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Let $A$ and $B$ be $n×n$ complex matrices.

If $A$ is a nilpotent matrix, and $A$ commute with $AB−BA$, show that $AB$ is nilpotent.

Equivalently, the question can be expressed as following description.

Let $A$ and $B$ be $n×n$ complex matrices.

Define the linear transformation $T$ as $T(B)=AB-BA$.

If $A$ is a nilpotent matrix, and $T^2(B)=0$ , show that $AB$ is nilpotent.

I've known that $AB-BA$ is nilpotent.

Furtherly, if $A^m=0$ , by considering $T^n(B)=\sum_{i=0}^n(-1)^iA^{n-i}BA^i$ , I found that $A^kBA^l=0$ when $k+l\geqslant m$.

But I don't know how to continue, thanks for any help.

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  • $\begingroup$ Compare also with this question. $\endgroup$
    – Dietrich Burde
    Commented Apr 17, 2019 at 19:27
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    $\begingroup$ This is actually the main theorem of an article by Kaplansky. The statement is even generalized there. $\endgroup$
    – Helmut
    Commented Apr 18, 2019 at 21:34
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    $\begingroup$ @Helmut: The link is dead. What article of Kaplansky's is it? Maybe "Jacobson's Lemma Revisited" (J. Algebra 62, 473-476 (1980)), cf. zbmath.org/?q=an%3A0425.16020 ? $\endgroup$
    – Torsten Schoeneberg
    Commented Apr 5, 2021 at 3:36
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    $\begingroup$ @TorstenSchoeneberg Yes, that is the article. By the way, the link is active again:-) $\endgroup$
    – Helmut
    Commented Apr 10, 2021 at 13:59

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