I found that I lack many basic knowledge about linear algebra, so read the wiki article about Bilinear Forms. Especially this Paragraph. I tried to proof of "Every alternating form is skew-symmetric." was quite easy. And I found a counter-example for the inverse if $char(F) = 2$.
However, I am currently trying to find a proof for this:
A bilinear form is alternating if and only if its coordinate matrix is skew-symmetric and the diagonal entries are all zero
I looked at the following equations, to understand that.
$\begin{pmatrix}v_1 & v_2\end{pmatrix} \cdot \begin{pmatrix}a & b \\ c & d\end{pmatrix} \cdot \begin{pmatrix}v_1 \\ v_2\end{pmatrix} = 0$
$\begin{pmatrix}v_1 & v_2\end{pmatrix} \cdot \begin{pmatrix}a & b \\ c & d\end{pmatrix} \cdot \begin{pmatrix}w_1 \\ w_2\end{pmatrix} = – \begin{pmatrix}w_1 & w_2\end{pmatrix} \cdot \begin{pmatrix}a & b \\ c & d\end{pmatrix} \cdot \begin{pmatrix}v_1 \\ v_2\end{pmatrix}$
And they are both solvable for $a=0, d=0, c=-b$, thus the matrix $\begin{pmatrix}0 & b \\ -b & 0\end{pmatrix}$.
However, I am not any nearer to a proof. Can someone please point me into the right direction?
Best Answer
Hint: Let $\beta$ be a bilinear form and $B$ its coordinate matrix. Applying $\beta$ to two basis vectors $e_i$, $e_j$ gives $$ \beta(e_i, e_j) = e_i^t \cdot B \cdot e_j = B_{ij} $$ that is the $(i,j)$-entry of $B$. Now if $\beta$ is alternating, then it is skew-symmetric, hence $\beta(e_i, e_j) = -\beta(e_j, e_i)$ and $\beta(e_i, e_i) = 0$. Can you relate this to $B$, using the above?
Edit: As you noted in your comment, this gives $B = -B^t$ with zeros on the diagonal, that is, one direction. For the other direction, note that for arbitrary $x,y$, we have $$ \beta(x,y) = x^t B y = -x^t B^t y = -y^t B x $$ and that $$ \beta(x,x) = \sum_{i=1}^n B_{ii}x_i^2$$