I'm trying to solve the following exercise:
Let $A$ be a subalgebra of $M_n(\mathbb{C})$ with $a^* := \bar{a}^t \ (\forall a \in A)$. We define a $C^*$-module over $A$ as an $A$-module $\mathbb{V}$ (hence also a $\mathbb{C}$ vector space) with a hermitian inner product $\langle ,\rangle$ such that $\langle av ,w\rangle = \langle v,a^*w\rangle$. Prove that any $C^*$-module over $A$ is semisimple.
I think I've come up with a partial solution, if $\mathbb{V}$ is finite dimensional, but I'm almost sure it cannot be extended to the infinite dimesional setting.
Suppose that $\dim_{\mathbb{C}}\mathbb{V} < +\infty$. I will prove that $\mathbb{V}$ is semisimple by showing that any submodule in $\mathbb{V}$ is a direct summand. Let $S \subseteq \mathbb{V}$ be a submodule. Since $S$ is in particular a subspace of $\mathbb{V}$, and we've defined an inner product on the space, we can consider
$$ S^\perp = \{v \in \mathbb{V} : \langle s, v \rangle = 0 \ (\forall s \in S)\} $$
This is indeed an $A$-submodule: if $a,b \in A$ and $v,w \in S^\perp$,
$$ \langle s, av + bw\rangle = \langle s, av\rangle + \langle s, bw\rangle = \langle a^*s, v\rangle + \langle b^*s, w\rangle = 0 \ (\forall s \in S) $$
and therefore $av+bw \in S^\perp$. Moreover, $S \cap S^\perp = 0$, since if $v \in S \cap S^\perp$ then $\langle v,v \rangle = 0$ and thus $v = 0$. This implies that the sum is, in effect, direct; so we're left to prove that $S \oplus S^\perp = \mathbb{V}$. But then again, since we are dealing with a finite dimensional vector space,
$$ \dim_{\mathbb{C}} S \oplus S^\perp = \dim_{\mathbb{C}} S + \dim_{\mathbb{C}}S^\perp = \dim_{\mathbb{C}}\mathbb{V} $$
which concludes the proof.
I'd appreciate if you could comment on my approach and give me any advice on how to tackle the general case. Thanks in advance.