Symmetric bilinear forms in characteristic 2

Question

This is a homework question:
Prove that: In field $K$ of characteristic $2$, for symmetric bilinear forms on $K^2$, there exist a basis where the matrix of the bilinear form is either diagonal or anti-diagonal.

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I tried to do this in $\mathbb{F}_2$ with $2\times 2$ matrices to get some intuition but I found it to be wrong. Any ideas?

Answer 1

but I found it to be wrong

Then you must have made a mistake somewhere. If you post your counterexample, we could find out which mistake.

Let $\beta$ be the symmetric bilinear form. Then there are two possibilities,

1. $\beta(v,v) = 0$ for all $v \in K^2$, or
2. $\bigl(\exists v_1\in K^2\bigr)\bigl(\beta(v_1,v_1)\neq 0\bigr)$.

In the first case, pick any basis. In the second case, consider the linear form $\lambda\colon w \mapsto \beta(v_1,w)$ and think about $\ker\lambda$.