6
$\begingroup$

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?

  • 0
    Don't worry, this is a tricky question!2011-10-01
  • 0
    $A=B=2I$ 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 :)2011-10-01
  • 0
    @percusse: There are infinitely many solutions. (You can construct a solution from any invertible matrix; see the answers below.)2011-10-02
  • 1
    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.2011-10-02
  • 0
    @HansLundmark ouch, of course! Sorry for my limitations :)2011-10-02

3 Answers 3