[proofplan]
We translate the [bilinear form](/page/Bilinear%20Form) into two induced linear maps, one from $V$ to $W^*$ and one from $W$ to $V^*$. The kernels of these maps are exactly the left and right radicals of $B$. In the chosen bases, the matrix of the second map is $A$, while the matrix of the first map is $A^\top$, so the standard determinant criterion for square matrices converts vanishing radicals into $\det A \neq 0$ and back.
[/proofplan]
[step:Identify the radicals as kernels of the induced dual maps]
Let $W^*=\operatorname{Hom}_k(W,k)$ and $V^*=\operatorname{Hom}_k(V,k)$ denote the dual vector spaces. Define the induced linear maps
\begin{align*}\Phi: V \to W^*\end{align*}
by requiring, for each $v \in V$, that $\Phi(v): W \to k$ is the linear functional
\begin{align*}
w \mapsto B(v,w),
\end{align*}
and
\begin{align*}
\Psi: W \to V^*
\end{align*}
by requiring, for each $w \in W$, that $\Psi(w): V \to k$ is the linear functional
\begin{align*}
v \mapsto B(v,w).
\end{align*}
Bilinearity of $B$ implies that $\Phi$ and $\Psi$ are $k$-linear maps.
The kernel of $\Phi$ is
\begin{align*}
\ker \Phi = \{v \in V : \Phi(v)=0 \in W^*\}.
\end{align*}
Since the zero element of $W^*$ is the zero functional on $W$, this is precisely
\begin{align*}
\ker \Phi = \{v \in V : B(v,w)=0 \text{ for all } w \in W\}.
\end{align*}
Thus $\ker \Phi$ is the left radical of $B$. Similarly,
\begin{align*}
\ker \Psi = \{w \in W : B(v,w)=0 \text{ for all } v \in V\},
\end{align*}
so $\ker \Psi$ is the right radical of $B$.
[guided]
The point of introducing $\Phi$ and $\Psi$ is to turn the two-sided nondegeneracy condition into an ordinary statement about kernels of linear maps. Define $W^*=\operatorname{Hom}_k(W,k)$ and $V^*=\operatorname{Hom}_k(V,k)$. For each vector $v \in V$, the rule
\begin{align*}
w \mapsto B(v,w)
\end{align*}
is a $k$-linear functional on $W$ because $B$ is linear in its second variable. Therefore we obtain a map
\begin{align*}
\Phi: V \to W^*,
\end{align*}
where $\Phi(v)$ is the functional $w \mapsto B(v,w)$. Linearity of $\Phi$ follows from linearity of $B$ in its first variable: for $\lambda,\mu \in k$ and $v_1,v_2 \in V$,
\begin{align*}
\Phi(\lambda v_1+\mu v_2)(w)=B(\lambda v_1+\mu v_2,w)=\lambda B(v_1,w)+\mu B(v_2,w)
\end{align*}
for every $w \in W$, so
\begin{align*}
\Phi(\lambda v_1+\mu v_2)=\lambda \Phi(v_1)+\mu \Phi(v_2).
\end{align*}
Now compute its kernel. A vector $v \in V$ lies in $\ker \Phi$ exactly when $\Phi(v)$ is the zero functional on $W$. By the definition of $\Phi(v)$, this means
\begin{align*}
B(v,w)=0 \text{ for all } w \in W.
\end{align*}
Hence
\begin{align*}
\ker \Phi = \{v \in V : B(v,w)=0 \text{ for all } w \in W\},
\end{align*}
which is exactly the left radical of $B$.
The same construction in the other variable gives
\begin{align*}
\Psi: W \to V^*,
\end{align*}
where $\Psi(w)$ is the functional $v \mapsto B(v,w)$. Bilinearity again gives linearity of $\Psi$, and the kernel condition says
\begin{align*}
\ker \Psi = \{w \in W : B(v,w)=0 \text{ for all } v \in V\}.
\end{align*}
Thus $\ker \Psi$ is exactly the right radical of $B$.
[/guided]
[/step]
[step:Compute the matrices of the induced maps in the chosen dual bases]
Let $(w_1^*,\dots,w_n^*)$ be the [dual basis](/theorems/414) of $W^*$ and let $(v_1^*,\dots,v_n^*)$ be the dual basis of $V^*$. For each $j \in \{1,\dots,n\}$ and each $i \in \{1,\dots,n\}$,
\begin{align*}
\Psi(w_j)(v_i)=B(v_i,w_j)=a_{ij}.
\end{align*}
Therefore
\begin{align*}
\Psi(w_j)=\sum_{i=1}^n a_{ij}v_i^*.
\end{align*}
So the matrix of $\Psi:W \to V^*$ with respect to the basis $(w_1,\dots,w_n)$ of $W$ and the dual basis $(v_1^*,\dots,v_n^*)$ of $V^*$ is $A$.
Likewise, for each $i,j \in \{1,\dots,n\}$,
\begin{align*}
\Phi(v_i)(w_j)=B(v_i,w_j)=a_{ij}.
\end{align*}
Hence
\begin{align*}
\Phi(v_i)=\sum_{j=1}^n a_{ij}w_j^*.
\end{align*}
Thus, with the usual column-coordinate convention, the matrix of $\Phi:V \to W^*$ is $A^\top$. Since $\det(A^\top)=\det A$, the determinant of the matrix of $\Phi$ is nonzero exactly when $\det A$ is nonzero.
[/step]
[step:Use the determinant criterion to prove nondegeneracy from $\det A \neq 0$]
Assume
\begin{align*}
\det A \neq 0.
\end{align*}
By the standard determinant criterion for square matrices over a field, the [linear map](/page/Linear%20Map) represented by $A$ is an isomorphism, and the linear map represented by $A^\top$ is also an isomorphism because $\det(A^\top)=\det A \neq 0$.
Since $\Psi$ is represented by $A$, the map $\Psi:W \to V^*$ is injective, so
\begin{align*}
\ker \Psi = \{0\}.
\end{align*}
Since $\Phi$ is represented by $A^\top$, the map $\Phi:V \to W^*$ is injective, so
\begin{align*}
\ker \Phi = \{0\}.
\end{align*}
By the kernel identifications above, both the left and right radicals of $B$ vanish. Therefore $B$ is nondegenerate.
[/step]
[step:Use nondegeneracy to force the determinant to be nonzero]
Assume $B$ is nondegenerate. Then both radicals of $B$ vanish. In particular, the right radical vanishes, and the kernel identification gives
\begin{align*}
\ker \Psi = \{0\}.
\end{align*}
Thus $\Psi:W \to V^*$ is injective.
Because $V$ is finite-dimensional and $\dim_k V=n$, its [dual space](/page/Dual%20Space) satisfies
\begin{align*}
\dim_k V^*=n.
\end{align*}
Also $\dim_k W=n$. Hence $\Psi$ is an injective linear map between two $n$-dimensional vector spaces over $k$, so $\Psi$ is an isomorphism. Its matrix in the chosen bases is $A$, so $A$ is invertible. By the standard determinant criterion for square matrices over a field,
\begin{align*}
\det A \neq 0.
\end{align*}
This proves the converse implication and completes the equivalence.
[/step]
