# Proving there is no matrix in $\mathbb{F}_2^{2\times2}$ that commutes with every invertible matrix

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).

• 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.
– plop
May 28 at 2:03
• Brute force? There are only $16$ elements of $\mathbb F_2^{2\times2}$ ($6$ invertible, $10$ not), right? May 28 at 2:46
• Yeah, but that's still around 20 matrix multiplications to do by hand, which is definitely doable, but doesn't feel particularly sophisticated. May 28 at 2:57

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 $$B \ne 0$$.