Decidability of matrix problem in ${\mathbb Z}/p{\mathbb Z}$

Question

Let $p$ be a prime number, $n$ be a positive integer, and let ${\mathbb Z}_p^{n\times n}$ denote the set of $n\times n$-matrices over ${\mathbb Z}/p{\mathbb Z}$.

Suppose we are given an integer $m>0$ and matrices ${\bf A}_1,\ldots, {\bf A}_m\in {\mathbb Z}_p^{n\times n}.$ I am looking at the following problem: if there are positive integers $n_1,\ldots n_m$ such that $${\bf A}_1^{n_1}{\bf A}_2^{n_2}\cdots{\bf A}_m^{n_m} = {\bf 0},$$ output YES, otherwise output NO.

Is this problem decidable for all primes $p$, for all sizes of matrixes $n\times n$, and for all numbers of input matrices $m\in\mathbb{N}$?

Answer 1

Yes. If we conider the sequence $A_i^n$, because it is a sequence inside a finite set, it must eventually repeat. After it repeats, the sequence won't take any new values, so we can assume $n_i$ is less than this first repetition. This means that there are only finitely many possibilities, which of course reduces the problem to a finite calculation.

Answer 2

Even more is decidable: we may get a list of all possible spaces which are images of the operator $A_1^{n_1}\dots A_m^{n_m}$.

Induction in $m$. Base $m=1$. Consider the spaces $X_k=A^k(\mathbb{Z}_p^n)$. Two of them coincide, since they take only finitely many possible values (and we know an a priori estimate for this number), and if $X_k=X_l$, $k<l$, we have $X_{k+1}=X_{l+1}$ and so on, thus the list of possible spaces is $\{X_0,\dots,X_{l-1}\}$.

Induction step is the same.