If $\eta$ is some nondegenerate, symmetric bilinear on $V\cong \mathbb{R}^n$, I'm trying to show that the signature is the same regardless of a choice of orthonormal basis (is that the correct term for it in this case?), and hence well defined.
An outline of my proof goes as follows:
In some basis, I can write $\eta$ as a symmetric matrix satisfying:
$$\eta(v,w)=v^T\eta w$$
Since $\eta$ is symmetric, I can diagonalize $\eta$ it's eigenbasis, $\{f_i\}$, and then, since $\eta$ is nondegenerate, scale each $f_i$ such that this new basis is orthonormal, and $\eta$ can be written as:
$$\begin{pmatrix} -I_t&0\\ 0&I_s \end{pmatrix}$$ where $I_t$ and $I_s$ are the identity matrices of size $i\times i$ and $s\times s$ respectively, and $s+t=n$. Let $P=span\{f_{t+1},\dots, f_{t+s}\}$, this subspace is then positive definite, and of dimension $s$. Let $W$ be any negative definite subspace of $V$, then $W\cap P=\{0\}$, as for any $v\in V$, if $v\in W\cap P$, we have that $\eta(v,v)>0$, and $\eta(v,v)<0$. This then implies that $\dim W+s\leq n$, so $\dim W\leq t$.
Now suppose that $\{b_i\}$ is any other basis which diagonalizes $\eta$ into the matrix: $$\begin{pmatrix} -I_{t+1}&0\\ 0&I_{s-1} \end{pmatrix}$$ Then $W=span\{b_1,\dots,b_{t+1}\}$ is a negative definite subspace of dimension $t+1$, which is a contradiction. So for any orthonormal basis, we see that $\eta$ must be, up to reordering: $$\begin{pmatrix} -I_{t}&0\\ 0&I_{s} \end{pmatrix}$$ so the signature of $\eta$, $(t,s)$ is well defined
Does that make sense? I feel kind of iffy about the middle part of the proof, where I'm trying to show that $\dim W\leq t$. Any feedback would be greatly appreciated.