If $A=AB-BA$, is $A$ nilpotent?

Question

Let matrix $A_{n\times n}$, be such that there exists a matrix $B$ for which $$AB-BA=A$$

Prove or disprove that there exists $m\in \mathbb N^{+}$such $$A^m=0,$$

I know $$tr(A)=tr(AB)-tr(BA)=0$$ then I can't.Thank you

Answer 1

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 $A^m=0$.

Answer 2

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 3

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 $A^m = 0$.

Answer 4

With the risk of being repetitive, let me just record here the (original?) proof of N. Jacobson in this paper, where you only need to assume that $[A,B]$ and $A$ commute to deduce $[A,B]$ is nilpotent.

Let $[A,B]=A'$, and consider $D(X) = [X,B]$. Then for any polynomial $F$, we have that $D(F(A)) = F'(A)A'$. Now pick a polynomial such that $F(A)=0$ (this can always be done, since the matrices $\{1,\ldots,A^N\}$ are linearly dependent if $N$ is large).

Since $A'$ commutes with $A$, iterating the above we see that if $F$ has degree $d$, then we have that $F^{(d)}(A)A'^{(2d-1)}=d! A'^{(2d-1)}$. Hence, over a field of zero characteristic we get $A'$ nilpotent.

Answer 5

Here is another proof (using characteristic zero).

The matrix $B$ defines a linear transformation on the space of $n\times n$ matrices by $X\mapsto \varphi(X)=XB-BX$. The equation you gave just means that $A$ is an eigenvector for the eigenvalue 1. Now suppose $X$ is an eigenvector for eigenvalue $\lambda$. Then $$\varphi(X^2)=X^2B-BX^2=X(BX+\lambda X)-BX^2=XBX+\lambda X^2-BX^2=(BX+\lambda X)X+\lambda X^2-BX^2=BX^2+\lambda X^2+\lambda X^2-BX^2=2\lambda X^2$$ So we get that $X^2$ is an eigenvector for an eigenvalue $2\lambda$. The endomorphism $\varphi$ has only finitely many distinct eigenvalues. Assuming the your field is of characteristic zero (so the $2^n \lambda$ is an infinite sequence), then you must have $X^n=0$ for some big enough $n$.