Proving orthonormal basis for linear operators given
\begin{equation*}
   E_{ij} \mathbf{e}_k = \langle \mathbf{e}_j,\mathbf{e}_k \rangle \mathbf{e}_i, \quad 1 \leq k \leq n.
  \end{equation*}
prove that for each $1 \leq i,j \leq n$ let $E_{ij} \in \mathrm{End}\mathbf{V}$
be the linear operator defined by
\begin{equation*}
   E_{ij} E_{kl} = \langle \mathbf{e}_j,\mathbf{e}_k \rangle E_{il}
  \end{equation*}
Deduce from this that $\mathcal{E} = \{ E_{ij} \colon 1 \leq i , j \leq n\}$
is an orthonormal basis of $\mathrm{End} \mathbf{V}$, where by definition the scalar product of
two operators $A,B \in \mathrm{End}\mathbf{V}$ is
$\langle A,B \rangle = \operatorname{Tr} A^*B$, with $\operatorname{Tr}$ the trace and $A^*$ the
adjoint (aka transpose) of $A$. What is $\dim \mathrm{End}\mathbf{V}$?
I know how to prove the first part but I'm confused on how to use the trace to prove that $\mathcal{E}$ is an orthonormal basis and obtain $\dim \mathrm{End}\mathbf{V}$
 A: It is clear that $E_{ij}$ defined in this way can be described as the matrix with entries $0$ but on the line $i$ and column $j$ where the entry is a "one".
The orthonormality propeerty you have to establish amounts to prove that:
$$Tr(E_{ij}^TE_{kl})=\begin{cases}1 & \text{if} \ i=k \ \text{and} \ j=l \\ 0 & \text{otherwise} \end{cases}\tag{1}$$
It will be established in the following way.
Using the relationship you have established:
$$E_{ij} E_{kl} = \langle \mathbf{e}_j,\mathbf{e}_k \rangle E_{il},$$
the LHS of (1) can be written:
$$Tr(E_{ji}E_{kl})=Tr(\langle \mathbf{e}_i,\mathbf{e}_k \rangle E_{jl})=\underbrace{\langle \mathbf{e}_i,\mathbf{e}_k \rangle}_a \underbrace{Tr(E_{jl})}_b$$
This product will be $0$ in general, unless its components $a$ and $b$ are both non zero, which happens iff

*

*on one hand: $a=1 \iff e_i=e_k \iff $ $ \ \color{red}{i=k}$ and


*on the other hand: $Tr(E_{jl})=1$ which happens if the single entry equals to "one" in $E_{jl}$ belongs to its diagonal, which is possible iff $\color{red}{j=l}$.
We have therefore obtained an equivalence with the RHS of (1).
