There exists a nonsingular matrix P such that 1AP

6、 BP.Let 1SP,TBP.Then ASTand BTS.#12.Let be a linear transformation on the vector space V of dimension n.If there exist a vector v such that 1()v0n and()v0n,show that(a)1,(),()vvvn are linearly independent.(b)there exists a basis E for V such that the matrix representing with respect to the basis E i

7、s 000010000010 Proof(a)Suppose that 1011()()vvv0nnkkk Then 11011()()vvv0nnnkkk That is,12210110()()()()vvvv0nnnnnkkkk Thus,0k must be zero since 1()v0n.211111()()()vvv0nnnnkkk This will imply that 1k must be zero since 1()v0n.By repeating the process above,we obtain that 011,nkkk must be all zero.Th

8、is proves that 1,(),()vvvn are linearly independent.(b)Since 1,(),()vvvn are n linearly independent,they form a basis for V.Denote 112,(),()v vvnn 12()23()5 .1()nn ()0n 12(),(),()n121,nn000010000010#13.If A is a nonzero square matrix and kAOfor some positive integer k,show that A can not be similar

9、to a diagonal matrix.Proof Suppose that A is similar to a diagonal matrix 12diag(,)n.Then for each i,there exists a nonzero vector xi such that xxiiiA xxx0kkiiiiiA since kAO.This will imply that 0i for 1,2,in.Thus,matrix A is similar to the zero matrix.Therefore,AOsince a matrix that is similar to the zero matrix must be the zero matrix,which contradicts the assumption.This contradiction shows that A can not be similar to a diagonal matrix.Or If 112diag(,)nAPP then 112diag(,)kkkknAPP.kAO implies that 0i for 1,2,in.Hence,BO.This will imply that AO.Contradiction!