I want to show that an $(n + k) \times (n + k)$ real matrix $A$ belongs to $O(n;k)$ iff $gA^Tg = A^{-1}$. I know that for all $\vec{x}, \vec{y} \in \mathbb{R}^{n+k}$, and for the matrix $$ g = \begin{pmatrix} I_n & 0 \\ 0 & -I_k \\ \end{pmatrix}, $$
if $[\cdot,\cdot]_{n,k}$ is the bilinear form on $\mathbb{R}^{n+k}$ defined as $$[\vec{x},\vec{y}]_{n,k} = x_1y_1 + \cdots x_ny_n - x_{n+1}y_{n+1} - \cdots - x_{n+k}y_{n+k},$$
it is true that $[\vec{x},\vec{y}]_{n,k} = \langle \vec{x}, g\vec{y} \rangle$. I know I want to show $[A\vec{x},A\vec{y}]_{n,k} = [\vec{x},\vec{y}]$. So, my first thinking is to consider $[A\vec{x},A\vec{y}]_{n,k} = \langle A\vec{x}, gA\vec{y} \rangle$ and somehow cleverly multiply by the identity $gA^TgA = I$.
However, I am sooo lost on doing so. Help would be super appreciated!
$\langle Ax,gAy\rangle =\langle x,A^TgAy\rangle$. $A^TgA=g^{-1}$ you deduce that $\langle Ax,Ay\rangle=\langle x,g^{-1}y\rangle=\langle x,gy\rangle$ since $g=g^{-1}$.