# Let $V=V_{1}⊕ ⋯ ⊕ V_{n}$ be semisimple. $U$ irreducible. Show that $\dim_{k} (Hom_A(U,V))$ is equal to the number of $V_i$ equivalent to $U$.

$\DeclareMathOperator{\End}{End} \DeclareMathOperator{\Ker}{Ker} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\Mat}{Mat} \DeclareMathOperator{\Irr}{Irr}$
Definition.
An $A$-module $V$ is called semisimple if $V$ can be decomposed as a direct sum of irreducible submodules.

Definition.
Let $\Irr(A)$ denote a complete set of representatives for the equivalence classes of irreducible representations of $A$.

Proposition.
Let $V=V_{1}\oplus \cdots \oplus V_{n}$ be a finite dimensional semisimple $A$-module. For each $U\in \Irr(A)$ let
$$n_U:=\dim_{k} (\Hom_A(U,V)) \in \Bbb{Z}_{\geq 0}.$$
Show that $n_U$ is equal to the number of $V_{i}$ that are equivalent to $U$.

I’ve been staring to this for quite some time, but nothing that really comes to my mind. My book says that it has to do with Schur’s Lemma. My main problem is that I don’t really get what the dim actually means in this context I guess. I would say that $\dim_{k} (\Hom_k(U,V))$ is just $\dim_k(U)\dim_k(V)$.

Edit: After staring a little bit more to it, I think I can reduce this problem by Schur’s lemma to showing that:

$$\dim_{k}(\Hom_A(U,V))=\sum _{i}\dim(\Hom_A(U,V_{i})$$

I’m not sure how to show that, but it seems to do with this proposition:

Proposition.
Let $U,V$ be finite dimensional $k$-vector spaces with direct sum decompositions $U=\bigoplus_{j=1}^nU_{j}$ and $V=\bigoplus_{i=1}^mV_{i}$ and with corresponding complete sets of idempotents $e_{j}\in \End_{k}(U)$ and $f_{j}\in \End_{k}(V)$.

We have a $k$-linear isomorphism $M$ between $\Hom_{k}(U,V)$ and the $k$-vector space consisting of $m\times n$ matrices
\begin{align*}
\begin{pmatrix}\phi _{1,1} &\cdots &\phi _{1,n} \\
\vdots & & \vdots \\
\phi _{m,1} &\cdots &\phi _{m.n} \end{pmatrix}
\end{align*}
with $\phi _{i,j}\in \Hom_{k}(U_{j},V_{i})$. This isomorphism $M$ is defined by $\phi \mapsto (\phi _{i,j})$ where $\phi _{i,j}=f_{i}\phi |_{U_{j}}$.