The tensor product of $M_{n}(\mathbb{C})$

Question

There is a quotation below:

Let $\{e_{i,j}\}_{1\leq i, j\leq n}$ be a system of matrix units fro $M_{n}(\mathbb{C})$ and consider $$\sum\limits_{i,j=1}^{n}e_{j, i}\otimes e_{j, i}.$$

A straightforward computation shows that this matrix is equal to $nP$ where $P$ is the one-dimensional projection onto the span of the vector $$v=\sum\limits_{k=1}^{n}\delta_{k} \otimes \delta_{k}.$$

The $\{\delta_{k}\}$ is an orthonormal basis of $\mathbb{C}^{n}.$

My question is: how to comprehend the "equal to" here? And how to verify it?

Answer 1

Let $V:=\Bbb C^n$, this is the canonical $n$ dimensional vector space over $\Bbb C$, and I guess, $(\delta_k)_k$ is the canonical (orthonormal) basis.

Now $M_n(\Bbb C)$ can be identified with the space ${\rm End}(V)$ of linear transformations $V\to V$.
Here $e_{ji}$ corresponds to the linear map which sends $\delta_i$ to $\delta_j$ and all other $\delta_k$ to $0$ (for $k\ne i$).

Finally, we have a canonical linear embedding ${\rm End}(V)\otimes{\rm End}(V)\ \hookrightarrow\ {\rm End}(V\otimes V)$ which maps $$\alpha\otimes\beta\ \mapsto\ \big((v\otimes w)\mapsto (\alpha v\otimes\beta w)\big)$$ and extends linearly to the formal sums of these tensors, where $\alpha,\beta\in{\rm End}(V)$ and $v,w\in V$.

Can you then verify that $\sum_{i,j}\,e_{ji}\otimes e_{ji}$ fixes $\nu=\sum_k\,\delta_k\otimes\delta_k$ and maps each element of $V\otimes V$ to $0$ which is orthogonal to $\nu$?