Linear Maps as Tensors Let $V$ and $W$ be finite dimensional vector spaces and let $V^{\ast}$ denote the dual $V$. I read that the space $V^{\ast}\otimes W$ may be thought of in four different ways: as the space of linear maps $V\to W$, as the space of linear maps $W^{\ast}\to V$, as the dual space to $V\otimes W^{\ast}$, and as the space of linear maps $V\times W^{\ast}\to \mathbb{C}$.
I am having trouble seeing why this statement is true because I am not comfortable with dual spaces yet. Can someone help me out? Is this all a simple consequence of the universal property of tensor products? Thanks.
 A: Remember: for finite dimensional spaces, two vectors spaces (over the same field) are isomorphic if and only if they have the same dimension.  So, for all these, it suffices to simply find a basis, and therefore conclude the dimension.  In particular, we have
$$
\dim(\mathcal L(V,W)) = \dim(V \otimes W) = \dim(V) \dim(W)
$$
When $V$ and $W$ are finite dimensional, their dual spaces have the same dimension.

The key to answering any of these in a more general setting is the find an explicit map that gives you the desired isomorphism.  Note that often, the "nicer" isomorphisms are so-called "natural" isomorphisms, whose construction does not require that you specify a basis.  The natural isomorphisms also tend to be the only ones that still hold for infinite-dimensional spaces or extend to the context of modules.
For the isomorphism $(V^* \otimes W) \to (V \to W)$: note that $f \in V^*$ is a linear map $f:V \to \Bbb F$, and that we can multiply elements in $W$ by scalars in $\Bbb F$.  So, for $f \in V^*$ and $w \in W$, consider the following linear map generated by the following map on simple tensors:
$$
f \otimes w \mapsto w\,f(\cdot)
$$
That is, given a functional $f$ and vector $w$, we map the tensor product thereof to the map $T:V \to W$ defined by $T(x) = w\,f(x) = f(x) \cdot w$.  Note that this multiplication makes sense because $f(x) \in \Bbb F$ and $w$ is in some $\Bbb F$ vector space.
In order to prove that this is an isomorphism, we must show that it is well defined and that it is bijective.
That this map is well defined can be proven nicely using the universal property.  If you know about the universal property, you probably already have a theorem that states that any linear map on the simple tensors extends to a linear map on the entire space.
To prove surjectivity, it helps to note that the simple tensors can be used to create all maps of rank $1$.  From there, it suffices to note that arbitrary linear transformations may be expressed as a sum of rank-1 maps.
To prove injectivity, it suffices to note that arbitrary tensors can be written as $\sum_i f_i \otimes w_i$, where the $w_i$ are selected from some basis.  From there, we can see that the non-zero maps taken from the simple tensors have linearly independent one-dimensional images.  We may also prove this using the universal property.
A: Here's a purely symbol-pushing approach.  It's a worthwhile exercise to work out exactly what all the following isomorphisms do to individual elements of the relevant spaces.
As a rule of thumb, if you have a space of functions from $X$ to $Y$, something like $A(X,Y)$, with a definition like "the space of functions from $X$ to $Y$ satisfying such-and-such a condition", and you tensor it with a third space $Z$, you often get
$$ A(X,Y)\otimes Z \cong A(X,Y\otimes Z) $$
where the isomorphism is given by sending elementary tensors $f\otimes z$ (where $f\colon X\to Y$) to $x\mapsto f(x)\otimes z$, and extending linearly.
Warning!  Rule of thumb only!
In your situation, though, with everything being a finite-dimensional vector space, this works exactly as written.  (It's helpful to identify $V^\ast = L(V,\mathbb C)$ with $\mathbb C^B$, the space of all functions from a finite set $B$ (a basis for $V$) to $\mathbb C$.)  Thus:
$$ V^\ast\otimes W = L(V,\mathbb C)\otimes W \cong L(V,\mathbb C\otimes W) \cong L(V,W) $$
which is your first identification.
Next, the universal property of the tensor product says that
$$ L(V\otimes W,X) \cong L(V,L(W,X)) \cong B(V,W;X) $$
(where $B(V,W;X)$ denotes the space of bilinear maps from $V\times W$ to $X$).  This gives the last of your identifications, assuming "linear" was an error for "bilinear".  Furthermore,
\begin{align*}
V^\ast \otimes W^\ast
&= L(V,\mathbb C)\otimes L(W,\mathbb C)
\cong L(V,\mathbb C\otimes L(W,\mathbb C)) \\
&\cong L(V,L(W,\mathbb C))
\cong L(V\otimes W,\mathbb C)
= (V\otimes W)^\ast
\end{align*}
and so $(V\otimes W^\ast)^\ast \cong V^\ast\otimes W^{\ast\ast} = V^\ast\otimes W$, your third identification.
Your second identification $W^\ast\to V$ comes from the identification of a (finite-dimensional) vector space with its dual.  I'll leave that one to you.
