Generalized eigenvectors of AB and BA proof Let $A \in R^{m\times k}, B\in R^{k \times m}$ be non-square matrices. A vector $v \in R^m$ is said to be a **generalized** eigenvector of $AB$ corresponding to $\lambda \neq 0$ if $(AB- \lambda I)^m v = 0$. I know that if $v$ is an eigenvector of $AB$, then $Bv$ is an eigenvector of $BA$. However, if $v$ be any **generalized** eigenvector of $AB$, i.e., $ (AB- \lambda I)^m v = 0$ then can I say that $Bv$ is a corresponding **generalized** eigenvector of $BA$ ? That is, is it true that $(BA-\lambda I)^k Bv = 0$ ? By expanding the binomial series,and pre-multiplying with B, I think am getting the result. However, since generalized eigenvectors include ordinary eigenvectors, am worried if my purported proof is correct.

Let $n = \max(k, m)$. We have $$ (BA - \lambda I)^nBv = \left(\sum_{i = 0}^n \binom ni\lambda^{n-i}(BA)^i\right)Bv\\\ = \left(\lambda^nI + \sum_{i = 1}^n\binom ni \lambda^{n-i}B(AB)^{i-1}A\right)Bv\\\ = \left(\lambda^n B + \sum_{i = 1}^n\binom ni \lambda^{n-i}B(AB)^{i-1}AB\right)v\\\ = B\left(\lambda^nI + \sum_{i = 1}^n\binom ni \lambda^{n-i}(AB)^{i}\right)v\\\ = B(AB-\lambda I)^nv = B0 = 0 $$ So yes, indeed, if $v$ is a generalized eigenvector of $AB$ with generalized eigenvalue $\lambda$ and generalized eigenvector $v$, and $Bv\
eq 0$, then $Bv$ is a generalized eigenvector of $BA$.