[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 $B$.[/step]

[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: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]