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Given matrices A and B, where that AB = A + B, prove AB = BA.

I keep coming up with AB = AB. It seems like basic algebra, but for the life of me, I'm getting nowhere :/. Someone help please?

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Don't worry, this is a tricky question! –  Evariste Oct 1 '11 at 20:59
    
$A=B=2I$ is a solution and $AB=BA=4I$. I suspect that this is the only solution but have no proof. I will try to come up with one later. And I can smell those matrix exponentials :) –  user13838 Oct 1 '11 at 22:46
    
@percusse: There are infinitely many solutions. (You can construct a solution from any invertible matrix; see the answers below.) –  Hans Lundmark Oct 2 '11 at 6:51
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This question really was a bit tricky. I admit that I had made it as far as proving that any eigenvector of $B$ is also an eigenvector of $A$, and, hence, the claim follows in the case that $B$ is diagonalizable, because $A$ and $B$ are then simultaneously diagonalizable. Luckily Hans' hint came quickly enough to save me from trying to figure out what to do with non-diagonal Jordan components (rolls eyes). Deleted the commutative-algebra tag, as it is misplaced here. –  Jyrki Lahtonen Oct 2 '11 at 7:19
    
@HansLundmark ouch, of course! Sorry for my limitations :) –  user13838 Oct 2 '11 at 9:34

2 Answers 2

Assume for now that 1 is not an eigenvalue of $B$. Then $B-I$ is invertible, so from the assumption we get $A=B (B-I)^{-1}$. $(B-I)^{-1}$ commutes with $B$ since it commutes with $B-I$ and with $I$.

EDIT: Now I claim that 1 can't be an eigenvalue of $B$. Indeed, suppose $Bv=v$ for some vector $v$. Then $ABv=Av+Bv$, hence $Av=Av+v$, hence $v=0$.

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It seems to me that rescaling $A$ and $B$ does hurt the hypotheses (except for scalars satisfying $\lambda^2=\lambda$, which aren't very useful for changing the eigenvalues)... On the other hand, it follows from the assumption that $B-I$ is invertible (see the hint in my answer). –  Hans Lundmark Oct 2 '11 at 6:55
    
$Hans: you're right. I edited my answer to include a different approach. Hopefully everything is OK now. –  Mark Oct 2 '11 at 13:26
    
Looks fine now. Nice argument! –  Hans Lundmark Oct 2 '11 at 16:27

Hint: Consider $(A-I)(B-I)$, where $I$ is the identity matrix.

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