Let $ V $ be a vector space of finite dimension on $\mathbb{C} $ and $ \dim(V) \gt 1 $. Show that for every quadratic form $ q : V \to \mathbb{C} $ there exists $ 0 \neq v \in V $ such that $ q(v) = 0 $.
It certainly has something to do with the fact that we are in $\mathbb{C} $ and not $ \mathbb{R} $, but I'm since $ q $ can be represented by any non-singular complex matrix, I don't know how this is even possible.
(edit: V is of finite dimension)
Consider the case $n=2$. Thus we have $$ q(\begin{bmatrix} c_1 & c_2 \\ \end{bmatrix}) = \begin{bmatrix} c_1 & c_2 \\ \end{bmatrix}
\begin{bmatrix} q_{11} & q_{12} \ q_{21} & q_{22} \ \end{bmatrix}
\begin{bmatrix} c_1 \ c_2 \ \end{bmatrix} = q_{11} c_1^2 + (q_{12} + q_{21}) c_1 c_2 + q_{22}c_2^2 $$ For every value of $c_2$, we can consider the polynomial in one variable $q_{c_2}(c_1) = q(\begin{bmatrix} c_1 & c_2 \ \end{bmatrix})$. Since $\mathbb C$ is algebraically closed, this polynomial always has two roots. Thus we have a solution by letting $c_2 \neq 0$.
If $n > 2$, note that $q( \begin{bmatrix} c_1 & c_2 & 0 & \dots & 0 \end{bmatrix} )$ has a non-zero solution since it is a quadratic form in two variables.
If $V $ has infinite dimension, it suffices to take a $2$-dimensional subspace of $V$ and apply the case $n=2$.
Note that this result holds whatever is the matrix that represents $q$, i.e. in the finite dimension case, you could consider singular matrices as well.
Hope that helps,
