首页 > 其他 > 详细

46

时间:2014-05-04 20:20:23      阅读:541      评论:0      收藏:0      [点我收藏+]

$\bf命题1:$任意方阵$A$均可分解为可逆阵$B$与幂等阵$C$之积

证明:设$r\left( A \right) = r$,则存在可逆阵$P,Q$,使得

PAQ=(Ebubuko.com,布布扣rbubuko.com,布布扣bubuko.com,布布扣0bubuko.com,布布扣bubuko.com,布布扣0bubuko.com,布布扣0bubuko.com,布布扣bubuko.com,布布扣)bubuko.com,布布扣

从而可知
Abubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣=Pbubuko.com,布布扣?1bubuko.com,布布扣(Ebubuko.com,布布扣rbubuko.com,布布扣bubuko.com,布布扣0bubuko.com,布布扣bubuko.com,布布扣0bubuko.com,布布扣0bubuko.com,布布扣bubuko.com,布布扣)Qbubuko.com,布布扣?1bubuko.com,布布扣bubuko.com,布布扣=Pbubuko.com,布布扣?1bubuko.com,布布扣Qbubuko.com,布布扣?1bubuko.com,布布扣.Q(Ebubuko.com,布布扣rbubuko.com,布布扣bubuko.com,布布扣0bubuko.com,布布扣bubuko.com,布布扣0bubuko.com,布布扣0bubuko.com,布布扣bubuko.com,布布扣)Qbubuko.com,布布扣?1bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣

取$B = {P^{ - 1}}{Q^{ - 1}}$,$C = Q\left( {
Ebubuko.com,布布扣rbubuko.com,布布扣bubuko.com,布布扣0bubuko.com,布布扣bubuko.com,布布扣0bubuko.com,布布扣0bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣
} \right){Q^{ - 1}}$,即证

46,布布扣,bubuko.com

46

原文:http://www.cnblogs.com/ly758241/p/3706354.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!