# Prove that: There is no Field for which $(\mathbb{F}^{2\times 2}, \cdot)$ is commutative.

Assuming it is about matrices in $$\mathbb{R}$$ and not taking edge cases into consideration.

Intuitively it is true that for $$(\mathbb{F}^{2\times 2}, \cdot)$$ commutative property doesn't hold true.

But I'm struggling to prove it with help of definition of Field. Help on it would be greatly appreciated.

• Well, how would you prove it for $\mathbb{R}$? Then you can see whether that proof works for a general field. Commented Nov 23, 2022 at 1:39
• every field has $0$ and $1.$ Try pairs of matrices with those entries. Commented Nov 23, 2022 at 1:45
• Do not use x for $\times$. Use \times. Commented Nov 23, 2022 at 1:58

Consider the matrices $$A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$$ and $$B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$. Their commutator $$[A, B] = AB - BA = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$$, which can only be $$0$$ if $$0 = 1$$. But that cannot happen in a field.