Find an isomorphism between $L(V,V)$ and the space of $n\times n$ matrices over $F$ with inner product $(A|B) = tr(AB^*)$ This problem is rather difficult with me and I can't find a solution for the last point. Please help me:


*

*Let $V$ be an $n$-dimensional inner product space over the field $F$ and let $L(V,V)$ be the space of linear operators on $V$. Show that ther is a unique inner product on $L(V,V)$ with the property that $||T_{\alpha, \beta} ||^{2}= ||\alpha||^{2} ||\beta||^{2}$ , here we suppose $T_{\alpha, \beta}(x) = (x|\beta)\alpha$.


*Find an isomorphism between $L(V,V)$ with this inner product and the space of $n\times n$ matrices over $F$, with the inner product $(A|B) = tr(AB^*)$

I can solve the first point using basis for $L(V,V)$ includes $E_{p,\,q} (\alpha_{i}) = \delta_{ip}\alpha_{q}$ because we can prove that $E_{p,\,q} = T_{\alpha_{q}, \, \alpha_{p}}$, here we have ${(\alpha_{i})}_{i = 1}^{n}$ is basis of $V$. But I can't find any way to solve the second point.
 A: Note that
$$
\beta\longmapsto (\cdot,\beta)
$$
is an isometry between $V$ and the dual $V^*$.
Then observe that there is a natural isomorphism
$$
L(V,V)\simeq  V\otimes V
$$
where $\alpha\otimes \beta$ is identified with $T_{\alpha,\beta}$.
So the first question amounts to showing that there exists a unique inner-product on $V\otimes V$ such that $\| \alpha\otimes \beta \|=\| \alpha \| \| \beta\|$. This is true. And it is actually the standard way to put an inner product on the tensor product of two inner spaces. The defining formula is
$$
(\alpha\otimes \beta,\alpha'\otimes \beta')=(\alpha,\alpha')(\beta,\beta')
$$
from which we extend by linearity. By polarization, we clearly have uniqueness. So it only remains to check that the above actually defines an inner product on $V\otimes V$. The only delicate part is to verify that the claim: extend by linearity yields a well-defined operation.
For the second question, I think you meant isometric isomorphism, for otherwise this is true by dimension and regardless of the norms.
Now fix an orthonormal basis $(e_1,\ldots,e_n)$ of $V$, so that $L(V,V)$ is identified with $M_n(F)$ and so
$$
M_n(F)\simeq V\otimes V.
$$
Now for every $\alpha,\beta$, 
$$
T_{\alpha,\beta}^*(x)=\overline{(\alpha,x)}\beta=(x,\alpha)\beta
$$
so
$$
T_{\alpha,\beta}T_{\alpha,\beta}^*(x)=(x,\alpha)\|\beta\|^2\alpha.
$$
Now 
$$
\mbox{trace}T_{\alpha,\beta}T_{\alpha,\beta}^*=\sum_{i=1}^n(T_{\alpha,\beta}T_{\alpha,\beta}^*(e_i),e_i)=\|\beta\|^2\sum_{i=1}^n|(\alpha,e_i)|^2=\|\alpha\|^2\|\beta\|^2.
$$
Clearly, $(S,T)\longmapsto \mbox{trace}(ST^*)$ is an inner-product on $L(V,V)$ and the corresponding norm coincides with the previous one on elements $T_{\alpha,\beta}$, i.e. $\alpha\otimes \beta$. 
The uniqueness of question 1 shows therefore that $L(V,V)\simeq V\otimes V$ is isometric to $M_n(F)$ equipped with $\mbox{trace}(ST^*)$.
A: Hints: let $\{e_1,\ldots,e_n\}$ be an orthonormal basis of $V$.


*

*Consider the inner product $(L_A|L_B) = \sum_{j=1}^n (L_Ae_j|L_Be_j)$. Verify that $\|T_{\alpha, \beta}\|^2 = \|\alpha\|^2 \|\beta\|^2$.

*Consider the mapping that maps $L_A\in L(V,V)$ to $A\in M_n(F)$ by $L_A e_i = \sum_{j=1}^n A_{ij} e_j$ for each $i$, i.e. $A_{ij} = (L_A e_i|e_j)$. Verify that this is indeed an isomorphism and $(L_A|L_B)=(A|B)$.

