Connection between dual maps and bilinear forms? Say $\xi$ is a linear map from $V$ to $W$. Then its dual map is the map $\xi^\star:W^\star \to V^\star$ such that $g\mapsto \xi^\star(g) = g\circ \xi \doteq f$. If $A_\mathcal V^\mathcal W$ is the matrix associated to $\xi$ in the bases $\mathcal V$ of $V$ and $\mathcal W$ of $W$, then its transpose $(A_\mathcal V ^ \mathcal W)^T$ is the matrix associated to $\xi^\star$ in the dual bases $\mathcal W^\star$ and $\mathcal V^\star$. This also means that if $V$ is one-dimensional, the matrix associated to $\xi^\star$ is a row vector (covector), the transpose of a column vector in $\mathbb K^m$.
Now suppose we have a bilinear form $\phi : V\times W \to \mathbb K$. Then I can represent it through the matrix
$$\Phi =  \begin{bmatrix}\phi(v_1,v_1) & \cdots & \phi(v_1,w_m) \\ \vdots & \ddots & \vdots \\ \phi(v_n,w_1) & \cdots& \phi (v_n,w_m) \end{bmatrix}$$
where the $v_i$'s are the elements of the basis $\mathcal V$ and the $w_j$'s are the elements of the basis $\mathcal W$. If $x=(x_1,\dots,x_n)$ and $y=(y_1,\dots,y_n)$ are the coordinates of $v$ and $w$ respectively in the bases $\mathcal V$ and $\mathcal W$, then I can write $\phi(v,w) = x^T\Phi y$.
This last step is justified in my notes by means of the proposition:

Proposition. The bilinear form $\phi$ is completely determined by the $nm$ scalars $a_{ij} \doteq \phi(v_i,w_j)$.
Proof. We can write $v = \sum_{i=1}^n x_iv_i$ and $w = \sum_{j=1}^m y_jw_j$, so, by iterating bilinearity, we have
$$\phi(v,w) = \phi\left(\sum_{i=1}^n x_iv_i,\sum_{j=1}^my_jw_j\right) = \sum_{i,j} a_{ij} x_iy_j $$

The last term corresponds to the expansion of $x^T\Phi y$, but this seems to fall relatively out of the blue. Indeed, seeing that a row vector is involved would have one think that some dual map is involved – and when learning about matrix congruence, this involvement seems even more evident. Yet nowhere in my notes is the connection directly mentioned and clarified (the two concepts belong to separate chapters).
So, is there a way to justify the transpose formula in terms of dual spaces? In other words, does this way of representing bilinear forms (once the bases are chosen) conceal a deeper connection to the concept of dual spaces and dual maps?
 A: Yes.  We can break bilinear form into a linear map by Currying it.  In particular, define
$$
\tilde \phi: V \to W^\star\\
[\tilde \phi(x)](y) = \phi(x,y)
$$
That is, $\tilde \phi$ maps the vector $x$ to the functional $\phi(x, \cdot)$.  As such, any bilinear form induces an linear transformation between the first vector space and the dual of the second.  When this map is an isomorphism, we say that $\phi$ is non-degenerate.
A: If $f:V\times W\to k$ is a bilinear map, there is a linear function $F:V\to W^*$ such that $F(v)(w)=f(v,w)$. The rule which maps $f$ to $F$ is a bijection, and is the standard way to connect bilinear maps and dual spaces. You should write down what the relation between the matrix of $f$ with respect to bases $B_V$ and $B_W$ of $V$ and $W$, and the matrix of $F$ with respect to $B_V$ and the basis of $W^*$ dual to $B_W$.
A: Is the following what you are looking for?
For fixed $v \in V$, consider the map
$$
\Psi(v) : W \to K, \qquad w \mapsto \Phi(v, w). 
$$
For each $v \in V$, this is a linear map, thus an element of $W^{\star}$.
Now note that the map
$$
V \to W^{\star} \qquad v \mapsto \Psi(v)
$$
is linear.
