If $A=AB-BA$, is $A$ nilpotent? [duplicate]

For any $k$, you have $A^{k+1}=AA^k=(AB-BA)A^k=ABA^k-BA^{k+1}$. So $$ \text{tr}(A^{k+1})=0,\ \ k=0,1,2,\ldots $$ We deduce that $\text{tr}(p(A))=0$ for every polynomial $p$ with $p(0)=0$. This implies that all eigenvalues of $A$ are zero (because otherwise we can get a polynomial $p$ that is 1 at all nonzero eigenvalues, and $p(A) $ would not have zero trace). So $A$ is nilpotent, i.e. there exists $m$ with $A^m=0$.

Denote by $p$ the characteristic of the field. The statement does not necessarily hold if $0<p\le n$. For a counterexample, consider $p=n=2$ and $A = \pmatrix{0&1\\ 1&0},\ B=\pmatrix{1&1\\ 1&0}$.

The statement is true, however, if $p=0$ or $p>n$. From the condition $AB-BA=A$, one can prove by mathematical induction that $A^k=ABA^{k-1}-BA^k$ and in turn $\operatorname{trace}(A^k)=0$ for $k=1,2,\ldots,n$. Now, it is know that this latter condition together with $p=0$ or $p>n$ imply that $A$ is nilpotent (for a proof, see achille hui's answer in this thread or my answer to another question).

Answer Changed Since the other answer (beat me by 10 minute) is already using eigenvalues, I've rewritten the answer to do it in an alternate way.

Let $A, B \in M_{m\times m}(\mathbb{K})$ such that $[A,B] = A B - BA = A$ and $\mathbb{K}$ is a field with characteristic $0$ or greater than $m$. For any $k \in \mathbb{Z}_{+}$, we have

$$[ A^k, B ] = A^{k-1} [ A, B ] + A^{k-2} [A, B ] A + \cdots + A [ A, B ] A^{k-2} + [A, B] A^{k-1} = kA^{k}$$

Let $\chi_{A}(\lambda)$ be the characteristic polynomial of $A$, i.e.

$$\chi_{A}(\lambda) = \det(\lambda I_m - A) = \lambda^m + \alpha_{m-1}\lambda^{m-1} + \cdots + \alpha_0 = 0$$

By Cayley-Hamilton theorem, we have $$\chi_{A}(A) = A^m + \alpha_{m-1}A^{m-1} + \cdots + \alpha_0 I_m.$$ Repeat apply the commutator $[ \cdot, B ]$ to it, we find: $$ \begin{array}{rcrcrcrcr} 0 &=& A^m &+& \alpha_{m-1}A^{m-1} &+ \cdots +& \alpha_1 A &+& \alpha_0 I_m\\ 0 &=& m A^m &+& (m-1)\alpha_{m-1}A^{m-1} &+ \cdots +& \alpha_1 A &+& 0\\ 0 &=& m^2 A^m &+& (m-1)^2\alpha_{m-1}A^{m-1} &+ \cdots +& \alpha_1 A &+& 0\\ &\vdots\\ 0 &=& m^m A^m &+& (m-1)^m\alpha_{m-1}A^{m-1} &+ \cdots +& \alpha_1 A &+& 0\\ \end{array} $$ This can be recasted in a matrix form: $$ \begin{pmatrix} 1 & 1 & \ldots & 1 & 1\\ m & m-1 & \ldots & 1 & 0\\ m^2 & (m-1)^2 & \ldots & 1 & 0\\ &\vdots\\ m^m & (m-1)^m & \ldots & 1 & 0\\ \end{pmatrix} \begin{pmatrix}A^m\\ \alpha_{m-1}A^{m-1} \\ \alpha_{m-2}A^{m-2} \\ \vdots \\ \alpha_0 I_m\end{pmatrix} = \begin{pmatrix}0 \\ 0 \\ 0 \\ \vdots \\ 0\end{pmatrix}$$

Since the characteristic of $\mathbb{K}$ is $0$ or greater than $m$, the Vandermonde matrix appear in LHS above is invertible. This allow us to conclude $A^m = 0$.