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,