Endomorphisms of $V$ and the dual space I was told that $V\otimes V^{*}\simeq\mbox{End}\left(V\right)$. I can't find the isomorphism itself though. Can anyone tell me what it is with a proof?
Thanks!
 A: We'll assume that $V$ is finite-dimensional. Let's think about what elements of $V \otimes V^*$ are, to understand how these can be thought of as elements in $\operatorname{End}(V)$. If you give me a gadget like $v \otimes f$, I need to be able to take in an arbitrary $w \in V$ and give you something else. There's really only one thing I can do in this situation: I can apply $f$ to $w$, and scale $v$ by that amount:
$$
v \otimes f: w \mapsto f(w)*v.
$$
You can instantly see that any such endomorphism has rank 1. To get endomorphisms with larger rank, we'll have to move away from pure tensors. Let's pick a basis $e_1, \dots, e_n$ that comes with a dual basis $e_1^*, \dots, e_n^*$. When we apply $A \in \operatorname{End}(V)$ to a basis vector $e_i$, we get the $i^{th}$ column of $A$, $Ae_i$. This is represented by the element $e_i^*\otimes Ae_i$. So we can write $A$ in this form as
$$
A \mapsto \sum_{i=1}^n Ae_i\otimes e_i^* 
$$
We can go one step further if we write $Ae_j$ in terms of the basis we choose. We have $Ae_i = \sum_{j =1}^n a_{ij}e_j$, so that in the basis $\{e_i^* \otimes e_j\}$ of $V^*\otimes V$ that we have constructed, we see that the coefficient on $e_i^* \otimes e_j$ of $A$ is just $a_{ij}$:
$$
A \mapsto \sum_{i,j}a_{ij} e_j \otimes e_i^*
$$
At this point, the fact that we have an isomorphism of vector spaces is very easy to show: we just observe that the above correspondence is a linear bijection. 
