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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?

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up vote 0 down vote accepted

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.

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You confuse $W$ with $V$. –  Atsushi Kanazawa Jun 14 '13 at 19:15
    
i think yes, also when Will Sawin writes $W+V$ i think he means $c+W$. Do you know a book where i can find the proof for the statement that the positive vectors in a quadratic space of signature $(1,n)$ form a solid bicone? –  Tony Harrison Jun 14 '13 at 20:49
    
It's an equation of the form $x_1^2-x_2^2 -x_3^2 - \dots x_{n+1}^2$. I consider this a solid bicone by definition. The point is that all quadratic forms of a given signature are equivalent. –  Will Sawin Jun 14 '13 at 20:51
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