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Let $A$ and $B$ be two real $n\times n$ matrices such that $AB-BA=A$

Prove that $A$ is not invertible and that $A$ is nilpotent.

My attempt is the following.

It holds that $AB=(B+I)A$

If $A$ were invertible, $B=A^{-1}(B+I)A$.

Taking trace on both sides yields $tr(B)=tr(B)+n$, hence $n=0$, which is a contradiction.

I can't prove that $A$ is nilpotent though.

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Gabriel Romon
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    This is a special case of http://math.stackexchange.com/questions/299640/ab-ba-is-a-nilpotent-matrix-if-it-commutes-with-a?rq=1 – J. J. May 27 '14 at 12:05
  • Since nilpotent implies being not invertible, that is the only part you actually need. –  May 27 '14 at 14:17

1 Answers1

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We prove by induction that

$$A^kB-BA^k=kA^k$$ in fact for the inductive step we multiply the above equality on the left by $A$ we get $$A^{k+1}B-\color{red}{AB}A^k=kA^{k+1}$$ and since $\color{red}{AB}=BA+A$ we get the desired result.

Now let the linear transformation:

$$\Phi(X)=XB-BX$$ If $A^k\ne0,\;\forall k$ then we see from the first equality that $k$ is an eigenvalue of $\Phi$ for every $k$ which's impossible (in finite dimensional space) then there's $m$ such that $A^m=0$ hence $A$ is a nilpotent matrix and then not invertible.

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    Nicely done! I like your proof. – 5xum May 27 '14 at 12:00
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    HMmm it works, it's beautiful... But how did you first derive the first formula ? I mean, it's not obvious to find it... – Gabriel Romon May 27 '14 at 18:19
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    @G.T.R To prove that $A$ is nilpotent we think to take the power of $A$. First we multiply the given equality by $A$ and we find $$A^2B-BA^2=2A^2$$ eurêka.... –  May 27 '14 at 19:09