Let $\mathbb{M}_n(\mathbb{C})$ be the $n\times n$ matrix, Let $\{e_{i,j}\}_{1<i,j<n}$ be a system of matrix units for $\mathbb{M}_n(\mathbb{C})$. the matrix $\sum_{i,j=1}^{n}e_{j,i}\otimes e_{j,i}$ is equal to $nP$ where $P$ is the one-dimensional projection on the span of the vector $\nu=\sum_{k=1}^{n} \delta _{k}\otimes \delta _{k}$.
my question: how to show the matrix $\sum_{i,j=1}^{n}e_{j,i}\otimes e_{j,i}$ is equal to $nP$ where $P$ is the one-dimensional projection on the span of the vector $\nu=\sum_{k=1}^{n} \delta _{k}\otimes \delta _{k}$.
Let $X=\sum_{k,j} e_{kj}\otimes e_{kj}$. We have \begin{align} X^2&=\sum_{k,j}\sum_{s,t}(e_{kj}\otimes e_{kj})(e_{st}\otimes e_{st}) =\sum_{k,j}\sum_{s,t} e_{kj}e_{st}\otimes e_{kj}e_{st}\\ \ \\ &=\sum_j\sum_{k,t}e_{kt}\otimes e_{kt} =\sum_j X=nX. \end{align} It follows that if $P=X/n$, then $P^2=P$. As $P^*=P$, we get that $P$ is a projection and $X=nP$.
If we denote by $\{e_k\}$ the canonical basis (I assume those are the $\delta_k$ in the question),
$$ X=\sum_{k,j}e_ke_j^*\otimes e_ke_j^*=\sum_{k,j} (e_k\otimes e_k)(e_j\otimes e_j)^* =\left(\sum_k e_k\otimes e_k\right)\left(\sum_k e_k\otimes e_k\right)^* $$ Note that $$ \left\|\sum_k e_k\otimes e_k\right\|^2=\sum_{k,j}(e_k^*e_j)( e_k^*e_j)=\sum_k1=n. $$ So $$P=xx^*,\ \ \ \ \ \ \text{ where } x=\frac1{\sqrt n}\,\sum_k e_k\otimes e_k.$$