Consider the set of scalar products:
$M(k) = \begin{bmatrix} 1 & 0 & 0&k^2 \\ 0 & 1 & k &0 \\ 0 & k & 1 &0 \\ k^2&0&0&1 \end{bmatrix}$ , $k \in \mathbb{R}$
- Determine the values of $k$ s.t. the scalar product is not degenerate, the values such that it has signature $(3,1)$, and the values such that it has signature $(2,2)$.
So, the values for which $M(k)$ is degenerate are $k=1$ and $k=-1$, struggling a bit with the determinant. How determine the signature with parameters? I tried with the characteristic polynomial - $((\lambda -1)^2 -k^2)((\lambda - 1)^2 - k^4)$ - but I don't think it's the right way (or the better one).
Well, $(3,1)$ must have negative determinant and "orthogonal" vectors of positive length. $(2,2)$ must have positive determinant and vectors of both positive and negative lengths. The determinant is $1-k^2-k^4+k^6=(1-k^2)(1-k^4)$, which is always nonnegative. So signature $(3,1)$ is impossible. Note also that this is an orthogonal direct sum of subspaces with forms $\begin{bmatrix} 1 & k \\ k & 1 \end{bmatrix}$ and $\begin{bmatrix} 1 & k^2 \\ k^2 & 1 \end{bmatrix}$. Now a vector $(a,b)$ in the first subspace has negative length if $a^2+2k~ab+b^2<0$. This is possible if and only if $|k|>1$. Simiarly for the other subspace if $|k^2|>1$ which holds exactly when $|k|>1$. Thus, you have signature $(2,2)$ when $|k|>1$, signature $(4,0)$ if $|k|<1$ and a degenerate form (of signature $(2,0)$) if $|k|=1$.