TITLE: Generalized eigenvector for product of commuting matrices
QUESTION [7 upvotes]: Suppose $A,B$ are commuting invertible matrices with a common generalized eigenvector $v$ with eigenvalues $a,b$ respectively. That is, suppose there exist positive integers $K,L$ such that $(A-aI)^K v= 0$ and $(B-bI)^Lv=0$. Is it true that there exists a positive integer $M$ such that $(AB-abI)^Mv = 0$?
A bit of context: I came across this while thinking about the analogous version of this for eigenvalues (when $K=1$ and $L=1$). This is easy to show. In fact, the above statement is also easy to show if only one of $K=1$ or $L=1$ is true (that is, $v$ is an eigenvector for one matrix and a generalized eigenvector for the other).
Proof: WLOG suppose $L=1$. Then we have
\begin{align}
(AB-abI)^Kv &= \sum_{j=0}^K {K \choose j} A^{K-j}B^{K-j}a^jb^jv\\
&= \sum_{j=0}^K {K \choose j} A^{K-j}a^jb^Kv\\
&= b^K(A-aI)^Kv\\
&= 0
\end{align}
A natural question: is it true if $v$ is a generalized eigenvector for both matrices?

REPLY [6 votes]: Here is a more abstract perspective.  Let $V$ be the smallest subspace of our vector space which contains $v$ and is closed under the actions of $A$ and $B$.  Let $R$ be the subring of the endomorphism ring of $V$ generated by the scalar matrices together with $A$ and $B$.  By assumption, this $R$ is commutative, and $(A-a)^K$ and $(B-b)^L$ are equal to $0$ in this ring (since they annihilate $v$ and $v$ generates $V$ under the action of $R$).  The question then becomes simply:

Let $R$ be a commutative ring and $A,B,a,b\in R$.  Suppose $A-a$ and $B-b$ are nilpotent.  Then must $AB-ab$ be nilpotent?

But now the answer is very easy using the fact that the set $N$ of nilpotent elements of $R$ is an ideal.  By assumption, $A=a$ and $B=b$ in the quotient ring $R/N$.  Thus $AB=ab$ in that quotient ring.  That is, $AB-ab\in N$.