复旦大学谢启鸿老师高等代数在线习题课思考题分析与解 ep.45

2022-02-01 19:52 作者:CharlesMa0606 0人读过 | 我要投稿

本文内容主要有关于Cayley-Hamilton定理的应用，在高代白皮书上对应第6.2.3节

题目来自于复旦大学谢启鸿教授在本站高等代数习题课的课后思考题，本文仅供学习交流

本人解题水平有限，可能会有错误，恳请斧正！

祝大家新年快乐！

练习题1(16级高代II每周一题第3题) 设 $A_1%2CA_2%2C%5Ccdots%2CA_m$ 为n阶方阵， $g%5Cleft(x%5Cright)%5Cin%20K%5Cleft%5Bx%5Cright%5D$ ，使得 $g%5Cleft(A_1%5Cright)%2Cg%5Cleft(A_2%5Cright)%2C%5Ccdots%2Cg%5Cleft(A_m%5Cright)$ 都是非异阵.证明：存在 $h%5Cleft(x%5Cright)%5Cin%20K%5Cleft%5Bx%5Cright%5D$ ，使得 $g%5Cleft(A_i%5Cright)%5E%7B-1%7D%3Dh%5Cleft(A_i%5Cright)%5Cleft(1%5Cle%20i%5Cle%20m%5Cright)$ .

证明注意到 $g%5Cleft(A_i%5Cright)$ 都是非异阵，取 $A_i$ 的任一特征值为 $%5Clambda$ ，则 $g%5Cleft(%5Clambda%5Cright)$ 都不为零.设分块对角阵 $A%3Ddiag%5C%7BA_1%2C%5Ccdots%2CA_m%5C%7D$ 的特征多项式为

$f%5Cleft(%5Clambda%5Cright)%3D%5Cleft%7C%5Clambda%20I_%7Bmn%7D-A%5Cright%7C%3D%5Cleft%7C%5Clambda%20I_n-A_1%5Cright%7C%5Ccdots%5Cleft%7C%5Clambda%20I_n-A_m%5Cright%7C%5C%5C%3D%5Cprod_%7Bi%3D1%7D%5E%7Bm%7D%7Bf_i%5Cleft(%5Clambda%5Cright)%7D%5Cin%20K%5Cleft%5Bx%5Cright%5D%2Cg%5Cleft(%5Clambda%5Cright)%3D%5Cleft(%5Clambda-x_1%5Cright)%5Ccdots%5Cleft(%5Clambda-x_k%5Cright)%2Cx_i%5Cin%20C$

从而 $f%5Cleft(x%5Cright)%2Cg%5Cleft(x%5Cright)$ 互素，因此存在 $u%5Cleft(x%5Cright)%2Cv%5Cleft(x%5Cright)%5Cin%20K%5Cleft%5Bx%5Cright%5D%2Cs.t.f%5Cleft(x%5Cright)u%5Cleft(x%5Cright)%2Bg%5Cleft(x%5Cright)v%5Cleft(x%5Cright)%3D1$ ，代入 $A_i$ ，由Cayley-Hamilton定理可知 $f%5Cleft(A_i%5Cright)%3D0%2Cg%5Cleft(A_i%5Cright)v%5Cleft(A_i%5Cright)%3DI_n$ ，这即是要找的 $h%5Cleft(x%5Cright)%5Cin%20K%5Cleft%5Bx%5Cright%5D$ ，使得 $g%5Cleft(A_i%5Cright)%5E%7B-1%7D%3Dh%5Cleft(A_i%5Cright)%5Cleft(1%5Cle%20i%5Cle%20m%5Cright)$ .

$%5BQ.E.D%5D$

练习题2(15级高代II每周一题第4题) 设n阶方阵A适合多项式 $f%5Cleft(x%5Cright)%3Da_mx%5Em%2Ba_%7Bm-1%7Dx%5E%7Bm-1%7D%2B%5Ccdots%2Ba_1x%2Ba_0$ ，其中 $%5Cleft%7Ca_m%5Cright%7C%3E%5Csum_%7Bi%3D0%7D%5E%7Bm-1%7D%5Cleft%7Ca_i%5Cright%7C$ .证明：矩阵方程 $2X%2BAX%3DXA%5E2$ 只有零解.

证明首先证明多项式方程 $f%5Cleft(x%5Cright)%3D0$ 的根的模长一定小于1，用反证法，设 $x_0%5Cgeq1$ 并且 $f%5Cleft(x_0%5Cright)%3D0$ ，则 $%5Cleft%7Ca_m%5Cright%7C%3D%5Cfrac%7B1%7D%7Bx_0%5Em%7D%5Cleft%7Ca_%7Bm-1%7D%7Bx_0%7D%5E%7Bm-1%7D%2B%5Ccdots%2Ba_1x_0%2Ba_0%5Cright%7C%5Cle%5Csum_%7Bi%3D0%7D%5E%7Bm-1%7D%5Cleft%7Ca_i%5Cright%7C$ ，矛盾.