Let $b$ denote the symmetric bilinear form $b(v,w) = v \cdot gw$ on $\mathbb R^{n+1}$ where $\cdot$ denotes the standard inner product on $\mathbb R^{n+1}$ and $g$ is the block diagonal matrix with a $n \times n$-block given by the unit matrix and in the position $(n+1) \times (n+1)$ we have a minus $1$. Let $O(b)$ be the isometry group of this form, that is, the group of all matrices $A$ that leave $b$ invariant: $b(Av,Aw) = b(v,w)$ for all $v,w \in \mathbb R^{n+1}$.
Why is $A \in O(b)$ equivalent to $A^T g A = g$?
I guess what I am asking is, if there is a neat way of seeing this, i.e. is it possible to avoid a messy calculation?
Well,
$b(Av,Aw) = b(v,w) \iff \langle Av,gAw \rangle=\langle v,gw \rangle \iff \langle v,A^TgAw \rangle=\langle v,gw \rangle \, \forall v,w \iff$
$\langle v,(A^TgA-g)w \rangle=0 \, \forall v,w$
Choosing $v=(A^TgA-g)w$ we get by the positivity of the inner product, that $(A^TgA-g)w=0$ for any $w$. Hence $A^TgA-g=0$.