# 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 $$\dim \mathrm{End}\mathbf{V}$$

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).