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?

  • $\begingroup$ Welcome! Note that askers are expected to provide context for their questions, as is explained here. For example, it would be helpful if you could edit your questions to address some of the following. What motivated this problem, or where did you encounter it? What are your thoughts on the problem? What have you tried so far? $\endgroup$ – Ben Grossmann Jun 22 at 16:20
  • $\begingroup$ A potentially helpful way to reframe the problem: let $P = A - aI$ and $Q = B - bI$. We could equivalently ask: if $PQ = QP$ and $K,L$ are such that $P^Kv = Q^Lv = 0$, then does there necessarily exist an $M$ such that $$ ([P + aI][Q + bI] - ab I)^Mv = 0? $$ $\endgroup$ – Ben Grossmann Jun 22 at 16:25
  • $\begingroup$ The answer is yes: the statement is true. Once you provide some context, I can leave a proof as an answer. $\endgroup$ – Ben Grossmann Jun 22 at 16:46
  • $\begingroup$ Just added some context. Thanks for pointing the rules out btw, I'm a new poster here. $\endgroup$ – Underbounded Jun 22 at 16:47
  • $\begingroup$ I figured as much, thanks for obliging. $\endgroup$ – Ben Grossmann Jun 22 at 16:48

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, $AB-ab\in N$.

  • $\begingroup$ This is a really neat approach! $\endgroup$ – Underbounded Jun 22 at 21:23

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.

  • 1
    $\begingroup$ Binomial theorem suffices if you write $AB-abI$ as $(A-aI)B+a(B-bI)$. This also lowers the value of $m$ to $K+L-1$. $\endgroup$ – user1551 Jun 22 at 17:45
  • $\begingroup$ @user1551 That's neat! Actually, I realize now that $M = K +L - 1$ suffices for me as well; we only need $k_1 + k_2 \geq K$ or $k_1 + k_3 \geq L$ $\endgroup$ – Ben Grossmann Jun 22 at 17:57

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