Constructing an explicit isomorphism $\widehat{\Phi}^{-1} : \operatorname{End}(V) \to V \otimes V^*$ Let $V$ be a finite-dimensional real vector space. Then it is well known, that 
$$\boxed{V \otimes V^* \cong \operatorname{End}(V)}$$ This is seen by considering the mapping
$$\Phi:\begin{cases} V \times V^* \to \operatorname{End}(V)\\
(v,f) \mapsto (u \mapsto f(u)v)\end{cases}$$ which descends to a linear mapping $\widehat{\Phi}: V \otimes V^* \to \operatorname{End}(V)$ which is actually an isomorphism. My question is know, how does $\widehat{\Phi}^{-1}$ look? Explicitly, given $g \in \operatorname{End}(V)$, what is the corresponding tensor?
 A: Let $f:V\to V$ be a linear map.
Let $(v_1,\dots,v_n)$ be a basis of $V$ and let $(\phi_1,\dots,\phi_n)$ be the corresponding dual basis of $V^*$. Consider the element $u=\sum_{i=1}^nf(v_i)\otimes\phi_i$: its image under your map $\Phi$ is equal to $f$. Indeed, for all $v\in V$ we have that $$\Phi(u)(v) = \sum_{i=1}^n\Phi(f(v_i)\otimes\phi_i)(v)=\sum_{i=1}^n\phi_i(v)f(v_i)=f\left(\sum_{i=1}^n\phi_i(v)v_i\right)=f(v).$$
A: It's easy to see what happens to rank-1 operators.  In particular: if $g(u) = f(u)v$ for some $v,f \in V \times V^*$, then $\Phi^{-1}(g) = v \otimes f$.
From there, it suffices to note that every operator can be written as a linear combination of rank-1 operators.  In terms of matrices: suppose that $V = \Bbb F^n$ $g$ is the $n \times n$ matrix $A$ with entries $a_{ij}$.  Then
$$
A = \sum_{i,j = 1}^n a_{ij} e_ie_j^T
$$
which is to say that $g(u) = \sum_{i,j = 1}^n a_{ij} e_i\langle u,e_j \rangle$.  Correspondingly, 
$$
\Phi^{-1}(g) = \sum_{i,j = 1}^n a_{ij} \left[e_i \otimes \langle \cdot ,e_j \rangle\right]
$$
A: In general, this is not a single tensor $v\otimes f$, but a sum of these.
To get a particular inverse for $\Phi$, one needs to fix a basis $e_1,\dots,e_n$ in $V$. The dual basis consists of $e_i^*$'s which map all $e_j$ to $0$ except for $e_i$. 
Then, $V\otimes V^*$ is spanned by $e_i\otimes e_j^*$ which corresponds to the basic matrix with all $0$'s but one $1$ at $(i,j)$.
Finally, an arbitrary $g\in {\rm End}(V)$ will be mapped to $\displaystyle\sum_{i,j}g_{i,j}\cdot e_i\otimes e_j^* $ where $(g_{i,j})_{i,j}$ is the matrix of $g$ in the given basis.
A: Not every element $\psi \in V \otimes V^*$ has the form $v \otimes f$ (such a tensor, by the way, is called simple). That statement is true only for those for which $\Phi(\psi)$ has rank $\leq 1$. A general element of $V \otimes V^*$ has the form
$$\sum_{i = 1}^m v_i \otimes f_i.$$
A basis $(E_a)$ of $V$ determines a dual basis $(e^a)$ of $V^*$, and this in turn gives a basis $(E_a \otimes e^b)$ of $V \otimes V^*$. Then, we can write $\smash{\widehat\Psi^{-1}}$ as
$$\widehat\Psi^{-1}(g) = \sum_{a, b} e^a(g(E_b)) E_a \otimes e^b .$$
Note that while our formula uses a basis, you have already shown that the isomorphism $\Psi^{-1} : \textrm{End}(V) \to V \otimes V^*$ is canonical, i.e., independent of the choice of basis.
