I'm trying to prove that, if $A$ is an Algebra over a field $F$, and $U,V$ are $A-$modules. Then any element in $Hom_{M_n(A)}(U^n, V^n)$ (recall that $U^n, V^n$ is the set of size $n$ vectors as an $M_n(A)$-module) takes the form $$\tilde{f}(u_1,...,u_n) = (f(u_1),...,f(u_n)),$$ for $(u_1,...,u_n)\in V^n$ and some $f\in Hom_{A}(U,V)$.
My thoughts: Is it true (by straightforward calculations) that $\tilde{f}\in Hom_{M_n(A)}(U^n, V^n)$ (defined as above) for every $f\in Hom_A(U,V)$, but how to prove the converse?
Every element $\varphi\in Hom_{M_n(A)}(U^n, V^n)$ must satisfy that, if $B\in M_n(A)$, then $$\varphi(Bu) = B\varphi(u),$$ for every $u\in U^n$, equivalently $\varphi\circ B = B \circ \varphi$. Making an analog with matrices on vector spaces, it is true that if a transformation $T$, $TS = ST$ for every $S$, then $T$ is a multiple of identity. Any hint to proceed?