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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 $Y$?

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  • $\begingroup$ "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 $YV\subset V$?" The zero subspace is invariant under the action of both $X$ and $Y$. $\endgroup$
    – Jason Starr
    Commented Jul 29, 2018 at 11:37
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    $\begingroup$ The question was about proper subspaces, i.e. not $0$ and not $k^n$. $\endgroup$
    – Serge Nikoforov
    Commented Jul 29, 2018 at 11:39
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    $\begingroup$ Thanks for clarifying. Please clarify, what is "$\mathbb{Z}/\mathbb{Z}_2$"? $\endgroup$
    – Jason Starr
    Commented Jul 29, 2018 at 12:45
  • $\begingroup$ Indeed if you mean the finite field with two elements, it would be better to denote it by $ \mathbb{F}_{2} $ . $\endgroup$
    – Sylvain JULIEN
    Commented Jul 29, 2018 at 13:28
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    $\begingroup$ @SergeNikoforov FYI, I think your definition of "proper subspace" is not universal, hence the confusion. (My default interpretation is a subspace which is a proper subset, i.e., just not the whole space.) $\endgroup$
    – Kimball
    Commented Jul 29, 2018 at 17:20

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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.

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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 $\{v,Xv \}.$

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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.

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