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let $A,B,C$ be $n\times n$ matrices with real entries such that $A$ is invertible. if $(A-B)CA=B$ show that $AC(A-B)=B$.

any Ideas??

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up vote 6 down vote accepted

Let $Q = A-B$. Since $QCA=B = A-Q$, $Q(CA+I)=A$, and therefore $Q$ is invertible. Now $C = Q^{-1} B A^{-1} = Q^{-1}(A-Q)A^{-1} = Q^{-1} - A^{-1}$, so $AC(A-B) = A(Q^{-1} - A^{-1}) Q = A - Q = B$.

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I could n't follow how Q−1(A−Q)A−1=Q−1−A−1 (last step in line 2 came to be). Cheers –  Hardy Feb 7 '12 at 17:57
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Just expand it out: $Q^{-1} (A - Q) A^{-1} = Q^{-1} A A^{-1} - Q^{-1} Q A^{-1}$ and then cancel: $Q^{-1} A A^{-1} = Q^{-1}$, $Q^{-1} Q A^{-1} = A^{-1}$. –  Robert Israel Feb 7 '12 at 18:51
    
It may be simple, but why does $Q(CA+I)=A$ make $Q$ invertible? –  Christian Rau Feb 7 '12 at 22:34
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$A$ was assumed invertible. $Q(CA+I)A^{-1} = I$. Any $n \times n$ matrix with a one-sided inverse has rank $n$ and therefore is inverible. –  Robert Israel Feb 8 '12 at 5:21
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