$*$-homomorphism between matrix algebras Let $\theta:M_3(\mathbb{C}) \to M_7(\mathbb{C})$ be a $*$-homomorphism. Since matrix algebras over a field are simple, we know that $\theta$ must be injective. After reading this I conclude that, up to unitary equivalence,
$$
\theta(A) = \left[\begin{array}{}
         A & 0_3 & 0 \\
         0_3 & A & 0 \\
         0 & 0 & 0 \\
  \end{array}\right]
$$
where $0_3$ denotes the $3 \times 3$ zero matrix. My question is the following. We have that $\theta':M_3(\mathbb{C}) \to M_7(\mathbb{C})$ defined by
$$
\theta'(A) = \left[\begin{array}{}
         A & 0_3 & 0 \\
         0_3 & 0_3 & 0 \\
         0 & 0 & 0 \\
  \end{array}\right]
$$
is also a $*$-homomorphism. So, how come $\theta(A)$ is unitary equivalent to $\theta'(A)$? In general, why is every possible $*$-homomorphism unitary equivalent to $\theta$? That is, how do you determine the unitary matrix $U \in M_7(\mathbb{C})$ such that $\theta(A)=U^*\theta'(A)U$?
 A: Complete rewrite
Let $\theta : M_m(\mathbb{C}) \to M_n(\mathbb{C})$ be a non-zero $\ast$-homomorphism, let $P = \theta(1_m)$, and let $V = P\mathbb{C}^n$. Then $P$ is the unit of $\theta(M_m(\mathbb{C}))$ as a $C^\ast$-algebra, and in particular, a central orthogonal projection in $\theta(M_m(\mathbb{C}) \subseteq M_n(\mathbb{C})$, so that for any $A \in M_m(\mathbb{C})$, $\theta(A) = P\theta(A)P$, and hence
$$
 M_m(\mathbb{C}) \to B(V), \quad  A \mapsto \theta(A)|_V,
$$
is a unital $\ast$-homomorphism. Thus, if $U_0$ is the change of coordinates matrix from the standard orthonormal basis on $\mathbb{C}^n$ to the extension of an orthonormal basis on $V$ to an orthonormal basis on $\mathbb{C}^n$, then
$$
 U_0 \theta(A) U_0^\ast = \theta_0(A) \oplus 0_{n-s}
$$
for a unital $\ast$-homomorphism $\theta_0 : M_m(\mathbb{C}) \to M_s(\mathbb{C})$, where
$$
 s := \dim V = \operatorname{rank}\theta(1_m).
$$
Now, $\theta_0$ defines a unital $\ast$-representation of $M_m(\mathbb{C})$ on $\mathbb{C}^s$. However, $M_m(\mathbb{C})$ is simple with unique (up to unitary equivalence) unital $\ast$-representation given by left matrix multiplication on $\mathbb{C}^m$. Hence, abstractly, there exists a non-zero integer $k$ and a unitary $S : \mathbb{C}^m \to (\mathbb{C}^m)^{\oplus k}$ such that
$$
 S \theta_0(A) S^\ast = A^{\oplus k};
$$
in particular, $s = km$, so that since $s = \operatorname{rank} \theta(1_m) \leq n$, $1 \leq k \leq q$, where $n = qm+r$ for the quotient $q$ and remainder $r$. In other words, by this unitary isomorphism, there exists an orthonormal basis of $\mathbb{C}^s = \mathbb{C}^{mk}$ such that if $U_1$ is the change of coordinates matrix from the standard orthonormal basis of $\mathbb{C}^{mk}$ to this new basis, then
$$
 U_1 \theta_0(A) U_1^\ast = A^{\oplus k}.
$$
Putting everything together, if $U := (U_1 \oplus 1_{n-km}) U_0$, then $U \in M_n(\mathbb{C})$ is a unitary such that
$$
 U \theta(A) U^\ast = (A^{\oplus k}) \oplus 0_{n-km}, \; \forall A \in M_m(\mathbb{C}).
$$
Note that by construction,
$$
 k = \frac{\operatorname{rank}\theta(1_m)}{m}
$$
is a complete unitary equivalence invariant of $\ast$-homomorphisms $\theta : M_m(\mathbb{C}) \to M_n(\mathbb{C})$. Indeed, this is precisely the label on the single edge of the Bratteli diagram of such a $\ast$-homomorphism between matrix algebras. Explicitly, let $\theta_1, \theta_2 : M_m(\mathbb{C}) \to M_n(\mathbb{C})$ be $\ast$-homomorphisms, and for $i=1,2$, let
$$
 k_i = \frac{\operatorname{rank}\theta_i(1_m)}{m}
$$
and let $U_i \in M_n(\mathbb{C})$ be the unitary such that
$$
 U_i \theta_i(A) U_i^\ast = A^{\oplus k_i} \oplus 0_{n - k_i m}.
$$
If $k_1 = k_2 =: k$, then
$$
 U_1 \theta_1(A) U_1^\ast = A^{\oplus k} \oplus 0_{n-km} = U_2 \theta_2(A) U_2^\ast,
$$
so that
$$
 \theta_2(A) = U \theta_1(A) U^\ast, \; \forall A \in M_m(\mathbb{C}), \quad U := U_2^\ast U_1 \in U(n),
$$
as required. If, instead, $k_1 \neq k_2$, then $\operatorname{rank}\theta_1(1_m) \neq \operatorname{rank}\theta_2(1_m)$, and hence, as Martin pointed out in his comment, $\theta_1(1_m)$ and $\theta_2(1_m)$ cannot possibly be similar, ruling out unitary equivalence of $\theta_1$ and $\theta_2$.
Finally, in the case that you are considering of $\ast$-homomorphisms $\theta : M_3(\mathbb{C}) \to M_7(\mathbb{C})$, since $7 = 2 \cdot 3 + 1$, the unitary equivalence classes of non-trivial such $\ast$-homomorphisms are given by:


*

*the case where $k=1$, so that there exists some unitary $U \in M_7(\mathbb{C})$ such that
$$
 U \theta(A) U^\ast = \begin{pmatrix} A & 0_{3 \times 4} \\ 0_{4 \times 3} & 0_{4}\end{pmatrix};
$$

*the case where $k=2$, so that there exists some unitary $U \in M_7(\mathbb{C})$ such that
$$
 U \theta(A) U^\ast = \begin{pmatrix} A & 0_3 & 0_{3 \times 1} \\ 0_3 & A & 0_{3 \times 1} \\ 0_{1 \times 3} & 0_{1 \times 3} & 0_1 \end{pmatrix}.
$$


In either case, the matrix $U$ is determined, as outlined above, by $P := \theta(1_m)$, and specifically by the unitary $S : P\mathbb{C}^n \to (\mathbb{C}^m)^{\oplus k}$ realising the induced unital $\ast$-representation of $M_m(\mathbb{C})$ on $P\mathbb{C}^n$ as a direct sum of irreducible representations, each of which is necessarily unitarily equivalent to the standard representation on $\mathbb{C}^m$.
