Why is the matrix representing a non-degenerate sesquilinear form invertible? Let's consider a finite-dimensional vector space $E$ on the field $\mathbb{K}$ (where $\mathbb{K}=\mathbb{C} \ \text{or}\ \mathbb{R}$) and a sesquilinear (or bilinear if $\mathbb{K}=\mathbb{R}$) form $q:E\times E \rightarrow \mathbb{K}$.
The definition for a non-degenerate form is that $q(x,y)=0\ \forall y\in E$ implies $x=0$.
Now if we represent $q(x,y)$ with a matrix, so $q(x,y) =x^HAy$, why does the condition that the form be non-degenerate impose that $A$ is non-singular?
I tried to see it using the dual space as $M(x,A)=x^HA\in E^*$, so that $M:E\times L(E,E)\rightarrow E^*$, where $L(E,E)$ is the vector space of all linear transformations from $E$ to $E$ and playing with the nullspace of $A$, but I just can't see it

$A$ is singular if and only if there exist $0 \ne y\in E$ such that $Ay=0$. If this is the case, then clearly $q(x, y)=0$ for all $x$ (modulo identification of $E$ with $\mathbb{K}^n$). Viceversa, if $q(x, y)=0$ for all $x\in E$ then the vector $Ay$ is orthogonal to the whole $\mathbb{K}^n$. In particular $\lVert Ay\rVert^2=(Ay)^HAy=0,$ so $Ay=0$.
