Irreducible representations and invariant subspaces 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 $Y$? 
 A: Suppose $X,Y$ are such that they coincide on a co-dimension one subspace. Then the representation is irreducible if and only if the characteristic polynomials of $X,Y$ are co-prime. This is a result due to Levelt and is proved in a paper by Beukers and Heckman (Inventiones math 1989)? This is the only general class of examples I know. 
A: I think it is hopeless to obtain a general solution to this question. One very special case where something can said is when $X$ and $Y$ both have quadratic minimum polynomial (and even then, you may have to work over a larger field). I think this result really originates with H. Blichfeldt.
Suppose for the sake of exposition that the matrix $X-Y$ has an eigenvector $v (\neq 0)$ associated to an eigenvalue $\lambda \in k.$ 
 Then note that the span of $\{v,Xv\}$ is the same as the span of $\{v,Yv\}$,
since $Yv = Xv -\lambda v.$  
But since $X$ has quadratic minimum polynomial, the span of $\{v,Xv\}$ is $X$-invariant. Likewise, the span of $\{v,Yv\}$ is $Y$-invariant.
     But since these subspaces are equal, we see that $V$ is both $X$-invariant and $Y$-invariant, where $V$ is the span of $\{v,Xv \}.$ 
A: 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.
