If $V$ over $\mathbb{C}$, $\dim V\ge 2$, let $q: V \rightarrow \Bbb{C}$ be a Quadratic form. s.t. exists $v \ne 0$ such that $q(v)=0$

Question

If $V$ is a vector space over $\mathbb{C}$ and $\dim V\ge 2$, prove that if $q: V \rightarrow \Bbb{C}$ is a Quadratic form then exists $v \ne 0$ such that $q(v)=0$

I'm not sure I even understand why this is true. Any tips?

Answer 1

Suppose

$$q(x)=q(x_1,...,x_n)=\sum_{k\le j=1}^na_{k,j}x_ix_j$$

Suppose some variable, say $\;x_1\;$ for simplicity, actually happens in the above with coefficient different from zero. Observe then that for any choice of complex number $\;a_2,...,a_n\;$ , we get that $\;f(x_1)=q(x_1,a_2,...,a_n)\;$ is a complex polynomial on $\;x_1\;$ and, since $\;\Bbb C\;$ is algebraically closed, it has a zero $\;t\in \Bbb C\;$ .

I'll leave it to you to check that we can always choose things so as to be sure the zero $\;(t,a_2,...,a_n)\;$ of $\;q\;$ isn't trivial...