[proofplan] We translate the [bilinear form](/page/Bilinear%20Form) into its two associated linear maps, one from $V$ into $W^*$ and one from $W$ into $V^*$. Left and right nondegeneracy are exactly the injectivity of these two maps. For finite-dimensional linear maps, injectivity is equivalent to the rank being the dimension of the domain, and applying this criterion to the two associated maps gives the two rank criteria. The final statement follows by combining the left and right criteria and using the convention that nondegenerate means both left and right nondegenerate. [/proofplan]

[step:Translate left nondegeneracy into injectivity of the map $V \to W^*$]Define the [linear map](/page/Linear%20Map) $L_B: V \to W^*$ by declaring that for each $v \in V$, the functional $L_B(v): W \to k$ is given by \begin{align*} L_B(v)(w)=B(v,w). \end{align*} The map $L_B(v)$ is linear because $B$ is linear in its second argument, and the map $L_B$ is linear because $B$ is linear in its first argument. By definition, $B$ is left nondegenerate exactly when the only vector $v \in V$ satisfying $B(v,w)=0$ for every $w \in W$ is $v=0$. This condition is equivalent to \begin{align*} \ker L_B=\{0\}. \end{align*} Thus $B$ is left nondegenerate if and only if $L_B$ is injective.[/step]

[guided]The purpose of $L_B$ is to package the first argument of $B$ into a linear map. For each fixed vector $v \in V$, define $L_B(v): W \to k$ by \begin{align*} L_B(v)(w)=B(v,w). \end{align*} Since $B$ is linear in the second variable, $L_B(v)$ is a linear functional on $W$, so $L_B(v) \in W^*$. Since $B$ is linear in the first variable, the assignment $v \mapsto L_B(v)$ is a linear map $L_B: V \to W^*$. Now compare this with left nondegeneracy. A vector $v \in V$ lies in $\ker L_B$ precisely when $L_B(v)$ is the zero functional on $W$. Written out, this means \begin{align*} L_B(v)(w)=0 \text{ for every } w \in W. \end{align*} By the definition of $L_B$, this is the same as \begin{align*} B(v,w)=0 \text{ for every } w \in W. \end{align*} Left nondegeneracy says that the only such vector is $v=0$. Therefore left nondegeneracy is exactly the assertion $\ker L_B=\{0\}$, which is exactly injectivity of $L_B$.[/guided]

[step:Convert injectivity of $L_B$ into the left rank criterion] We use the finite-dimensional rank-nullity identity: if $T: X \to Y$ is a linear map between finite-dimensional vector spaces, then \begin{align*} \dim X=\dim \ker T+\operatorname{rank} T. \end{align*} Applying this to $L_B: V \to W^*$ gives \begin{align*} \dim V=\dim \ker L_B+\operatorname{rank} L_B. \end{align*} Hence $\ker L_B=\{0\}$ if and only if $\operatorname{rank} L_B=\dim V$. By the definition of the rank of the bilinear form, $\operatorname{rank} L_B=\operatorname{rank}(B)$. Therefore $B$ is left nondegenerate if and only if \begin{align*} \operatorname{rank}(B)=\dim V. \end{align*} [/step]

[step:Translate right nondegeneracy into the rank of the map $W \to V^*$] Define the linear map $R_B: W \to V^*$ by declaring that for each $w \in W$, the functional $R_B(w): V \to k$ is given by \begin{align*} R_B(w)(v)=B(v,w). \end{align*} The map $R_B(w)$ is linear because $B$ is linear in its first argument, and $R_B$ is linear because $B$ is linear in its second argument. By definition, $B$ is right nondegenerate exactly when the only vector $w \in W$ satisfying $B(v,w)=0$ for every $v \in V$ is $w=0$. This is equivalent to $\ker R_B=\{0\}$, hence to injectivity of $R_B$. Applying the finite-dimensional rank-nullity identity to $R_B: W \to V^*$ gives \begin{align*} \dim W=\dim \ker R_B+\operatorname{rank} R_B. \end{align*} Thus $R_B$ is injective if and only if $\operatorname{rank} R_B=\dim W$. Since $\operatorname{rank} R_B=\operatorname{rank}(B)$ by definition of the rank of $B$, the form $B$ is right nondegenerate if and only if \begin{align*} \operatorname{rank}(B)=\dim W. \end{align*} [/step]

[step:Combine the one-sided criteria to obtain the nondegeneracy criterion] By convention, $B$ is nondegenerate if and only if it is both left nondegenerate and right nondegenerate. By the two criteria proved above, this is equivalent to the simultaneous equalities \begin{align*} \operatorname{rank}(B)=\dim V \end{align*} and \begin{align*} \operatorname{rank}(B)=\dim W. \end{align*} These two equalities hold simultaneously if and only if \begin{align*} \dim V=\dim W \end{align*} and \begin{align*} \operatorname{rank}(B)=\dim V=\dim W. \end{align*} This is precisely the stated criterion for nondegeneracy. [/step]