TITLE: Irreducible representations and invariant subspaces
QUESTION [5 upvotes]: Consider two invertible n-by-n matrices, $n>2$, $X$ and $Y$ over a finite field $k$ (say for simplicity $k=\mathbb Z/ \mathbb Z_2$). Is there any reasonable way to check that there is no proper subspace $V\subset k^n$ such that $V$ is invariant under the action of both $X$ and $Y$, i.e. $XV\subset V$ and $YV \subset V$?  
Any known classes of examples of such $X$ and $Y$ that do not have a proper
subspace $V\subset k^n$ such that $V$ is invariant under the action of both $X$ and $Y$?

REPLY [2 votes]: Firstly, about "known classes" of examples. Most obviously, if $X$ itself has irreducible characteristic polynomial, in which case it does not admit invariant subspaces.
A slightly more interesting example is where $X$ and $Y$ are both diagonalizable, but the eigenbasis of $Y$ consists entirely of vectors will all coordinates non-zero with respect to the eigenbasis of $X$.
Another interesting example is the matrix $X$ whose $(i,i+1)$th entry is $1$ and all other entries are $0$, and $Y=X^T$ (the transpose of $X$).
If you are interested in small values of $n$ (up to $4$), it may still be possible to classify all such pairs. This would be based on the fact that the invariant subspaces for $X$ depend only on its similarity class type, which has a classification (for each $n$) independent of the finite field.