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

Let $q$ be a sesquilinear form on a vector space $E$, given by a matrix $A$.  The following statements are equivalent:


*

*$q$ is degenerate.

*There exists a nonzero vector $x\in E$ so that $q(x,y)=0$ for all $y\in E$.

*There exists a nonzero vector $x\in E$ so that $x^H A y = 0$ for all $y\in E$.

*There exists a nonzero vector $x\in E$ so that $x^H A$ is the zero (row) vector.

*The left nullspace of $A$ is non-trivial.

*The matrix $A$ is singular.
It should be clear that $(1)\Leftrightarrow(2)\Leftrightarrow(3)\Leftrightarrow(4)\Leftrightarrow(5)\Leftrightarrow(6)$.
