TITLE: *-homomorphisms between matrix algebras
QUESTION [8 upvotes]: Edited question:
Are there any other non-trivial *-homomorphisms between matrix algebras apart from the unitary homomorphisms? 
Original question:
Does there exist a surjective (but not bijective) *-homomorphism between matrix algebras over the complex numbers? If so, are there any nice examples?
(I had not realized the matrix algebras were simple but since they are, the answer to the original question is indeed obvious)

REPLY [2 votes]: I do understand this is an old question but, considering that:
1. you might still be interested in a simpler (and surely less elegant) proof;
2. I had the same problem, so this might be helpful for others in the future
I'll tell you how I solved this problem
Take your *-homomorphism $\lambda:M_n\to B$, where $B$ is any other C*-algebra. Set
$$F_{ij}:=\lambda(E_{ij}),\qquad\forall i,j=1,\ldots,n,$$
where the $E_{ij}$ is the canonical basis of the underlying vector space of $M_n$ (then the $F_{ij}$ satisfy the very same algebra, i.e. $F_{ij}F_{mn}=\delta_{jm}F_{in}$), and suppose that the kernel of $\lambda$ is non-trivial. Then there exists $a\in M_n\smallsetminus \{0\}$ s.t. $\lambda(a)=0$. This is a constraint between all the $F_{ij}$, namely there are coefficients $\alpha_{ij}$ such that
$$\sum \alpha_{ij}F_{ij}=0.$$
Note that if just one of the $F_{ij}$ for some $(i,j)$ is 0, then $\lambda$ is 0, because $F_{kk}$ are all Murray-von Neumann equivalent projections, and the "off-diagonal" elements $F_{ij}$, $i\neq j$ are partial isometries linking them. Therefore sandwich the above linear combination between $F_{kk}$ and $F_{mm}$ to obtain
$$0=\sum_{ij}\alpha_{ij}F_{kk}F_{ij}F_{mm} = \alpha_{km}F_{km},$$
which implies
$$\alpha_{km}=0$$
for arbitrary $(k,m)$. Hence all the $F_{ij}$ are linearly independent, meaning that, as a vector space, $B$ must have enough space to accommodate at least a copy of $M_n$. But taking into account once again that all the projections $F_{kk}$ are Murray-von Neumann equivalent, we can conclude that there must be a subalgebra $C$ of $B$ such that $M_n\otimes C\subset B$ and a projection $P\in C$ such that, up to unitary equivalence
$$\lambda(E_{ij})=E_{ij}\otimes P\in M_n\otimes C.$$
If the rank of $P$ is $k$, then $B$ should be large enough to accommodate $k$ copies of $M_n$, i.e. the underlying vector space of B must have dimension greater than $kn$ (if it is not exactly $kn$ the you'll have some 0-padding).
Hope this helped like it did for me!