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Consider $\mathbb{F}_2^{2\times2}$, the $2\times2$-matrices over the finite field $\mathbb{F}_2$. It seems to me (by trial and intuition, if I'm being honest), that there should be no matrix (besides $\mathbf{1}, \mathbf{0}$) that would commute with every invertible matrix in $\mathbb{F}_2^{2\times2}$.

Note that I am not requiring that this matrix commute with all other matrices in $\mathbb{F}_2^{2\times2}$, only with the invertible ones. For one, we do know that the group of invertible matrices in $\mathbb{F}_2^{2\times2}$ has trivial center, so I would only need to check singular matrices.

I tried to prove this by brute-force calculation, but since that is rather tedious, I would be interested to know a more analytical approach to this problem (or if I'm mistaken entirely).

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    $\begingroup$ The elementary matrices are invertible. Their multiplication from the right and the left are either row or column transformations. Using row/column transposition shows that a matrix that commutes with them must be symmetric and have the same value along the diagonal. Then using adding row/column to another shows that outside of diagonal it is zero. $\endgroup$
    – plop
    May 28 at 2:03
  • $\begingroup$ Brute force? There are only $16$ elements of $\mathbb F_2^{2\times2}$ ($6$ invertible, $10$ not), right? $\endgroup$ May 28 at 2:46
  • $\begingroup$ Yeah, but that's still around 20 matrix multiplications to do by hand, which is definitely doable, but doesn't feel particularly sophisticated. $\endgroup$
    – tolUene
    May 28 at 2:57
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Fix $B$ non-invertible. Let $\text{im }B = \text{span } v$ for some nonzero $v$.

Then if $(v,w)$ is basis for $\Bbb F_2$ define $A$ by $v \mapsto w$, $w \mapsto v$.

Clearly $\text{im }BA = \text{im }B = \text{span } v \ne \text{span } w = \text{im }AB$.

So we can always find an invertible $A$ that doesn't commute with a given non-invertible $B \ne 0$.

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