Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question

2 Answers 2

up vote 3 down vote accepted

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 $2$.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.