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:
- $[q]_\beta = [Q]_{\beta^\ast\beta}$, so that $[q]_\beta$ is invertible if and only if $Q$ is invertible.
- $\ker q = \ker Q$, so that $q$ is non-degenerate if and only if $Q$ is injective.
- $\dim V = \dim V^\ast$, 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?
Let me offer a short proof using dual bases. Let $v_1, \dots, v_n$ be some basis of $V$ and $v^1, \dots, v^n$ the associated dual basis of $V^{*}$. Define the map $\phi \colon \mathcal{L}(V,V^{*}) \rightarrow \operatorname{Bi}(V \times V, \mathbb{F})$ by $\phi(T) := g_T$ where $g_T(u,v) = T(u)(v) = \varepsilon(T(u), v)$. It is readily verified that this map is linear so it is enough to show that $\phi$ is one-to-one and onto. Since
$$ T(v_i) = \sum_{i=1}^n \varepsilon(T(v_i), v_j) v^j $$
we see that the map $T$ is determined uniquely by the $n \times n$ scalars $\varepsilon(T(v_i), v_j)$ (since the matrix $(\varepsilon(T(v_i),v_j))$ is the matrix representing $T$ with respect to the bases $v_1,\dots,v_n$ of $V$ and $v^1,\dots,v^n$ of $V^{*}$). Hence, if $\phi(T) = 0$ then all the scalars $\varepsilon(T(v_i), v_j)$ are zero and $T = 0$. This shows the injectivity.
To show surjectivity, let $g$ be a bilinear form on $V$ and define a linear map $T$ by requiring that
$$ T(v_i) = \sum_{i=1}^n g(v_i, v_j)v^j $$
for all $1 \leq i \leq n$. Then a direct calculation shows that $\phi(T) = g$.
Best Answer
We can obtain a linear map $V \to W^*$ by sending $x \in V$ to the functional $y \mapsto F(x, y)$ on $W$. That $F$ is non-degenerate implies that this map is injective, so $\dim V \leq \dim W^*$. Since $W$ is finite-dimensional, $W^*$ has the same dimension as $W$ and hence $\dim V \leq \dim W$. Using the analogous map $W \to V^*$, we get the reverse inequality.