矩阵论作业13,14,15讲
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2025-02-09 14:49:41
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(AH)+=(A+)H
证明:
{1}逆:AH(A+)HAH=(AA+A)H=AH
{2}逆:(A+)HAH(A+)H=(A+AA+)H=(A+)H
{3}逆:(AH(A+)H)H=((A+A)H)H=(A+A)H=AH(A+)H
{4}逆:((A+)HAH)H=(A(A+)H)H=(AA+)H=(A+)HAH
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(AHA)+=A+(AH)+
证明:
{1}逆:AHAA+(AH)+AHA=AHAA+(A+)HAHA=AHAA+(AA+)HA=AHAA+(AA+)A=AHA
后面证明省略啦

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H3H=HHH H(前三个可运用{1,2}逆性质)=H2,认真看就能看出来

证明难点:
证明(1):
P2=(P1+P2)2=P12+P1P2+P2P1+P22=P1+P1P2+P2P1+P2,则P1P2+P2P1=0
P1P2+P1P2P1=0
P1P2P1+P2P1=0
两式相减可得:P1P2−P2P1=0,和P1P2+P2P1=0联立可得:P1P2=P2P1=0
后面的证明方法类似