Simple question about quadratic form Suppose $V$ is a vector space over $\mathbb{C}$ and $\dim V \geq 2$. I must prove that if $q: V \to \mathbb{C}$ is a quadratic form then there exists $v\neq 0$ such that $q(v)=0$. Also, how the answer would be different if $V$ was over $\mathbb{R}$? 
The question seems very simple but I simply can't find a way to prove it. Any suggestions?
 A: Suppose
$$q(z_1,z_2,\ldots,z_n)=\sum_{i,j}{a_{ij}z_iz_j}$$
with $a_{ij}\in \mathbb{C}$. Consider the matrix $M=(a_{ij})$. By changing basis we may assume that $M$ is in Jordan canonical form, so that $M=D+N$, with $D$ diagonal and $N$ nilpotent, with eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_n$, so that
$$q(z_1,z_2,\ldots,z_n)=\sum_i{\lambda_iz_i^2}+\sum_{j=i+1}{N_{ij}z_iz_j}$$
If any eigenvalue, say the $i$th, is $0$, then choosing the vector with $0$s everywhere and a $1$ in position $i$ will work. Otherwise consider
$$q(z,z^2,z^3,\ldots,z^n)=\sum_i{\lambda_iz^{2i}}+\sum_{j=i+1}{N_{ij}z^{2i+1}}$$
Since no terms cancel and there exist terms with nonzero coefficients of different nonzero degrees, this polynomial has a nonzero root $r$, so the vector $v=(r,r^2,\ldots,r^n)$ satisfies $q(v)=0$.
Over $\mathbb{R}$ there may not be such a $v$. For example, consider
$$q(x,y)=x^2+y^2$$
Then if $q(x,y)=0$ we must have that $x=y=0$.
A: Note that there exists a basis $(e)$ (with at least two vectors) where $q$ is diagonal, i.e. $$q=\sum_{i=1}^{n} \alpha_i x_i^2$$
Now if there eixsts $\alpha_i=0$ then $q(e_i)=0$. Otherwise $q(e_1) = \alpha_1 \neq 0$ and $q(e_2) = \alpha_2 \neq 0$. You can verify that $$q(i e_1 + \left( \sqrt{\alpha_1 / \alpha_2} \right) e_2)=0$$
So you can always find some $v \neq 0$ such that $q(v)=0$.
