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Problem: Suppose that $A$ and $B$ are linear transformations of a finite dimensional complex vector space such that $AB−BA=A$. If $v$ is an eigenvector of $B$ with eigenvalue $\lambda$, show that $Av$ is zero or an eigenvector of $B$ and find its eigenvalue. Prove that $A$ is nilpotent.

This problem has already been discussed many times here. But my doubt is in the official solution. After proving that $Av$ is zero or an eigenvector of $B$ with eigenvalue $\lambda-1$, they prove that $A^nv=0$, which I understand. But then they do this:

Suppose $V$ is the vector space, and if it is nonzero pick a non-zero eigenvector $v$ of $B$. By induction on the dimension of $V$, $A$ is nilpotent on $V/\mathbb Cv$, and is nilpotent on $v$, so is nilpotent on $V$.

How did they conclude that $A$ is nilpotent on $V/\mathbb{C}v$ and why does it imply $A$ is nilpotent on $V$ as well? Thanks.

Takamoto Yuji
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    $V/\mathbb Cv$ is a vector space of smaller dimension, so you use the inductive hypothesis on it. Then $V=(V/\mathbb Cv)\oplus \mathbb Cv$, so you get the conclusion. – Dave Jul 30 '23 at 19:31
  • An alternative solution is showing by induction $A^nB-BA^n=nA^n$ so that $\operatorname{tr}(A^n)=0$ and so $A$ is nilpotent (many proofs on this site). – zwim Jul 30 '23 at 20:15
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    Your second-to-last sentence should finish with $V$, not with $A$. Secondly, you should please directly refer to what you denote as "official solution". Also, you may look at https://math.stackexchange.com/a/3062183/316749 . – Hanno Jul 30 '23 at 20:22
  • If $AB-BA=A$, and we write $E_{\lambda}$ for the $\lambda$-eigenspace of $B$, then if $v \in E_{\lambda}$ we have $B(A(v)) = (\lambda-1)A(v)$, so that $A(E_{\lambda}) \subseteq E_{\lambda-1}$. Can you see why this implies that there is an eigenspace of $B$ which is contained in the kernel of $A$? – krm2233 Jul 30 '23 at 21:42
  • @Dave I had a similar thought, but for induction hypothesis to hold $A,B$ should satisfy the same equation over $V/\mathbb{C}v$. How do we deal with this? – Takamoto Yuji Jul 31 '23 at 03:28
  • They satisfy the equation as operators on $V$, so when we pass to the quotient $V\to V/\mathbb Cv$ they still satisfy the equation. – Dave Jul 31 '23 at 15:27
  • @Dave I still quite don't get it. For A to induce a map on $V/\mathbb C v$, we should have $v$ is an eigenvector of of $A$, right? But we have not proved anything like this. Here I am inducing the map $\overline{A}(\overline{v}):=\overline{A(v)}$. Again, I am sorry for this chaos. – Takamoto Yuji Jul 31 '23 at 16:11
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    That's a good question. One way to do this concretely is the following. Make $v$ be an eigenvector for the smallest eigenvalue $\lambda_0$ of $B$ (smallest in real part). Then $Av=0$, because otherwise $Av$ is a $\lambda_0-1$-eigenvector of $B$, but $B$ has no $\lambda_0-1$ eigenvalue. Then $A$ induces an operator on $V/\mathbb Cv$. – Dave Jul 31 '23 at 17:28

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