Generalized eigenvector for product of commuting matrices 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?
 A: 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, $AB-ab\in N$.
A: Yes. We can proceed as follows.
Begin by reframing the question: if $PQ = QP$ and $K,L$ are such that $P^Kv = Q^Lv = 0$ and if $a,b$ are arbitrary scalars, then does there necessarily exist an $M$ such that
$$
([P + aI][Q + bI] - ab I)^Mv = 0?
$$
Note that
$$
[P + aI][Q + bI] - ab I = PQ + bP + aQ.
$$
Now, take $m = \max\{K,L\}$ and $M = 2m-1$; note that $P^mv = Q^mv = 0$. Apply the multinomial theorem:
$$
(PQ + bP + aQ)^Mv = \sum_{k_1 + k_2 + k_3 = M} \binom{M}{k_1,k_2,k_3} a^{k_3}b^{k_2}P^{k_1+k_2}Q^{k_1+k_3}v \\= 
\sum_{k_1 + k_2 + k_3 = M} \binom{M}{k_1,k_2,k_3} 
a^{k_3}b^{k_2}
Q^{k_1+k_3}P^{k_1+k_2}v.
$$
Now, for all triples of non-negative integers $k_1,k_2,k_3$ with $k_1 + k_2 + k_3 = 2m-1$, we see that $k_1 + k_2$ and $k_1 + k_3$ are non-negative integers with sum greater than or equal to $2m-1$. It follows that either $k_1 + k_2 \geq m$ or $k_1 + k_3 \geq m$. In either case, we find that
$$
P^{k_1+k_2}[Q^{k_1 + k_3}v] = 
Q^{k_1+k_3}[P^{k_1+k_2}v] = 0.
$$
Thus, the above expansion of $(PQ + bP + aQ)^Mv$ must be equal to zero.
