Say, I have a non-degenerate symmetric bilinear form on a vector space with Dim = $4$. Suppose I have a set of basis $\{v_1,v_2,v_3,v_4\}$ and I know for any $i$, $<v_i,v_i>$ is positive. So what are the possible signatures?
I know $(4,0)$ is trivially a possible signature, just consider an Identity matrix.
However, about $(0,4)$, I do not know how to prove it is not possible. (I guess it is not)
So generally how to solve such kind of question, are there any useful tools?
It should be obvious $(0,4)$ isn't possible, since every vector has negative norm in that signature.
For the rest, it suffices to check $\mathbb{R}^{p,q}$ for every other possible signature $(p,q)$. (In theory, this might be useless idea if this is hard to do, but it turns out not to be hard to do.) Indeed, if you play around a bit you should be able to generalize to the following fact:
Proposition. A pseudo-Euclidean vector space admits a basis of vectors with positive quadratic form so long as it is not negative-definite signature.
Proof. Without loss of generality (by Sylvester's Theorem), we may as well consider $\mathbb{R}^{p,q}$ with $p>0$. If we pick any $v\in\mathbb{R}^p$ with $v\cdot v>1$ then there is such a basis
$$ \{e_1,\cdots,e_p,v+e_{p+1},\cdots,v+e_{p+q}\}. $$