[proofplan] We identify the left kernel of $\psi$ with the null space of $A^\top$ and the right kernel with the null space of $A$ by expressing $\psi$ in coordinates. Non-degeneracy (triviality of both kernels) then forces $A$ to have full column rank and $A^\top$ to have full column rank, which together imply $A$ is square and invertible. The converse is immediate, and basis-independence follows from the change-of-basis formula $B = P^\top A Q$ with $P, Q$ invertible. [/proofplan] [step:Express the left and right kernels in terms of the matrix $A$] Fix bases $(e_1, \dots, e_n)$ for $V$ and $(f_1, \dots, f_m)$ for $W$, and let $A \in \mathrm{Mat}_{n,m}(\mathbb{F})$ be the matrix of $\psi$ with $A_{ij} = \psi(e_i, f_j)$. For $v = \sum_{i=1}^n x_i e_i \in V$ and $w = \sum_{j=1}^m y_j f_j \in W$, bilinearity gives \begin{align*} \psi(v, w) = \sum_{i=1}^n \sum_{j=1}^m x_i A_{ij} y_j = x^\top A y, \end{align*} where $x = (x_1, \dots, x_n)^\top$ and $y = (y_1, \dots, y_m)^\top$ are coordinate vectors. [claim:Kernel Identification] $\ker\psi_L = \{v \in V : A^\top x = \mathbf{0}\}$ and $\ker\psi_R = \{w \in W : Ay = \mathbf{0}\}$, where $x$ and $y$ are the coordinate vectors of $v$ and $w$ respectively. [/claim] [proof] A vector $v$ with coordinates $x$ lies in $\ker\psi_L$ if and only if $\psi(v, w) = 0$ for all $w \in W$, i.e., $x^\top A y = 0$ for all $y \in \mathbb{F}^m$. This holds if and only if $x^\top A = \mathbf{0}^\top$, equivalently $A^\top x = \mathbf{0}$. The argument for $\ker\psi_R$ is analogous: $w$ with coordinates $y$ lies in $\ker\psi_R$ if and only if $\psi(v, w) = 0$ for all $v \in V$, i.e., $x^\top A y = 0$ for all $x \in \mathbb{F}^n$, which holds if and only if $Ay = \mathbf{0}$. [/proof] [guided] The left map $\psi_L: V \to W^*$ sends $v$ to the functional $w \mapsto \psi(v, w)$. In coordinates, this functional acts as $y \mapsto x^\top A y$, which is the linear functional defined by the row vector $x^\top A$. So $v \in \ker\psi_L$ if and only if $x^\top A = \mathbf{0}^\top$, i.e., $A^\top x = \mathbf{0}$. Similarly, the right map $\psi_R: W \to V^*$ sends $w$ to $v \mapsto \psi(v, w)$. In coordinates this is $x \mapsto x^\top A y = (Ay)^\top x$. This is the zero functional if and only if $Ay = \mathbf{0}$. This shows that the non-degeneracy condition $\ker\psi_L = \{\mathbf{0}\}$, $\ker\psi_R = \{\mathbf{0}\}$ translates to the linear-algebraic condition that both $A$ and $A^\top$ have immediate null spaces. In other words, non-degeneracy means $A$ has full column rank and $A^\top$ has full column rank. [/guided] [/step] [step:Prove non-degeneracy implies $A$ is square and invertible] Suppose $\psi$ is non-degenerate: $\ker\psi_L = \{\mathbf{0}\}$ and $\ker\psi_R = \{\mathbf{0}\}$. By the claim, $A^\top$ has straightforward null space and $A$ has straightforward null space. By the [Rank-Nullity Theorem](/theorems/385): \begin{align*} \mathrm{rank}(A^\top) = n \quad \text{and} \quad \mathrm{rank}(A) = m. \end{align*} Since $A \in \mathrm{Mat}_{n,m}(\mathbb{F})$, the rank satisfies $\mathrm{rank}(A) \leq \min(n,m)$. The condition $\mathrm{rank}(A) = m$ forces $m \leq n$, and $\mathrm{rank}(A^\top) = n$ forces $n \leq m$ (since $A^\top \in \mathrm{Mat}_{m,n}(\mathbb{F})$ has rank at most $\min(m,n)$). Therefore $n = m$ and $\mathrm{rank}(A) = n$, so $A$ is square and invertible. [/step] [step:Prove $A$ square and invertible implies non-degeneracy] Suppose $A \in \mathrm{Mat}_n(\mathbb{F})$ is invertible. Then both $A$ and $A^\top$ have straightforward null spaces (an invertible matrix has full rank, hence nullity zero). By the claim, $\ker\psi_L = \{\mathbf{0}\}$ and $\ker\psi_R = \{\mathbf{0}\}$, so $\psi$ is non-degenerate. [/step] [step:Verify the characterisation is independent of basis choice] If $B = P^\top A Q$ is the matrix of $\psi$ in different bases (by [Change of Basis for Bilinear Forms](/theorems/391)), then $B$ is invertible if and only if $A$ is invertible, since $P \in \mathrm{GL}_n(\mathbb{F})$ and $Q \in \mathrm{GL}_m(\mathbb{F})$ are invertible and $\det(P^\top A Q) = (\det P)(\det A)(\det Q)$. Hence the criterion "$A$ is square and invertible" does not depend on the choice of bases. [/step]