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关于等价标准形的专题讨论

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$\bf命题1:$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
A=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

关于等价标准形的专题讨论,布布扣,bubuko.com

关于等价标准形的专题讨论

标签:style   blog   http   ar   html   log   sp   amp   c   

原文地址:http://www.cnblogs.com/ly142857/p/3672719.html

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