復(fù)旦大學(xué)謝啟鴻老師高等代數(shù)在線習(xí)題課思考題分析與解 ep.45

2022-02-01 19:52 作者:CharlesMa0606 0人讀過 | 我要投稿

本文內(nèi)容主要有關(guān)于Cayley-Hamilton定理的應(yīng)用，在高代白皮書上對應(yīng)第6.2.3節(jié)

題目來自于復(fù)旦大學(xué)謝啟鴻教授在本站高等代數(shù)習(xí)題課的課后思考題，本文僅供學(xué)習(xí)交流

習(xí)題課視頻鏈接：復(fù)旦大學(xué)謝啟鴻高等代數(shù)習(xí)題課_嗶哩嗶哩_bilibili

本人解題水平有限，可能會有錯誤，懇請斧正！

祝大家新年快樂！?

練習(xí)題1(16級高代II每周一題第3題)? 設(shè) $A_1%2CA_2%2C%5Ccdots%2CA_m$ 為n階方陣， $g%5Cleft(x%5Cright)%5Cin%20K%5Cleft%5Bx%5Cright%5D$ ，使得 $g%5Cleft(A_1%5Cright)%2Cg%5Cleft(A_2%5Cright)%2C%5Ccdots%2Cg%5Cleft(A_m%5Cright)$ 都是非異陣.證明：存在 $h%5Cleft(x%5Cright)%5Cin%20K%5Cleft%5Bx%5Cright%5D$ ，使得 $g%5Cleft(A_i%5Cright)%5E%7B-1%7D%3Dh%5Cleft(A_i%5Cright)%5Cleft(1%5Cle%20i%5Cle%20m%5Cright)$ .

證明? 注意到 $g%5Cleft(A_i%5Cright)$ 都是非異陣，取 $A_i$ 的任一特征值為 $%5Clambda$ ，則 $g%5Cleft(%5Clambda%5Cright)$ 都不為零.設(shè)分塊對角陣 $A%3Ddiag%5C%7BA_1%2C%5Ccdots%2CA_m%5C%7D$ 的特征多項式為

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

從而 $f%5Cleft(x%5Cright)%2Cg%5Cleft(x%5Cright)$ 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 $f%5Cleft(A_i%5Cright)%3D0%2Cg%5Cleft(A_i%5Cright)v%5Cleft(A_i%5Cright)%3DI_n$ ，這即是要找的 $h%5Cleft(x%5Cright)%5Cin%20K%5Cleft%5Bx%5Cright%5D$ ，使得 $g%5Cleft(A_i%5Cright)%5E%7B-1%7D%3Dh%5Cleft(A_i%5Cright)%5Cleft(1%5Cle%20i%5Cle%20m%5Cright)$ .

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

練習(xí)題2(15級高代II每周一題第4題) 設(shè)n階方陣A適合多項式 $f%5Cleft(x%5Cright)%3Da_mx%5Em%2Ba_%7Bm-1%7Dx%5E%7Bm-1%7D%2B%5Ccdots%2Ba_1x%2Ba_0$ ，其中 $%5Cleft%7Ca_m%5Cright%7C%3E%5Csum_%7Bi%3D0%7D%5E%7Bm-1%7D%5Cleft%7Ca_i%5Cright%7C$ .證明：矩陣方程 $2X%2BAX%3DXA%5E2$ 只有零解.

證明? 首先證明多項式方程 $f%5Cleft(x%5Cright)%3D0$ 的根的模長一定小于1，用反證法，設(shè) $x_0%5Cgeq1$ 并且 $f%5Cleft(x_0%5Cright)%3D0$ ，則 $%5Cleft%7Ca_m%5Cright%7C%3D%5Cfrac%7B1%7D%7Bx_0%5Em%7D%5Cleft%7Ca_%7Bm-1%7D%7Bx_0%7D%5E%7Bm-1%7D%2B%5Ccdots%2Ba_1x_0%2Ba_0%5Cright%7C%5Cle%5Csum_%7Bi%3D0%7D%5E%7Bm-1%7D%5Cleft%7Ca_i%5Cright%7C$ ，矛盾.