[proofplan] Associate to the [bilinear form](/page/Bilinear%20Form) $B$ two linear maps: one map $\Phi: V \to W^*$ obtained by fixing the left argument, and one map $\Psi: W \to V^*$ obtained by fixing the right argument. The left radical is exactly $\ker \Phi$, and the right radical is exactly $\ker \Psi$. Applying the [Rank-Nullity Theorem](/theorems/916) to these finite-dimensional linear maps gives the two dimension formulas, after identifying $\operatorname{rank}(B)$ with the rank of either associated map. [/proofplan] [step:Realize the left radical as the kernel of the map into $W^*$] Let $W^* := \operatorname{Hom}_k(W,k)$ denote the dual [vector space](/page/Vector%20Space) of $W$. Define the map \begin{align*} \Phi: V \to W^*, \qquad \Phi(v)(w) := B(v,w). \end{align*} For $v_1,v_2 \in V$, $a,b \in k$, and $w \in W$, bilinearity of $B$ gives \begin{align*} \Phi(av_1+bv_2)(w)=B(av_1+bv_2,w)=aB(v_1,w)+bB(v_2,w)=(a\Phi(v_1)+b\Phi(v_2))(w). \end{align*} Since this equality holds for every $w \in W$, the map $\Phi: V \to W^*$ is $k$-linear. By definition of kernel, \begin{align*} \ker \Phi=\{v \in V : \Phi(v)=0_{W^*}\}. \end{align*} The zero functional $0_{W^*}: W \to k$ satisfies $0_{W^*}(w)=0$ for every $w \in W$, so \begin{align*} \ker \Phi=\{v \in V : B(v,w)=0 \text{ for every } w \in W\}=\operatorname{rad}_L(B). \end{align*} [guided] The purpose of introducing $\Phi$ is to turn the bilinear object $B$ into an ordinary [linear map](/page/Linear%20Map), because rank-nullity applies to linear maps. Let $W^* := \operatorname{Hom}_k(W,k)$ be the dual vector space. For each $v \in V$, fixing the first argument of $B$ gives a function $W \to k$, namely $w \mapsto B(v,w)$. This function is linear in $w$ because $B$ is bilinear. Thus it is an element of $W^*$, and we may define \begin{align*} \Phi: V \to W^*, \qquad \Phi(v)(w) := B(v,w). \end{align*} We now verify that $\Phi$ is a linear map. Let $v_1,v_2 \in V$, let $a,b \in k$, and let $w \in W$. Using bilinearity of $B$ in the first argument, \begin{align*} \Phi(av_1+bv_2)(w)=B(av_1+bv_2,w)=aB(v_1,w)+bB(v_2,w). \end{align*} By the definition of $\Phi(v_1)$ and $\Phi(v_2)$, the last expression is \begin{align*} a\Phi(v_1)(w)+b\Phi(v_2)(w)=(a\Phi(v_1)+b\Phi(v_2))(w). \end{align*} Since the two functionals agree at every $w \in W$, they are equal as elements of $W^*$. Therefore \begin{align*} \Phi(av_1+bv_2)=a\Phi(v_1)+b\Phi(v_2), \end{align*} so $\Phi$ is $k$-linear. Now we identify its kernel. A vector $v \in V$ lies in $\ker \Phi$ exactly when $\Phi(v)$ is the zero functional on $W$. This means precisely that $\Phi(v)(w)=0$ for every $w \in W$. Substituting the definition of $\Phi$ gives $B(v,w)=0$ for every $w \in W$. Hence \begin{align*} \ker \Phi=\{v \in V : B(v,w)=0 \text{ for every } w \in W\}=\operatorname{rad}_L(B). \end{align*} [/guided] [/step] [step:Apply rank-nullity to obtain the left radical formula] The vector space $V$ is finite-dimensional by hypothesis, and $\Phi: V \to W^*$ is linear. By the Rank-Nullity Theorem (citing a result not yet in the wiki: Rank-Nullity Theorem), \begin{align*} \dim_k V=\operatorname{rank}(\Phi)+\dim_k \ker \Phi. \end{align*} By definition, $\operatorname{rank}(B)=\operatorname{rank}(\Phi)$, and the previous step gives $\ker \Phi=\operatorname{rad}_L(B)$. Therefore \begin{align*} \operatorname{rank}(B)=\dim_k V-\dim_k \operatorname{rad}_L(B). \end{align*} [/step] [step:Realize the right radical as the kernel of the map into $V^*$] Let $V^* := \operatorname{Hom}_k(V,k)$ denote the dual vector space of $V$. Define the map \begin{align*} \Psi: W \to V^*, \qquad \Psi(w)(v) := B(v,w). \end{align*} For $w_1,w_2 \in W$, $a,b \in k$, and $v \in V$, bilinearity of $B$ gives \begin{align*} \Psi(aw_1+bw_2)(v)=B(v,aw_1+bw_2)=aB(v,w_1)+bB(v,w_2)=(a\Psi(w_1)+b\Psi(w_2))(v). \end{align*} Since this equality holds for every $v \in V$, the map $\Psi: W \to V^*$ is $k$-linear. The kernel of $\Psi$ is \begin{align*} \ker \Psi=\{w \in W : \Psi(w)=0_{V^*}\}. \end{align*} Since $0_{V^*}: V \to k$ is the zero functional, this is equivalent to \begin{align*} \ker \Psi=\{w \in W : B(v,w)=0 \text{ for every } v \in V\}=\operatorname{rad}_R(B). \end{align*} [/step] [step:Apply rank-nullity to obtain the right radical formula] The vector space $W$ is finite-dimensional by hypothesis, and $\Psi: W \to V^*$ is linear. By the Rank-Nullity Theorem (citing a result not yet in the wiki: Rank-Nullity Theorem), \begin{align*} \dim_k W=\operatorname{rank}(\Psi)+\dim_k \ker \Psi. \end{align*} The rank of the bilinear form is the common rank of the two associated maps $\Phi: V \to W^*$ and $\Psi: W \to V^*$, so $\operatorname{rank}(B)=\operatorname{rank}(\Psi)$. Using $\ker \Psi=\operatorname{rad}_R(B)$ from the previous step, we obtain \begin{align*} \operatorname{rank}(B)=\dim_k W-\dim_k \operatorname{rad}_R(B). \end{align*} Together with the left radical formula, this proves both asserted identities. [/step]