$AB-BA=A$ implies $A$ is singular and $A$ is nilpotent. [duplicate]

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.

Solution 1:

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.