[proofplan]
We compare the [rank of a bilinear form](/page/Rank%20of%20a%20Bilinear%20Form) with the dimension of the image of its associated [linear map](/page/Linear%20Map) into the [dual space](/page/Dual%20Space). The restricted form has an associated map $V_0 \to W_0^*$, and this map factors through the original associated map $V \to W^*$ followed by restriction of functionals from $W$ to $W_0$. Since the image of the restricted associated map is the image of a subspace of $\operatorname{im}(\Phi_B)$ under a linear map, its dimension cannot exceed $\dim \operatorname{im}(\Phi_B)$.
[/proofplan]
[step:Represent the bilinear form by its associated map into the dual space]
Let $W^* := \operatorname{Hom}_k(W,k)$ denote the dual [vector space](/page/Vector%20Space) of $W$. Define the associated linear map
\begin{align*}
\Phi_B: V \to W^*
\end{align*}
by requiring that, for each $v \in V$, the functional $\Phi_B(v): W \to k$ is given by
\begin{align*}
\Phi_B(v)(w) = B(v,w)
\end{align*}
for all $w \in W$. The bilinearity of $B$ implies that each $\Phi_B(v)$ is $k$-linear in $w$, and also implies that $\Phi_B$ is $k$-linear in $v$.
By the definition of the rank of a [bilinear form](/page/Bilinear%20Form) through its associated linear map,
\begin{align*}
\operatorname{rank}(B) = \dim_k \operatorname{im}(\Phi_B).
\end{align*}
[/step]
[step:Factor the associated map of the restricted form through the original associated map]
Let $W_0^* := \operatorname{Hom}_k(W_0,k)$ denote the dual vector space of $W_0$. Define the associated map of the restricted bilinear form
\begin{align*}
\Phi_0: V_0 \to W_0^*
\end{align*}
by requiring that, for each $v_0 \in V_0$, the functional $\Phi_0(v_0): W_0 \to k$ is given by
\begin{align*}
\Phi_0(v_0)(w_0) = B(v_0,w_0)
\end{align*}
for all $w_0 \in W_0$.
Let
\begin{align*}
\iota: V_0 \to V
\end{align*}
be the inclusion map, and let
\begin{align*}
\rho: W^* \to W_0^*
\end{align*}
be the restriction map defined by
\begin{align*}
\rho(\lambda) = \lambda|_{W_0}
\end{align*}
for each $\lambda \in W^*$. Both $\iota$ and $\rho$ are $k$-linear maps. For every $v_0 \in V_0$ and every $w_0 \in W_0$, we have
\begin{align*}
(\rho \circ \Phi_B \circ \iota)(v_0)(w_0) = \rho(\Phi_B(v_0))(w_0).
\end{align*}
By the definition of $\rho$, this equals
\begin{align*}
\Phi_B(v_0)(w_0).
\end{align*}
By the definition of $\Phi_B$, this equals
\begin{align*}
B(v_0,w_0).
\end{align*}
By the definition of $\Phi_0$, this equals
\begin{align*}
\Phi_0(v_0)(w_0).
\end{align*}
Since the two functionals agree on every $w_0 \in W_0$, we have
\begin{align*}
\Phi_0 = \rho \circ \Phi_B \circ \iota.
\end{align*}
[guided]
The point of introducing $\Phi_B$ and $\Phi_0$ is that ranks of bilinear forms become dimensions of images of linear maps. We now show that the map for the restricted form is obtained by first using the original map and then restricting the resulting functional to the smaller space $W_0$.
Let
\begin{align*}
\iota: V_0 \to V
\end{align*}
be the inclusion map, so $\iota(v_0)=v_0$ for each $v_0 \in V_0$. Let
\begin{align*}
\rho: W^* \to W_0^*
\end{align*}
be the restriction map defined by
\begin{align*}
\rho(\lambda)=\lambda|_{W_0}
\end{align*}
for each $\lambda \in W^*$. The map $\rho$ is $k$-linear because restriction preserves addition and scalar multiplication of linear functionals.
We claim that
\begin{align*}
\Phi_0 = \rho \circ \Phi_B \circ \iota.
\end{align*}
To verify equality of two maps into $W_0^*$, we evaluate both sides on an arbitrary vector $v_0 \in V_0$ and then evaluate the resulting functionals on an arbitrary vector $w_0 \in W_0$. For such $v_0$ and $w_0$,
\begin{align*}
(\rho \circ \Phi_B \circ \iota)(v_0)(w_0) = \rho(\Phi_B(v_0))(w_0).
\end{align*}
Since $\rho$ restricts a functional on $W$ to $W_0$, this is
\begin{align*}
\Phi_B(v_0)(w_0).
\end{align*}
By the defining property of the associated map $\Phi_B$, this is
\begin{align*}
B(v_0,w_0).
\end{align*}
But the restricted form $B|_{V_0 \times W_0}$ is defined by the same scalar value $B(v_0,w_0)$ on pairs $(v_0,w_0) \in V_0 \times W_0$, so the associated map $\Phi_0$ satisfies
\begin{align*}
\Phi_0(v_0)(w_0)=B(v_0,w_0).
\end{align*}
Thus $(\rho \circ \Phi_B \circ \iota)(v_0)$ and $\Phi_0(v_0)$ agree on every element of $W_0$, so they are the same element of $W_0^*$. Since $v_0 \in V_0$ was arbitrary, the desired factorization follows.
[/guided]
[/step]
[step:Compare the image dimensions produced by the factorization]
From the factorization
\begin{align*}
\Phi_0 = \rho \circ \Phi_B \circ \iota,
\end{align*}
we obtain
\begin{align*}
\operatorname{im}(\Phi_0) = \rho(\Phi_B(V_0)).
\end{align*}
Since $V_0 \subset V$, we have $\Phi_B(V_0) \subset \Phi_B(V)=\operatorname{im}(\Phi_B)$. Therefore
\begin{align*}
\operatorname{im}(\Phi_0) \subset \rho(\operatorname{im}(\Phi_B)).
\end{align*}
The restriction of $\rho$ to the finite-dimensional vector space $\operatorname{im}(\Phi_B)$ is a linear map
\begin{align*}
\rho|_{\operatorname{im}(\Phi_B)}: \operatorname{im}(\Phi_B) \to W_0^*.
\end{align*}
For any linear map between finite-dimensional vector spaces, the dimension of the image is at most the dimension of the domain. Hence
\begin{align*}
\dim_k \rho(\operatorname{im}(\Phi_B)) \leq \dim_k \operatorname{im}(\Phi_B).
\end{align*}
Combining the inclusion with this dimension inequality gives
\begin{align*}
\dim_k \operatorname{im}(\Phi_0) \leq \dim_k \operatorname{im}(\Phi_B).
\end{align*}
[/step]
[step:Translate the image dimension inequality back into ranks]
By the definition of rank for the restricted bilinear form,
\begin{align*}
\operatorname{rank}(B|_{V_0 \times W_0}) = \dim_k \operatorname{im}(\Phi_0).
\end{align*}
By the definition of rank for $B$,
\begin{align*}
\operatorname{rank}(B) = \dim_k \operatorname{im}(\Phi_B).
\end{align*}
The image dimension inequality obtained above therefore gives
\begin{align*}
\operatorname{rank}(B|_{V_0 \times W_0}) \leq \operatorname{rank}(B).
\end{align*}
This is the desired conclusion.
[/step]
