here's my question:


Let $V$ be a k-dimensional vector space over $\mathbb{R}$ and $q$ a quadratic form on $V$ of signature $(m,n)$ , $m+n=k$.


We have $W\subset V$ a positive (with respect to the quadratic form, of course) subspace of dimension $m-1$. Is it always possible to find a vector $c\in V$ such that $\langle W,c\rangle$ is a positive subspace of dimension $m$? If yes, which kind of space (in particular of which dimension) is $\Lambda:=\{c\in V| \langle W,c\rangle\text{is a positive subspace of dimension }m\}$? 


I was trying to understand the problem working on the example with $m=3,n=1$. So we get the quadratic form $q$ on $\mathbb{R}^4$ which is $x\_1^2+x\_2^2+x\_3^2-x\_4^2$ after diagonalisation. So $W=\langle a,b\rangle$ and saying $W$ is a positive subspace is equivalent to say that


$\alpha^2(a\_1^2+a\_2^2+a\_3^2-a\_4^2)+\beta^2(b\_1^2+b\_2^2+b\_3^2-b\_4^4)+2\alpha\beta(a\_1b\_1+a\_2b\_2+a\_3b\_3-a\_4b\_4)>0$
for every $\alpha,\beta \in \mathbb{R}$.


But then the vector $c$ i am searching must fulfill the condition 


$\alpha^2(a\_1^2+a\_2^2+a\_3^2-a\_4^2)+\beta^2(b\_1^2+b\_2^2+b\_3^2-b\_4^4)+2\alpha\beta(a\_1b\_1+a\_2b\_2+a\_3b\_3-a\_4b\_4)+$ $\gamma^2(c\_1^2+c\_2^2+c\_3^2-c\_4^2)+2\alpha\gamma(a\_1c\_1+a\_2c\_2+a\_3c\_3-a\_4c\_4)+2\beta\gamma(b\_1c\_1+b\_2c\_2+b\_3c\_3-b\_4c\_4)>0$ for every $\alpha,\beta,\gamma\in \mathbb{R}$.


I can't see which kind of space this equation defines: it is an equation in $c\_1,c\_2,c\_3,c\_4$ which varies with $\alpha,\beta,\gamma$ (i must valuate the "worst" case? but which is it?) and the positive term $\alpha^2(a\_1^2+a\_2^2+a\_3^2-a\_4^2)+\beta^2(b\_1^2+b\_2^2+b\_3^2-b\_4^4)+2\alpha\beta(a\_1b\_1+a\_2b\_2+a\_3b\_3-a\_4b\_4)$ also varies with $\alpha$,$\beta$.


So please help me figure out how this situation works. Also do you know a book which analyses this kind of problems on quadratic forms?



Positive subspaces of quadratic forms
Write $V= W + W^{\perp}$. Any vector $c$ can be decomposed into $c\_{W}+ c\_{W^\perp}$. $W+c$ is a positive subspace if and only if $c\_{W^\perp}$ is positive. The positive vectors in $W^{\perp}$, a quadratic space of signature $(1,n)$, form a (nonempty) solid bicone. So there is always such a vector, and the space of such vectors is a $1+n$-dimensional solid bicone times a $m-1$-dimensional linear space.


