Intuition behind tensor expansions of linear maps Given finite-dimensional vector spaces $V,W$, there is an isomorphism $\text{Hom}(V,W) \rightarrow V^* \otimes W$. In particular, any linear map $\phi : V \rightarrow W$ has a tensor expansion $\sum v^*_i \otimes w_i$ where $v^*_i \in V^*, w_i \in W$.
For example, if one chooses dual bases $\{x_i\}, \{e_i\}$ of $V^*$ and $V$, then $\sum x_i \otimes e_i \in V^* \otimes V$ is a tensor expansion for the identity map on $V$.
What's the best way to intuitively understand tensor expansions of linear maps?
 A: Here's a very "low-tech" answer, in terms of coordinates. With respect to the dual bases (I take $\dim V=\dim W=2$ for simplicity), the linear transformation with matrix $$\begin{pmatrix} a&b \\ c&d \end{pmatrix}$$ can be expanded as
$$a \begin{pmatrix} 1 \\ 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix}
+ b \begin{pmatrix} 1 \\ 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix}
+ c \begin{pmatrix} 0 \\ 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix}
+ d \begin{pmatrix} 0 \\ 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix}.$$
Here the column vectors are coordinate vectors for elements in $W$, while the row vectors are coordinate vectors for elements in $V^*$.
A: Ok, let's try to incorporate your question in a more gneral setting: at least one of the vector spaces in the following is finite-dimensional.
Let $f\colon U\to U'$ and $g\colon V\to V'$ be linear maps. We easily define the "tensor product" of $f$ and $g$ to be the map $f\otimes g$ such that
$$ (f\otimes g)(u\otimes v)=f(u)\otimes g(v) $$
for all $u\in U$ and $v\in V$. So, we obtain a linear map
$$ \lambda\colon\hom(U,U')\otimes \hom(V,V')\cong \hom(V\otimes U, V'\otimes U'),$$
which is easily checked to be an isomorphism, provided at least one of the pairs $(U,U')$, $(V,V')$, $(U,V)$ consists of finite-dimensional vector spaces.
With an intelligent choice of the spaces, the isomorphism $\lambda$ allows us to prove every sort of useful identity relating tensor product of spaces and their duals: Try putting $U'=V'=\mathbb{K}$ (the underlying field), you'll obtain $(V\otimes U)^\star\cong U^\star\otimes V^\star $ (don't worry for the twist). Try putting $U=V'=\mathbb{K}$, you'll obtain... $\lambda_{UV}\colon V\otimes U^\star\cong \hom(U,V)$.
Now, let $f\colon U\to V$ be a linear map. Using bases for $U$ and $V$, we have
$$f(u_j) = \sum_i f_i^j v_i$$
for a family $(f_i^j)$ of scalars. Try proving that
$$  f = \lambda_{UV}\Big(\sum_{ij} f_i^j v_i\otimes u^j\Big). $$
A: Not entirely sure what you're looking for so I'll just throw out some thoughts:
Recall that $V^{*}$ is the set of linear maps from $V$ to $\mathbb{R}$.  Tensoring with $W$ effectively replaces the $\mathbb{R}$ with $W$.  So the star on $V$ represents that the maps are coming from $V$; the lack of one on $W$ represents that you're getting elements of $W$.  Likewise, I'd expect that elements of $\text{Hom}(V,W)$ would covary with a change of basis of $V$ and contravary with one of $W$ (if you're unfamiliar with these terms, here's the Wikipedia article).
For a bit more rigor, recall that you have a natural pairing between $V$ and $V^{*}$, which is just a map $$\langle -,- \rangle: V\times V^*\rightarrow\mathbb{R}$$ defined by $\langle v,\alpha\rangle=\alpha(v)$.  As you can check, this is invariant under change of basis (hence the "natural") -- this actually follows from $V$ and $V$ varying oppositely.  This also has the property that if $v_i,\alpha^i$ are dual bases for $V$ and $V^{*}$, then $\langle v_i,\alpha^i\rangle=1$, and $\langle v_i,\alpha^j\rangle=0$ when $i\ne j$.
So now choose bases $v_i,w_j$ for $V$ and $W$ and a dual basis $\alpha^i$ for $V^{*}$.  Given an element $v=\sum b^iv_i$ of $V$ and an element $f=\sum a_i^jv^i\otimes w_j$ of $V^*\otimes W$, we get
$$f(v)=\sum_{i,j,k}(a_i^j\alpha^i\otimes w_j)(b^kv_k)=\sum_{i,j,k}(a_i^jb^k\langle v_k,\alpha^i\rangle\otimes w_j).$$
Since the pairing is only nonzero for $k=i$, this reduces to
$$f(v)=\sum_{i,j}a_i^jb^i\langle v_i,\alpha^i\rangle\otimes w_j=\sum_{i,j}a_i^jb^iw_j$$
where the tensor product has disappeared because the output of the pairing is just a real number.
But this is just a vector in $W$!  And conversely, given a linear map $V\rightarrow W$, we can pick bases and write the map as a matrix $(a_i^j)$, which then transforms to the above tensor product.
I found these notes very helpful when I was learning this stuff.  You can change the 5 in the URL to other numbers to read all the notes, though I believe they stop at 9.
