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.43

2022-01-30 17:39 作者:CharlesMa0606 0人读过 | 我要投稿

本文内容主要有关于可对角化的判定，在高代白皮书上对应第6.2.2节、第7.2.3节、第9.2.6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！

练习题1(例6.35) 设A,B,C是n阶方阵，A,B都有n个不同的特征值.设 $f%5Cleft(%5Clambda%5Cright)$ 是A的特征多项式，并且 $f%5Cleft(B%5Cright)$ 是可逆矩阵.求证：矩阵 $M%3D%5Cleft(%5Cbegin%7Bmatrix%7DA%26C%5C%5CO%26B%5C%5C%5Cend%7Bmatrix%7D%5Cright)$ 可对角化.

证明由A,B都有n个不同的特征值可知，A,B均可对角化，注意到对A,B做相似变换不改变 $f%5Cleft(%5Clambda%5Cright)$ 是A的特征多项式，也不改变 $f%5Cleft(Q%5E%7B-1%7DBQ%5Cright)%3DQ%5E%7B-1%7Df%5Cleft(B%5Cright)Q$ 是非异阵的事实，要证明的内容中C是任意的，并且做使得A,B化为对角阵的相似变换也不改变M的分块上三角性，因此也不改变证明的结论，从而不妨设A,B均为对角阵，并且各自的对角元不相同.

注意到 $f%5Cleft(B%5Cright)$ 是可逆矩阵，并且B是对角元互不相同的对角阵，从而 $f%5Cleft(B%5Cright)$ 是一个对角阵，并且对角元为B的特征值代入 $f%5Cleft(%5Clambda%5Cright)$ 中的函数值 $f%5Cleft(%5Cmu_1%5Cright)%2C%5Ccdots%2Cf%5Cleft(%5Cmu_n%5Cright)$ ，从而 $%5Cmu_i$ 不是 $f%5Cleft(%5Clambda%5Cright)$ 的根，这说明A,B的特征值互不相同，从而矩阵M有2n个互不相同的特征值，于是矩阵M可对角化.

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

练习题2(19级高代II每周一题第4题) 设A为数域K上的n阶方阵，f(x),g(x)为K上互素的多项式，且它们在C中均无重根.证明：若 $r%5Cleft(f%5Cleft(A%5Cright)%5Cright)%2Br%5Cleft(g%5Cleft(A%5Cright)%5Cright)%3Dn$ ，则A可对角化.

证明注意到f(x),g(x)在K上互素，从而

$%5Cexists%20u%5Cleft(x%5Cright)%2Cv%5Cleft(x%5Cright)%5Cin%20K%5Cleft%5Bx%5Cright%5D%2Cs.t.f%5Cleft(x%5Cright)u%5Cleft(x%5Cright)%2Bg%5Cleft(x%5Cright)v%5Cleft(x%5Cright)%3D1$

代入A，则 $f%5Cleft(A%5Cright)u%5Cleft(A%5Cright)%2Bg%5Cleft(A%5Cright)v%5Cleft(A%5Cright)%3DI_n$ .

考虑分块矩阵

$%5Cleft(%5Cbegin%7Bmatrix%7Df%5Cleft(A%5Cright)%26O%5C%5CO%26g%5Cleft(A%5Cright)%5C%5C%5Cend%7Bmatrix%7D%5Cright)%5Crightarrow%5Cleft(%5Cbegin%7Bmatrix%7Df%5Cleft(A%5Cright)%26g%5Cleft(A%5Cright)v%5Cleft(A%5Cright)%5C%5CO%26g%5Cleft(A%5Cright)%5C%5C%5Cend%7Bmatrix%7D%5Cright)%5Crightarrow%5Cleft(%5Cbegin%7Bmatrix%7DO%26I_n%5C%5C-f%5Cleft(A%5Cright)g%5Cleft(A%5Cright)%26O%5C%5C%5Cend%7Bmatrix%7D%5Cright)$

这说明f(A)g(A)=0，从而A适合首一多项式 $m%5Cleft(x%5Cright)%3Dcf%5Cleft(x%5Cright)g%5Cleft(x%5Cright)$ ，其中c是常数.由f(x),g(x)为K上互素的多项式，且它们在C中均无重根可知，m(x)在C上没有重根，原因是f(x),g(x)在K上互素，满足

$%5Cexists%20u%5Cleft(x%5Cright)%2Cv%5Cleft(x%5Cright)%5Cin%20K%5Cleft%5Bx%5Cright%5D%2Cs.t.f%5Cleft(x%5Cright)u%5Cleft(x%5Cright)%2Bg%5Cleft(x%5Cright)v%5Cleft(x%5Cright)%3D1$

将它们看作C上的多项式，则由多项式互素的充要条件可知 $f%5Cleft(x%5Cright)%2Cg%5Cleft(x%5Cright)$ 在C上互素.

设 $p%5Cleft(x%5Cright)%3D%5Cleft(x-a_1%5Cright)%5Ccdots%5Cleft(x-a_m%5Cright)$ 为p(x)在C上的因式分解，其中 $a_i$ 是互异的复数.我们现在先证明 $C%5En%3DKer%5Cleft(A-a_1I_n%5Cright)%5Coplus%5Ccdots%5Coplus%20Ker%5Cleft(A-a_mI_n%5Cright)$ .

可设 $p_i%5Cleft(x%5Cright)%3D%5Cleft(x-a_1%5Cright)%5Ccdots%5Cleft(x-a_%7Bi-1%7D%5Cright)%5Cleft(x-a_%7Bi%2B1%7D%5Cright)%5Ccdots%5Cleft(x-a_m%5Cright)$ ，则 $%5Cleft(p_1%5Cleft(x%5Cright)%2Cp_2%5Cleft(x%5Cright)%2C%5Ccdots%2Cp_m%5Cleft(x%5Cright)%5Cright)%3D1$ ，故存在 $u_i%5Cleft(x%5Cright)$ ，使得 $u_1%5Cleft(x%5Cright)p_1%5Cleft(x%5Cright)%2B%5Ccdots%2Bu_m(x)p_m(x)%3D1$ ，代入x=A，可得恒等式

$u_1%5Cleft(A%5Cright)p_1%5Cleft(A%5Cright)%2B%5Ccdots%2Bu_m%5Cleft(A%5Cright)p_m%5Cleft(A%5Cright)%3DI_n$

对任意 $%5Calpha%5Cin%20C%5En$ ，由上式可知 $%5Calpha%3D%5Csum_%7Bi%3D1%7D%5E%7Bm%7D%5Cleft(p_i%5Cleft(A%5Cright)u_i%5Cleft(A%5Cright)%5Cleft(%5Calpha%5Cright)%5Cright)$ ，其中 $p_i%5Cleft(A%5Cright)u_i%5Cleft(A%5Cright)%5Calpha%5Cin%20Ker%5Cleft(A-a_iI_n%5Cright)$ ，于是 $C%5En%3D%5Csum_%7Bi%3D1%7D%5E%7Bm%7DKer%5Cleft(A-a_iI_n%5Cright)$ .任取 $%5Calpha%5Cin%20Ker%5Cleft(A-a_1I_n%5Cright)%5Ccap%5Cleft(Ker%5Cleft(A-a_2I_n%5Cright)%2B%5Ccdots%2BKer%5Cleft(A-a_mI_n%5Cright)%5Cright)$ ，则 $%5Calpha%3D%5Calpha_2%2B%5Ccdots%2B%5Calpha_m%EF%BC%8C%E5%85%B6%E4%B8%AD%5Calpha_i%5Cin%20Ker%5Cleft(A-a_iI_n%5Cright)%5Cleft(i%5Cgeq2%5Cright)$ .于是

$%5Calpha%3Du_1%5Cleft(A%5Cright)p_1%5Cleft(A%5Cright)%5Cleft(%5Calpha_2%2B%5Ccdots%2B%5Calpha_m%5Cright)%2Bu_2%5Cleft(A%5Cright)p_2%5Cleft(A%5Cright)%5Cleft(%5Calpha%5Cright)%2B%5Ccdots%2Bu_m%5Cleft(A%5Cright)p_m%5Cleft(A%5Cright)%5Cleft(%5Calpha%5Cright)%3D0$ 由于指标的任意性，因此 $C%5En%3D%5Csum_%7Bi%3D1%7D%5E%7Bm%7DKer%5Cleft(A-a_iI_n%5Cright)$ 是直和.注意到A的特征值一定属于集合 $%5C%7Ba_1%2C%5Ccdots%2Ca_m%5C%7D$ ，从而剔除 $Ker%5Cleft(A-a_iI_n%5Cright)$ 中非特征值的部分，即得全空间等于特征子空间的直和，从而A可对角化.