Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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??

share|cite|improve this question
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 $AC(A-B) = A(Q^{-1} - A^{-1}) Q = A - Q = B$.

share|cite|improve this answer
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 – Comic Book Guy Feb 7 '12 at 17:57
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
$A$ 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.