What is the Matrix of a Non-Degenerate Non-Symmetric Bilinear Form?

[Dmak]

Hello I was reading through some research and I came across the proof of a lemma which I did not wholly understand. The problem statement is as follows:

Let F be a non-degenerate non-symmetic bilinear form in V. Then there exists a basis in V with respect to which F has one of the following matrices:

$$H_{\phi} = \[ \left( \begin{array}{ccc} cos \phi & sin \phi & 0 \\ -sin \phi & cos \phi & 0 \\ 0 & 0 & 1 \end{array} \right)\], J = \[ \left( \begin{array}{ccc} 1& 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right)\], K = \[ \left( \begin{array}{ccc} 1 & 1 & 0 \\ -1 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right)\]$$.

First we decompose F into the sum of a symmetric and a skew symmetric form, $$F = F_{+} + F_{-}$$. The proof then makes the assumption that $$dim \; ker \; F_{-}= 1$$. It seems that the dimension could equally be 2. Any ideas why this is the case? Or better yet can anyone offer another proof of this lemma? Thank you for your help.

 

[fresh_42]

I guess $\dim V =3$. In this case, if $\dim \ker F_->1$, then $F$ will be symmetric, which is excluded.

Let $\{\,u,v,w\,\}$ be a basis with $v,w \in \ker F_-$. Then $u^\tau Fv=u^\tau F_+v=v^\tau Fu$ and similar for $w$. If all three are in the kernel, or for $v^\tau Fw$ we have zero anyway.