Properties of an alternating bilinear form its coordinate matrix 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?
 A: 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 
$$ \beta(x,x) = \sum_{i=1}^n B_{ii}x_i^2$$
A: If $B(x,y)$ is alternating, clearly the coordinate matrix of $B$ has a zero diagonal. Also, from the identity
$B(x+y,x+y) = B(x,x) + B(y,y) + B(x,y) + B(y,x)$, we see that $B$ is skew-symmetric. So is its coordinate matrix (why?).
Conversely, if the coordinate matrix $(b_{ij})$ of $B$ is skew-symmetric, then $b_{ij}x_ix_j+b_{ji}x_jx_i=0$ and hence $B(x,x)=\sum_i b_{ii}x_i^2$. So, if $(b_{ij})$ also has a zero diagonal, then $B(x,x)=0$ for all $x$, regardless of $\operatorname{char}(\mathbb{F})$.
