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关于可交换阵的专题讨论

时间:2014-04-24 04:15:05      阅读:717      评论:0      收藏:0      [点我收藏+]

$\bf命题1:$设$A,B \in {M_n}\left( F\right)$且矩阵$A$各特征值互异,若$AB=BA$,则

$(1)$$A,B$可同时相似对角化

$(2)$$A,B$有公共的特征向量

$(3)$存在唯一的次数不超过$n-1$的多项式$f\left( x \right) \in F\left[ x \right]$,使得$B=f(A)$

证明:$(1)$由矩阵$A$各特征值互异知,存在可逆阵$R$,使得

Rbubuko.com,布布扣?1bubuko.com,布布扣AR=diag(λbubuko.com,布布扣1bubuko.com,布布扣,?,λbubuko.com,布布扣nbubuko.com,布布扣)bubuko.com,布布扣

其中${{\lambda _1}, \cdots ,{\lambda _n}}$为$A$互异的特征值

由$AB=BA$知,${R^{ - 1}}AR \cdot {R^{ - 1}}BR = {R^{ - 1}}BR \cdot {R^{ - 1}}AR$,从而可知

Rbubuko.com,布布扣?1bubuko.com,布布扣BR=diag(μbubuko.com,布布扣1bubuko.com,布布扣,?,μbubuko.com,布布扣nbubuko.com,布布扣)bubuko.com,布布扣

即$A,B$可同时相似对角化

$(2)$由$(1)$知,存在可逆阵$R = \left( {{\alpha _1}, \cdots ,{\alpha _n}} \right)$,使得

AR=Rdiag(λbubuko.com,布布扣1bubuko.com,布布扣,?,λbubuko.com,布布扣nbubuko.com,布布扣),BR=Rdiag(μbubuko.com,布布扣1bubuko.com,布布扣,?,μbubuko.com,布布扣nbubuko.com,布布扣)bubuko.com,布布扣

Aαbubuko.com,布布扣ibubuko.com,布布扣=λbubuko.com,布布扣ibubuko.com,布布扣αbubuko.com,布布扣ibubuko.com,布布扣,Bαbubuko.com,布布扣ibubuko.com,布布扣=μbubuko.com,布布扣ibubuko.com,布布扣αbubuko.com,布布扣ibubuko.com,布布扣,i=1,2,?,nbubuko.com,布布扣

$(3)$由于
bubuko.com,布布扣?bubuko.com,布布扣?bubuko.com,布布扣bubuko.com,布布扣B=abubuko.com,布布扣0bubuko.com,布布扣E+abubuko.com,布布扣1bubuko.com,布布扣A+abubuko.com,布布扣2bubuko.com,布布扣Abubuko.com,布布扣2bubuko.com,布布扣+?+abubuko.com,布布扣n?1bubuko.com,布布扣Abubuko.com,布布扣n?1bubuko.com,布布扣bubuko.com,布布扣Rbubuko.com,布布扣?1bubuko.com,布布扣BR=abubuko.com,布布扣0bubuko.com,布布扣E+abubuko.com,布布扣1bubuko.com,布布扣(Rbubuko.com,布布扣?1bubuko.com,布布扣AR)+abubuko.com,布布扣2bubuko.com,布布扣(Rbubuko.com,布布扣?1bubuko.com,布布扣AR)bubuko.com,布布扣2bubuko.com,布布扣+?+abubuko.com,布布扣n?1bubuko.com,布布扣(Rbubuko.com,布布扣?1bubuko.com,布布扣AR)bubuko.com,布布扣n?1bubuko.com,布布扣bubuko.com,布布扣?bubuko.com,布布扣?bubuko.com,布布扣?bubuko.com,布布扣?bubuko.com,布布扣?bubuko.com,布布扣abubuko.com,布布扣0bubuko.com,布布扣λbubuko.com,布布扣1bubuko.com,布布扣+abubuko.com,布布扣1bubuko.com,布布扣λbubuko.com,布布扣1bubuko.com,布布扣+abubuko.com,布布扣2bubuko.com,布布扣λbubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣2bubuko.com,布布扣+?+abubuko.com,布布扣n?1bubuko.com,布布扣λbubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣n?1bubuko.com,布布扣=μbubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣?bubuko.com,布布扣abubuko.com,布布扣0bubuko.com,布布扣λbubuko.com,布布扣nbubuko.com,布布扣+abubuko.com,布布扣1bubuko.com,布布扣λbubuko.com,布布扣1bubuko.com,布布扣+abubuko.com,布布扣2bubuko.com,布布扣λbubuko.com,布布扣nbubuko.com,布布扣bubuko.com,布布扣2bubuko.com,布布扣+?+abubuko.com,布布扣n?1bubuko.com,布布扣λbubuko.com,布布扣nbubuko.com,布布扣bubuko.com,布布扣n?1bubuko.com,布布扣=μbubuko.com,布布扣nbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣

而上述线性方程组的系数矩阵的$\bf{Vandermonde行列式}

D=bubuko.com,布布扣
\prod\limits_{1 \le j < i \le n} {\left( {{\lambda _i} - {\lambda _j}} \right)} \ne 0$,故存在唯一解${{a_0},{a_1},{a_2}, \cdots ,{a_{n - 1}}}$,即存在唯一的多项式$f\left( x \right) = {a_0} + {a_1}x + {a_2}{x^2} + \cdots + {a_{n - 1}}{x^{n - 1}}$满足$B=f(A)$

$\bf命题2:$设$A,B \in {M_n}\left( F\right)$且$A,B$均可对角化,若$AB=BA$,则$A,B$可同时相似对角化

关于可交换阵的专题讨论,布布扣,bubuko.com

关于可交换阵的专题讨论

原文:http://www.cnblogs.com/ly142857/p/3683207.html

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