Non-degenerate bilinear forms and invertible matrices

Let me assume that you have a bilinear form $q : V \times V \to F$, where $V$ is a finite-dimensional vector space over the field $F$. Fix a basis $\beta = \{e_1,\dotsc,e_n\}$ of $V$, which defines the corresponding dual basis $\beta^\ast = \{e_1^\ast,\dotsc,e_n^\ast\}$ of $V^\ast$.

First, check that $q$ defines a linear transformation $Q : V \to V^\ast$ by $Q(v) := q(v,\cdot)$ satisfying $\ker Q = \ker q$. Now, since for any $\phi \in V^\ast$, $\phi = \sum_k \phi(e_k) e_k^\ast$, $$ Q(e_k) = \sum_j Q(e_k)(e_j) e_j^\ast = \sum_j q(e_k,e_j) e_j^\ast = \sum_j q(e_j,e_k)e_j^\ast $$ for each $k$, and hence the matrix $[Q]_{\beta^\ast\beta}$ of $Q$ with respect to the bases $\beta$ and $\beta^\ast$ is precisely the matrix $[q]_\beta$ of $q$ with respect to $\beta$. As a result, your matrix of interest $[q]_\beta$ is invertible if and only if the linear transformation $Q$ is invertible.

So, let's collect what we know:

1. $[q]_\beta = [Q]_{\beta^\ast\beta}$, so that $[q]_\beta$ is invertible if and only if $Q$ is invertible.
2. $\ker q = \ker Q$, so that $q$ is non-degenerate if and only if $Q$ is injective.
3. $\dim V = \dim V^\ast$, so that by the rank-nullity theorem, $Q : V \to V^\ast$ is injective if and only if it is surjective.

What can you therefore conclude?

I also proposed the question, but (I think) solved it myself before reading any of this. We know that any bilinear form can be represented as $X^T * (n \times n \text{ matrix}) * Y$ where $X^T$ and $Y$ are $n$-dimensional vector spaces.

The definition I have of a nonDegenerate bilinear form is that $B ( X, Y) = 0$ for all $X\Longrightarrow Y = 0$ ; somewhat convoluted no?

Prove the contrapositive -- There exists a non-zero vector $Y$ such that for all $X, B (X, Y) = 0$

Start with non-zero $Y$ and extend it to a new basis for a vector space. This makes $Y$ equal the column vector $( 1, 0, . . . , 0 )$.

Now consider the row vector $X^T$ * matrix of $B$ in the new basis, to get $B ( X , Y)$ to be zero, all $X$ must transform to a row vector with 0 in the first position. This implies that the matrix has a zero row, hence not invertible.