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I'm having trouble with the following problem:

Let $\tau_A: F^2\times F^2 \rightarrow F$ be a symmetric bilinear form given by $\tau_A (v,w)=v^tAw$, $\forall v,w\in F^2$ and $A=\begin{pmatrix} 1 & 0 \\ 0 & -1 \\ \end{pmatrix}$, where $F$ is a field.

Suppose the characteristic of $F$ is not equal to $2$. Prove that there exists a basis $\mathcal{B}=\{v_1,v_2\}$ of $F^2$ such that $\tau_A (v_1,v_1)=\tau_A (v_2,v_2)=0$

So far I've tried seeing if I could milk anything out of the non-degeneracy of $\tau_A$ (so that $(F^2,\ \tau_A)$ is an inner product space), but got stuck. I also split this problem into two cases: $Char(F)=0$ and $Char(F)=p$, but wasn't able to get anywhere. I have no experience dealing with the characteristic of a field, so I think conceptually I'm having a hard time understanding why it would matter in a problem like this.

Any tips or solutions (preferably as elementary as possible) would be appreciated! Thanks in advance!

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We can construct the vectors $v_1,v_2$ explicitly: $$v_1=(1,1),\qquad v_2=(1,-1).$$

It is clear that $\tau_A(v_1,v_1)= \tau_A(v_2, v_2)=0$. But when are they a basis?

They are a basis exactly when $1$ and $-1$ are different, i.e. everytime $\mathrm{char}(F)\neq 2$.

On the other hand if $\mathrm{char}(F)=2$ then $A=I$ the identity matrix. So $\tau_A((x,y),(x,y))=0$ implies $x^2+y^2=(x+y)^2=0$, so $x=-y=y$.

Then those vectors form a vector space of dimension $1$, so you cannot find a basis for your vector space of dimension $2$.

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Just proceed straightforwardly: if $v=(x,y)$ then $\tau_A(v)=0$ means $x^2=y^2$, so just take as basis $\{(1,1),(1,-1)\}$. Note that this is not a basis if the characteristic is 2.

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