[proofplan] We compare the [bilinear form](/page/Bilinear%20Form) with its associated [linear map](/page/Linear%20Map) $\Psi_B:W\to V^*$. In the chosen basis of $W$ and the [dual basis](/theorems/414) of $V^*$, the coordinate matrix of $\Psi_B$ is exactly the matrix $A$ whose entries are $B(v_i,w_j)$. Therefore the rank of the bilinear form, defined as the dimension of the image of $\Psi_B$, is the same as the column rank of $A$. [/proofplan] [step:Pass from the bilinear form to its associated linear map] Define the map \begin{align*} \Psi_B: W \to V^* \end{align*} by declaring that, for each $w \in W$, the functional $\Psi_B(w):V\to k$ is \begin{align*} \Psi_B(w)(v)=B(v,w). \end{align*} For fixed $w \in W$, the map $v \mapsto B(v,w)$ is linear because $B$ is linear in its first variable, so $\Psi_B(w)\in V^*$. For $w,w' \in W$ and $\lambda \in k$, bilinearity in the second variable gives \begin{align*} \Psi_B(w+\lambda w')(v)=B(v,w+\lambda w')=B(v,w)+\lambda B(v,w')=(\Psi_B(w)+\lambda\Psi_B(w'))(v) \end{align*} for every $v\in V$. Hence $\Psi_B$ is a linear map $W\to V^*$. By definition of the [rank of a bilinear form](/page/Rank%20of%20a%20Bilinear%20Form), \begin{align*} \operatorname{rank}(B)=\operatorname{rank}(\Psi_B)=\dim_k \operatorname{im}(\Psi_B). \end{align*} [/step] [step:Identify the coordinate matrix of $\Psi_B$] Let $(v_1^*,\dots,v_m^*)$ denote the dual basis of $V^*$, so $v_i^*:V\to k$ is the linear functional satisfying $v_i^*(v_\ell)=1$ if $i=\ell$ and $v_i^*(v_\ell)=0$ otherwise. For each $j\in\{1,\dots,n\}$, write $\Psi_B(w_j)$ in the dual basis: \begin{align*} \Psi_B(w_j)=\sum_{i=1}^m c_{ij}v_i^* \end{align*} for uniquely determined scalars $c_{ij}\in k$. Evaluating both sides at $v_\ell$ gives \begin{align*} c_{\ell j}=\Psi_B(w_j)(v_\ell)=B(v_\ell,w_j)=A_{\ell j}. \end{align*} Thus \begin{align*} \Psi_B(w_j)=\sum_{i=1}^m A_{ij}v_i^*. \end{align*} Therefore the matrix of the linear map $\Psi_B:W\to V^*$ with respect to the basis $(w_1,\dots,w_n)$ of $W$ and the dual basis $(v_1^*,\dots,v_m^*)$ of $V^*$ is precisely $A$. [guided] The point of introducing $\Psi_B$ is that a bilinear form has two inputs, while matrix rank is naturally attached to a linear map. We freeze the first input as the variable of a functional and send each $w\in W$ to the functional $v\mapsto B(v,w)$. Let $(v_1^*,\dots,v_m^*)$ be the dual basis of $V^*$. To find the coordinate matrix of $\Psi_B$, we must compute the coordinates of each image vector $\Psi_B(w_j)$ in this dual basis. So write \begin{align*} \Psi_B(w_j)=\sum_{i=1}^m c_{ij}v_i^*. \end{align*} The coefficient $c_{\ell j}$ is recovered by evaluating at $v_\ell$, because the dual basis satisfies $v_i^*(v_\ell)=0$ unless $i=\ell$, and $v_\ell^*(v_\ell)=1$. Hence \begin{align*} c_{\ell j}=\Psi_B(w_j)(v_\ell). \end{align*} By the definition of $\Psi_B$, this is \begin{align*} \Psi_B(w_j)(v_\ell)=B(v_\ell,w_j). \end{align*} By the definition of the matrix $A$ of the bilinear form in the bases $(v_1,\dots,v_m)$ and $(w_1,\dots,w_n)$, we have \begin{align*} B(v_\ell,w_j)=A_{\ell j}. \end{align*} Therefore every column of the coordinate matrix of $\Psi_B$ is exactly the corresponding column of $A$. This verifies that the coordinate matrix of $\Psi_B$ is $A$ itself. [/guided] [/step] [step:Compare the rank of the map with the rank of its coordinate matrix] The rank of a matrix is the dimension of the span of its columns. Since the columns of $A$ are exactly the coordinate vectors of \begin{align*} \Psi_B(w_1),\dots,\Psi_B(w_n) \end{align*} in the basis $(v_1^*,\dots,v_m^*)$ of $V^*$, the column span of $A$ is the coordinate representation of the subspace \begin{align*} \operatorname{span}_k\{\Psi_B(w_1),\dots,\Psi_B(w_n)\}\subseteq V^*. \end{align*} Because $(w_1,\dots,w_n)$ is a basis of $W$, every $w\in W$ is a $k$-linear combination of $w_1,\dots,w_n$, and linearity of $\Psi_B$ gives \begin{align*} \operatorname{im}(\Psi_B)=\operatorname{span}_k\{\Psi_B(w_1),\dots,\Psi_B(w_n)\}. \end{align*} Passing to coordinates in a basis preserves [dimension of subspaces](/theorems/375). Therefore \begin{align*} \operatorname{rank}(A)=\dim_k\operatorname{im}(\Psi_B)=\operatorname{rank}(\Psi_B)=\operatorname{rank}(B). \end{align*} This proves the claimed equality. [/step]