*-homomorphisms $M_m(\mathbb{C})\rightarrow M_n(\mathbb{C})$ I've heard that every *-homomorphism $\phi:M_m(\mathbb{C})\rightarrow M_n(\mathbb{C})$ is unitarily equivalent to some *-homomorphism of the form
$A\in M_m(\mathbb{C})\mapsto\left(\oplus_{k}A\right)\oplus0_{n-km}\in M_n(\mathbb{C})$ for some $k$, that is, the matrix $A$ is copied $k$ times in the diagonal as below:
$$A\in M_m(\mathbb{C})\mapsto\left[\begin{array}{}
A\\
&\ddots\\
&&A\\
&&&0_{n-km}
\end{array}\right]\in M_n(\mathbb{C})$$
Do someone have a (maybe not so) simple proof of this fact?
 A: If $m<n$, then no homomorphism exists. That can be seen from the fact that $\phi$ is necessarily injective (details below) so it cannot decrease dimension. So we may assume $m\geq n$. Since $\phi(I_m)$ is an identity in its image, it is a projection in $M_n(\mathbb C)$. This shows us that $M_n(\mathbb C)=\phi(M_m(\mathbb C))\oplus 0_{n-km}$; the value of $k$ will be explained below. But the point is that we may assume that $\phi$ is unital by restricting its codomain.
Let $\{e_{ij}\}$ be the matrix units in $M_m(\mathbb C)$.
First note that if $\phi$ is nonzero, then $\phi(e_{ij})\ne0$ for all $i,j$. This follows from the fact that $M_m(\mathbb C)$ is simple; but to see more explicitly that it is the case, consider for example that $\phi(e_{12})=0$. Then
$$
\phi(e_{11})=\phi(e_{12}e_{21})=\phi(e_{12})\phi(e_{21})=0.
$$
Then $\phi(e_{1j})=\phi(e_{11}e_{1j})=\phi(e_{11})\phi(e_{1j})=0$ for any $j$. And then
$$
\phi(e_{ij})=\phi(e_{i1}e_{1j})=\phi(e_{i1})\phi(e_{1j})=0.
$$
So all matrix units would map to zero, and so $\phi$ would be zero.
This can be done for any other choice than 1,2 with minimal notation changes.
So $\phi(e_{11}),\ldots,\phi(e_{mm})$ are pairwise orthogonal projections that add to the identity, with equal trace. Indeed,
\begin{align}
\operatorname{tr}(\phi(e_{11}))&=\operatorname{tr}(\phi(e_{1j}e_{j1}))=\operatorname{tr}(\phi(e_{1j})\phi(e_{j1}))=\mbox{tr}(\phi(e_{j1})\phi(e_{1j}))\\[0.3cm]
&=\mbox{tr}(\phi(e_{j1}e_{1j}))=\mbox{tr}(\phi(e_{jj}).
\end{align}
Let $k$ denote this trace, that is $k=\operatorname{tr}(\phi(e_{11}))$; this is the "multiplicity". Note that the argument above shows, by taking the trace of the sum, that $n\geq km$. In particular this implies that no nonzero homomorphism exists if $m>n$.
Now construct matrix units (i.e. choose an appropriate orthonormal basis) $\{f_{gh}\}$ for $M_n(\mathbb C)$ such that
$$
\phi(e_{jj})=\sum_{l=1}^kf_{k(j-1)+l,k(j-1)+l},\ \ \ \ j=1,\ldots,m
$$
If we reorder the basis (which in terms of matrices means conjugating by a unitary) according to the permutation
\begin{align}
1,2,3,\ldots,km\mapsto &1,k+1,\ldots,k(m-1)+1,\\[0.3cm]
& 2,k+2,\ldots,k(m-1)+2,\\[0.3cm]
&\ldots,\\[0.3cm]
&k,k+k,\ldots,k(m-1)+k,
\end{align}
we get the desired representation (what is happening with the reordering is that each projection $\phi(e_{jj})$ contains $k$ copies of each matrix unit; then we take all the "first" matrix unit in each $\phi(e_{jj})$ to get the first copy of $A$; then the second, etc.). That is, in this bases the map $\phi$ is precisely the map described in the question.
