Proof for $A,B \in M_n(\mathbb{F})$ that if $[A,B]=tA$ for $0\neq t\in\mathbb{F}$, then $A^n=0$

Question

Statement. Suppose we have a square matrices $A,B$ of order $n$ over a field $\mathbb{F}$ of characteristics $0$ or $p>n$. If $[A,B]=AB-BA=tA$ for some nonzero $t\in\mathbb{F}$, then $A^n=0$. The problem is to prove this statement.

Where I've got so far. There is an idea to take trace of both sides of equation: $$ \mathrm{tr}\,(AB-BA)=\mathrm{tr}\,(tA) $$ $$ \mathrm{tr}\,(AB)-\mathrm{tr}\,(BA)=t\times\mathrm{tr}\,(A) $$ since $\mathrm{tr}\,(AB) = \mathrm{tr}\,(BA)$ (by properties of the trace) we have $$ t\times A = \mathrm{tr}\,(A) $$ and since $t\neq 0$, we have $\mathrm{tr}\,(A) = 0$. That's all I could get form it. What sould I do next?

@user1551 I wouldn't say it's a duplicate. While the question you linked certainly imply the result here (and thus one of the possible answers is "If $[A,B] = t A$ then $[A,B]$ and $A$ commute and so by this other question we're done"), there are simpler answers in this case. — Najib Idrissi, Sep 18 '15 at 08:43
@NajibIdrissi I agree that it's not an exact duplicate, but several similar questions have been asked on this site. So, I picked one with the weakest condition. It's fine if the community wants to reopen this question. — user1551, Sep 18 '15 at 08:53
@user1551 Do you know if there's another question with this condition as a hypothesis? — Najib Idrissi, Sep 18 '15 at 08:54
@NajibIdrissi There is a very similar one, with $t=1$: if $A=AB−BA$ then $A^n=0$?, and that question has its own duplicates. — user1551, Sep 18 '15 at 08:59
@user1551 I think closing as a duplicate of that one could be a good compromise, no? — Najib Idrissi, Sep 18 '15 at 09:01
@NajibIdrissi Oh, actually if you let $B=t\widetilde{B}$, the two questions are the same. Maybe I should close this question using that one. — user1551, Sep 18 '15 at 09:03

score 3 · Accepted Answer · edited Apr 13 '17 at 12:20

Hints.

Prove (by mathematical induction or otherwise) that $A^kB-BA^k=ktA^k$ for $k=1,2,\ldots$
The trace of a product of matrices is invariant under cyclical permutation of the multiplicands.
See the question "$\DeclareMathOperator{\tr}{tr}\tr(A)=\tr(A^{2})= \ldots = \tr(A^{n})=0$ implies $A$ is nilpotent". Several answers to this question are applicable when the underlying field has characteristic 0 or characteristic $p>n$.

score 2 · Answer 2 · answered Sep 18 '15 at 08:26

As I mentioned in the comments, we need to impose $t\neq0$ in order for the statement to hold. As Najib, says, this statement can be proved using the previous statement you asked about.

We have$$t\cdot\mathrm{tr}(A)=\mathrm{tr}(tA)=\mathrm{tr}(AB-BA)=0,$$which yields$$\mathrm{tr}(A)=0.$$Likewise, for any $k$, we have $tA^k=A^kB-A^{k-1}BA.$ Using an argument similar to the one above, we deduce that for any $k$, $\mathrm{tr}(A^k)=0$. Hence, $A$ is nilpotent.

Proof for $A,B \in M_n(\mathbb{F})$ that if $[A,B]=tA$ for $0\neq t\in\mathbb{F}$, then $A^n=0$

2 Answers2

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