What can be the rank and signature of the restriction of a non-degenerate real quadratic form $q$ of signature $(,)$ to a vector subspace $U$ of codimension $1$?
My attempt. The Gram matrix $Q$ of $q$ is diagonable. $$ Q = \begin{pmatrix}\lambda_{1} \\ & \lambda_{2} \\ & & \ddots \\ & & & \lambda_{p} \\ & & & & \mu_{1}\\ &&&&&\mu_{2} \\ & &&&&& \ddots\\&&&&&&& \mu_{m} \end{pmatrix}, \lambda_{i}>0,\mu_{j}<0.$$ $\exists ! F_{q}: q(v)=(v,F_{q}(v)) $.
- Let $v$ be an an eigenvector $v:F_{q} (v)=\lambda_{i}v$ then $rk(q|_{v^{\bot}})=p+m-1$ and the signature is $(p-1,m)$ or $(p,m-1)$.
- $U \neq v^{\bot}$ for any eigenvector of $F_{q}$. $$Q \sim Q_{u} = \begin{pmatrix} a_{1}& 0 &\cdots & 0 & a_{1n} \\ 0 & a_{2} &&\vdots & a_{2n}\\ \vdots & & \ddots & 0 & \vdots \\ 0 & \cdots & 0 & a_{n-1}\\a_{1n} &a_{2n}& \cdots && a_{n} \end{pmatrix}.$$
I tried to solve the problem when $a_{1}\cdot a_{2}\cdots \cdots a_{n-1} \ne 0$. I used Gaussian elimination for quadratic forms to diagonalize $ Q_{u}$ to see the connection between $a_{i}$ and $\lambda,\mu$.
The result: $\{a_{1},...,a_{n-1}, a_{n}-a_{1n}^{2}/a_{11}-...-a_{1,n-1}^2/a_{n-1}\}=\{\lambda_{1},...,\lambda_{p},\mu_{1},..., \mu_{m}\}$. Something wrong with it...
Thank you in advance.
I don't understand at all what you tried to do, the answer is much simpler.
Take $H=U^\perp$ the line orthogonal to $U$ with respect to $q$, and diagonalize $q$ on both $U$ and $H$: $q_{|U}\simeq \langle a_1,\dots,a_{n-1}\rangle$ and $q_{|H}\simeq \langle b\rangle$ where $n$ is the dimension of the whole space. Since $q=q_{|U}\perp q_{|H}$ You can clearly see the link between the signatures of $q$ and $q_{|U}$ depending on the sign of $b$.