Let $V$ be a $\mathbb C$-vector space, $\dim V>1$. Then for any bilinear form $\phi$ there is $v\in V$ s.t. $\phi(v,v)=0$ (so there is no positiv definite bilinear form on $V$)
If one takes the usual multiplication then $\begin{pmatrix}i\\0\end{pmatrix}\cdot \begin{pmatrix}i\\0\end{pmatrix}=-1$, so for a particular bilinear form, positive definiteness fails, but I need some example which gives $0$ and for any bilinear form.
On
Let $v = (1,\dots,v_n)$. We can write $$ \phi(v,v) = p(v_1,v_2,\dots,v_n) $$ That is, $\phi(v,v)$ gives us a polynomial on the entries of $v$.
Suppose that $\phi$ is constant over $v$. By bilinearity, $\phi = 0$.
Now, suppose that $\phi(v,v)$ is not constant. Then there is some $v_i$ such that $p(v_1,v_2,\dots,v_n)$ is a non-constant polynomial on $v_i$.
For any fixed values of $v_j$ (for $j\neq i$), $p$ gives us a quadratic polynomial on $v_i$. Thus, there is a value of $v_i$ that makes $p$ equal to zero.
In either case, there exists a $v$ such that $\phi(v,v) = 0$.
Show first that if $\phi$ is a non-trivial bilinear form on a vector space $V$ over a field of characteristic $\neq 2$, then there is $v \in V$ such that $\phi(v, v) \neq 0$.
If $\phi \equiv 0$, any $v \in V$ will do. Otherwise, choose $v \in V$ with $\phi(v,v) \neq 0$. By replacing $v$ with $\frac{v}{\sqrt{\phi(v,v)}}$ (where $\sqrt{\phi(v,v)}$ is any, possibly complex, root of $\phi(v,v)$), we can assume that $\phi(v,v) = 1$. Consider $U = \operatorname{span} \{ v \}^{\perp}$. Since $\dim V > 1$, we must have $\dim U \geq 1$. Choose $0 \neq u \in U$. If $\phi(u,u) = 0$, we are done. If $\phi(u,u) \neq 0$, multiply $u$ by a scalar and make it that $\phi(u,u) = -1$. Then $v + u \neq 0$ (as $u$ and $v$ are linearly independent) and $\phi(v+u, v+u) = 0$.