$F$ is indefinite if and only if $K$ is not a subspace I get the following linear algebra problem in my class.
Let $F: \mathbb{R}^N \times \mathbb{R}^N \to \mathbb{R}$ be bilinear and symmetric, and let $K = \{ v \in \mathbb{R}^N \mid F(v,v) = 0 \} $. Prove that $F$ is indefinite if and only if $K$ is not a vector subspace of $\mathbb{R}^N$.
I am having troubles proving both directions. Suppose $F$ is positive-semidefinite or negative-semidefinite, and let $v,v' \in K$. Then $F(v+v',v+v')=2F(v,v')$, then how can I show that $v+v' \in K?$ Conversely, if $F$ is indefinite, then there exist $v_1,v_2$ such that $F(v_1,v_1) < 0$ and $F(v_2,v_2)>0$, so $F(v_1+\lambda v_2, v_1+\lambda v_2) = 0$, but I don't know how to proceed. Any help is appreciated.
 A: Write $F(v,w) = v^T A w$ with a symmetric matrix $A$. Choose an orthonormal basis $v_1,...,v_n$ of eigenvectors with eigenvalues $\lambda_1,...,\lambda_n$; then $$F(v,v) = \sum_{i=1}^n \lambda_i a_i^2\; \mathrm{if} \; v = \sum_{i=1}^n a_i v_i.$$
If all the $\lambda$ have the same sign then you see that $K$ is the eigenspace of $0$.
When $F$ is indefinite, you can fix norm-1 eigenvectors $v$ and $w$ for positive resp. negative eigenvalues, say $Av = \lambda v$ and $Aw = \mu w$. Then for any $a,b \in \mathbb{R}$, $$F(av+bw,av+bw) = \lambda a^2 + \mu b^2.$$ There are different combinations of these that will give you zero ($a,b$ can be positive or negative); if $K$ were a subspace, then this would imply that $v$ and $w$ are in $K$, which is not true.
A: In case you don't know that any symmetric bilinear form can be represented by a symmetric matrix, here is another answer (which also works in infinite dimensions).
Assume that $F$ is semi-definite. Then it satisfies the Cauchy inequality. So, if $v\in K$ then $|F(v,w)|^2\le F(v,v)F(w,w) = 0$ for all $w\in\mathbb R^N$. So, $K = (\mathbb R^N)^{\perp_F}$ which is trivially seen to be a subspace.
If $F$ is indefinite, then there exist $u,w\in\mathbb R^N$ such that $F(u,u) = 1$ and $F(w,w) = -1$. Now set $v := (1+f^2)^{-1/2}(u+fw)$, where $f = F(u,w)$, and easily check that $F(v,v) = 1$ and $F(v,w) = 0$. Then we have $F(v+w,v+w) = F(v-w,v-w) = 0$. So, the vectors $v\pm w$ are in $K$ but its sum $2v$ is not. Hence, $K$ is not a subspace.
