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时间:2014-05-04 20:10:21      阅读:332      评论:0      收藏:0      [点我收藏+]

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$\bf命题:$设$A \in {M_{m \times n}}\left( F \right),B \in {M_{n \times m}}\left( F \right),m \ge n,\lambda \ne 0$,则

|λEbubuko.com,布布扣mbubuko.com,布布扣?AB|=λbubuko.com,布布扣m?nbubuko.com,布布扣|λEbubuko.com,布布扣nbubuko.com,布布扣?BA|bubuko.com,布布扣

方法二:等价标准形

由$r\left( A \right) = r$知,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
Qbubuko.com,布布扣?1bubuko.com,布布扣BPbubuko.com,布布扣?1bubuko.com,布布扣=(Bbubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣Bbubuko.com,布布扣4bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣Bbubuko.com,布布扣3bubuko.com,布布扣bubuko.com,布布扣Bbubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣)bubuko.com,布布扣

其中${B_1}$为$r \times r$矩阵,于是有
PABPbubuko.com,布布扣?1bubuko.com,布布扣=PAQ?Qbubuko.com,布布扣?1bubuko.com,布布扣BPbubuko.com,布布扣?1bubuko.com,布布扣=(Bbubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣0bubuko.com,布布扣bubuko.com,布布扣Bbubuko.com,布布扣3bubuko.com,布布扣bubuko.com,布布扣0bubuko.com,布布扣bubuko.com,布布扣)bubuko.com,布布扣

Qbubuko.com,布布扣?1bubuko.com,布布扣BAQ=Qbubuko.com,布布扣?1bubuko.com,布布扣BPbubuko.com,布布扣?1bubuko.com,布布扣?PAQ=(Bbubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣Bbubuko.com,布布扣4bubuko.com,布布扣bubuko.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
|λEbubuko.com,布布扣mbubuko.com,布布扣?AB|=bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣λEbubuko.com,布布扣rbubuko.com,布布扣?Bbubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣0bubuko.com,布布扣bubuko.com,布布扣?Bbubuko.com,布布扣3bubuko.com,布布扣bubuko.com,布布扣λEbubuko.com,布布扣m?rbubuko.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=λbubuko.com,布布扣m?rbubuko.com,布布扣|λEbubuko.com,布布扣rbubuko.com,布布扣?Bbubuko.com,布布扣1bubuko.com,布布扣|bubuko.com,布布扣

|λEbubuko.com,布布扣nbubuko.com,布布扣?BA|=bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣λEbubuko.com,布布扣rbubuko.com,布布扣?Bbubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣?Bbubuko.com,布布扣4bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣0bubuko.com,布布扣λEbubuko.com,布布扣n?rbubuko.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=λbubuko.com,布布扣n?rbubuko.com,布布扣|λEbubuko.com,布布扣rbubuko.com,布布扣?Bbubuko.com,布布扣1bubuko.com,布布扣|bubuko.com,布布扣

故结论成立

$\bf注1:$命题中的结论称为$\bf行列式的降阶公式$

$\bf注2:$$\left| {\lambda {E_m} - AB} \right| = \left| {{P^{ - 1}}} \right| \cdot \left| {\lambda {E_m} - AB} \right| \cdot \left| P \right| = \left| {{P^{ - 1}}\lambda {E_m}P - {P^{ - 1}}ABP} \right| = \left| {\lambda {E_m} - {P^{ - 1}}ABP} \right|$

99,布布扣,bubuko.com

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原文地址:http://www.cnblogs.com/ly758241/p/3706332.html

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