finite dimensional representation of $C^*$ representation According to the standard definition of representation of a $C^*$ algebra $A$ ,we need to construct a $*$ homomorphism $\pi$ from $A$ to $B(H)$,where $H$ is a Hilbert space.My question is : if we have a $*$ homomorphism $\phi$ from $A$ to $B$,where B is a finite dimensional $C^*$ algebra.Can we call $\phi$ the finite dimensional representation of $A$?
 A: The answer is absolutely yes. Suppose that $\phi : A \to B$ is a $\ast$-homomorphism, where $B$ is finite-dimensional. By the structure theorem for finite-dimensional $C^\ast$-algebras (i.e., the appropriate refinement of Wedderburn's theorem), the $C^\ast$-algebra $B$ necessarily takes the form
$$
 B \cong M_{n_1}(\mathbb{C}) \oplus \cdots \oplus M_{n_k}(\mathbb{C}) \subseteq M_{n_1 + \cdots + n_k}(\mathbb{C})
$$
for some positive integers $n_1,\dotsc,n_k$. Now, let $\operatorname{tr}$ be the normalised matrix trace on $M_{n_1 + \cdots + n_k}(\mathbb{C})$, which is, in fact, a faithful trace, satisfying $\operatorname{Tr}(x^\ast x) \geq 0$ for any matrix $x$, with
$$
 \forall x \in M_{n_1 + \cdots + n_k}(\mathbb{C}), \quad \operatorname{tr}(x^\ast x) = 0 \iff x = 0.
$$
We can therefore restrict $\operatorname{tr}$ to a faithful trace on $B$, and use it to construct a positive-definite inner product on the finite-dimensional complex vector space $B$, by
$$
 \forall b_1, b_2 \in B, \quad \langle b_1,b_2 \rangle := \operatorname{tr}(b_1^\ast b_2).
$$
Since $\operatorname{tr}$ is faithful and $B$ is finite-dimensional,
$$
 L^2(B,\operatorname{tr}) := (B,\langle \cdot,\cdot \rangle)
$$
becomes a faithful finite-dimensional $\ast$-representation of $B$ by left multiplication—indeed, it's precisely the GNS representation of $B$ induced by the faithful tracial state $\operatorname{tr}\vert_{B}$. As a result, the $\ast$-homomorphism $\phi : A \to B \subseteq B(L^2(B,\operatorname{tr}))$ really does definite a finite-dimensional $\ast$-representation on $L^2(B,\operatorname{tr})$.
A: The answer is yes, because if $B$ is finite-dimensional, then it admits a representation $B\subset M_n(\mathbb C)$. So via this representation for $B$, we may see $\phi:A\to M_n(\mathbb C)=B(\mathbb C^n)$. 
