Let $A, B, C$ be some complex matrices. If $AB - BA = C$ and $AC = CA$, then $C^k = 0$ for some $k$.

Asked Jun 05 '19 at 16:47

Active Jun 05 '19 at 16:47

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Let $A, B, C$ be some complex matrices. Suppose that $AB - BA = C$ and $AC = CA$. Prove that: $C^k = 0$ for some $k$.

It is an exercise in the section of "Jordan Canonical Form of Nilpotent Transforms". But I am not sure how to prove the question. Could anyone give me a hint?

asked Jun 05 '19 at 16:47

mathdoge

One can infer this, but I think it should be explicitly stated that these matrices are square. – paulinho Jun 05 '19 at 16:54
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Possible duplicate of $C=AB-BA$ is a nilpotent matrix if it commutes with $A$ – Dietrich Burde Jun 05 '19 at 16:56
The idea is to prove that the trace of $C$ is zero. If this is true, then it is similar to a Jordan form matrix with trace 0, which is nilpotent of degree $n$ (where $n$ is the size of the matrix). – paulinho Jun 05 '19 at 17:03

Let $A, B, C$ be some complex matrices. If $AB - BA = C$ and $AC = CA$, then $C^k = 0$ for some $k$.

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