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Suppose $A$, $B$ are real $n \times n$ matrices with $A + B = I$ and rank A + rank B = n. How can one show that $AB = BA = 0$?

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What have you tried? First, did you manage to get the easy part, that is, $AB=BA$? – Harald Hanche-Olsen Oct 28 '12 at 9:14

The problem reduces to show that if $A$ is a $n\times n$ real matrix, and $\operatorname{rank}(A)+\operatorname{rank}(I-A)=n$, then $A-A^2=0$.

First, using the formula $\operatorname{rank}(M)+\dim\ker(M)=n$ for any $n\times n$ matrix $M$, we have $\dim \ker (A)+\dim\ker(I-A)=n$.

So the eigenspaces $\ker A$ and $\ker (I-A)$ span $\Bbb R^n$ (as their intersection is $\{0\}$).

For $x\in\Bbb R^n$, write $x=y+z$, where $y\in \ker A$ and $z\in \ker(I-A)$, and compute $A(I-A)x$.

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Methinks perhaps you provided too much detail for a homework problem. – Harald Hanche-Olsen Oct 28 '12 at 9:38
Thank you Davide, this is just the right help. The computation gives 0 and that solves the problem. – Martin Oct 28 '12 at 15:16

Hint: Try to prove that the $\Bbb R^n=im A\oplus imB$, including $im A\cap im B=\{0\}$.

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