For a real bilinear form $Q:V\times V\to\mathbb{R}$. $V$ is of dimension $2n$, for a basis $\{…,v_i,…\}$, we define a matrix $[Q_{ij}]$ with $Q_{ij}=Q(v_i,v_j)$.
How to see that if $Q$ is skew-symmetric, then there exists a basis, $\{…,v_i,…\}$, s.t.
$ [Q_{ij}]=
\bigg[
\begin{matrix}
0&I_g\\-I_g&0
\end{matrix}
\bigg]
$
and if $Q$ is symmetric, then there exists a basis, $\{…,v_i,…\}$, s.t.
$[Q_{ij}]=$
$ [Q_{ij}]=
\bigg[
\begin{matrix}
I_h&0\\0&-I_k
\end{matrix}
\bigg]
$
Best Answer
Suppose $Q$ is skew-symmetric. Pick a nonzero $v \in V$, and assuming $Q$ is nondegenerate, then there is some $w \in V$ so that $Q(v,w) = \lambda \neq 0$. Rescale $w \mapsto \frac{1}{\lambda}w$ so that $Q(v,w)=1$. Rename the vectors $v=x_1$ and $w = y_1$. Now look at the subspace $\mathrm{Span}(x_1,y_1)^\perp$ (where $\perp$ is with repsect to $Q$), which is of dimension $2n-2 = 2(n-1)$ and repeat, to build a basis $x_1,\dots,x_n,y_1,\dots,y_n$. Then the matrix of $Q$ in this basis is by construction $$ Q = \left(\begin{array}{cc} 0 & \mathrm{Id}_n \\ -\mathrm{Id}_n & 0 \end{array}\right) $$
The idea is similar for the symmetric case. Start with nonzero $v$ for which $Q(v,v) \neq 0$ and rescale $v$ so that $Q(v,v) = \pm 1$. Now look at $v^\perp$, which is of dimension one less, and repeat to build a basis $v_1,\dots,v_n$ where $Q(v_i,v_i) = \pm 1$.