Representing a nonzero bilinear alternating form on a two-dimensional space by $\bigl(\begin{smallmatrix} 0&1\cr-1&0\end{smallmatrix}\bigr)$.

Question

So I am having a little bit struggle with some question.

I have a bilenear form $B:V\times V\to \Bbb F$.

$B$ is not the zero form.

$B$ is alternating meaning it is also skew-symmetric.

I know that the $\dim(V)=2$

How can I show that there exists a basis for $V$ such that the matrix represent $B$ looks like that :

\begin{pmatrix} 0 & 1\\ -1 & 0\\ \end{pmatrix}

I do not really understand how can I show that if I do not know what $B$ does.

I am guessing that because of the alternating and the skew-symmetric I can show that the matrix represent $B$ looks like:

\begin{pmatrix} 0 & a\\ -a & 0\\ \end{pmatrix}

But how can I show that $a=1$?

Answer 1

Let $(v,w)$ be any basis of $V$. Since $B$ alternates, $B(v,v)=B(w,w)=0$. Hence $B(v,w)\neq 0$, otherwise $B$ would be the zero form. Now define $w^\prime = \frac{w}{B(v,w)}$. You should be able to show that $B=\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ with respect to the basis $(v,w^\prime)$.