Proof that q is a semidefinite quadratic form I've been stuck on a question in my linear algebra course for quite some time:
Let V be a finite dimensional vector space over $R$ and $q:V \to R$ a quadratic form.
Prove that if $$L=\{v|q(v)\ge0\}$$ is a subspace of $V$, then q is semidefinite. Not sure how to start, I'd appreciate a hint.
 A: this is my first answer on this site. I hope I'm not writing a disaster.
Suppose that $L = \{v|q(v) \geq 0\}$ is a proper subspace of $V$, so $V = L \bigoplus U$, where $U$ is a subspace of $V$ such that $U \cap L = \{ 0 \}$(this is important) and $U + L = V$. Now if $v \in U-\{ 0 \}$ then how $q(v)$ could be?
If $q(v) \geq 0$ then $v \in L \cap U = \{ 0 \}$, absurd. So if $v \in U$ then $q(v) < 0$. 
Let $w = v + u \in V$ where $v \in L$ and $u \in U$, what can you say about $q(u + v)$?
Hope this helps. 
A: The assertion is plainly wrong. Take the form $q(v)=v_1^2$ on the two-dimensional space $V$ with coordinates $(v_1,v_2)$; then $L=V$ is a subspace of $V$, but the form is not definite.
Now the amended statement (the form is semidefinite) is true. Proof: If $q\equiv 0$ on $L$, then $q$ is clearly negative semidefinite. If this is not the case, then let $v_0\in L$ be a point such that $q(v_0)>0$. Since $q$ is continuous, there exists a ball $B$ with center $v_0$ on which $q>0$. Hence $B\subset L$, and we conclude that $L$ is the entire space $V$; i.e., $q$ is positive semidefinite. 
