Let $V$ be a finite-dimensional real vector space and $$B:V\times V\to\Bbb R$$ a bilinear form. Suppose that $B$ is non-degenerate and positive semi-definite, i.e.
Does that imply that $B$ is positive definite?
First I thought it would, but then I couldn't prove it. If $B(X,X)=0$ then I would like to prove that $B(X,Y)=0$ for all $Y\in V$ so that $X=0$. But maybe this is not necessarily true?
Actually, I think this is false. Consider the standard anti-symmetric form on $\mathbb{R}^2$ given by $B(e_i, e_i) = 0$, $B(e_1,e_2) = 1$, $B(e_2,e_1) = -1$.
First, I claim that $B$ is non-degenerate:
Given nonzero $ae_1 + b e_2\in \mathbb{R}^2$, note that $$ B(ae_1 + be_2, -b e_1 + ae_2) = a^2 + b^2 > 0. $$
Second, I claim that $B$ is positive semi-definite. In fact, we have $$ B(ae_1 + be_2, ae_1 + be_2) = ab - ba = 0.$$
This gives counterexamples for all higher $\mathbb{R}^n$ by declaring the standard $\mathbb{R}^2$ to be orthogonal to the standard $\mathbb{R}^{n-2}$, and then using the usual Euclidean inner product on $\mathbb{R}^{n-2}$.
Finally, note that for $\mathbb{R}^1$, there all no counterxamples: $B(e_1, e_1)\geq 0$ and non degenerate implies $B(e_1, e_1) > 0$.
Edit This wasn't asked for, but I've just kept thinking about it. In the example above, on $\mathbb{R}^2$, no vector has positive length. Of course, one can copy this idea on an even dimensional $\mathbb{R}^n$ by viewing it as a direct sum of $\mathbb{R}^2$s each equipped with $B$.
On the other hand, in odd dimensions, $B$ must always have at least one vector $x$ with $B(x,x) > 0$. That is, suppose $n$ is odd and $B$ is a bilinear form which is positive-semidefinite and non-degenerate. Then there is a vector $x$ for which $B(x,x) > 0$.
Proof: Suppose for a contradiction that $B(x,x) = 0$ for all $x$. First, we claim $B$ is antisymmetric (independent of the parity of $n$). To see this, just note that $$0 = B(x+y,x+y) = B(x,y) + B(y,x)$$ since $B(x,x) = B(y,y) = 0$.
Now, we finish the proof (this part requires $n$ to be odd). If $A$ is the matrix associated to $B(\cdot,\cdot)$ in the standard basis, then $-A^t = A$. Then $\det(A) = (-1)^n \det(A^t) = (-1)^n\det(A)$. Thus, if $n$ is odd, we coclude $\det(A) = 0$, contradicting the fact that $B$ is positive definite.
