Isomorphism between End$(V)\otimes A$ and End$_A(V\otimes A)$. Let $K$ be a field and let $V,A$ be a finite dimensional vector space and algebra respectivly. I was told that there is a canonically isomorphism between  End$(V)\otimes A$ and End$_A(V\otimes A)$ where $A$ is considerd as a vector space in $V\otimes A$. I am stuck since I have truble even understand the structure of End$_A(V\otimes A)$ (unlike the clear notion End$_K(V\otimes A)$). Hope to get some help.
 A: I think that the most natural way to build an isomorphism is to take the following bilinear map $\operatorname{End}(V) \times A \longrightarrow \operatorname{End}_A(V \otimes A)$
$$
(\varphi, x) \mapsto \left(
\begin{matrix}
V \otimes A & \longrightarrow & V \otimes A \\
(v, y) & \longmapsto & \varphi(v) \otimes xy
\end{matrix}
\right)
$$
and consider the induced morphism $\operatorname{End}(V) \otimes A \longrightarrow \operatorname{End}_A(V \otimes A)$.
A: Most important fact to realize: $K^n\otimes_K A \cong A^n$. To see this, consider
\begin{align*}
 K^n\times A &\longrightarrow A^n \\
 ((v_1,\ldots,v_n),a) &\longmapsto (v_1a,\ldots,v_na)
\end{align*}
This is a $K$-bilinear map and therefore induces a linear map $\psi:K^n\otimes_K A \to A^n$. Let $e_i\in K^n$ be the $i$-th standard basis vector. If $B\subseteq A$ is a $K$-basis of $A$, then $B^n$ is a basis of $A^n$. On the other hand, the tensors $e_i\otimes b$ for $b\in B$ are a basis of $K^n\otimes A$. Clearly, $\psi$ maps this basis bijectively onto the basis $B^n$ of $A$, so $\psi$ is an isomorphism.
Now, examples help. Let $K=K$, $V=K^n$ and let's say $A=K[X]$ the polynomial ring in one variable over $K$. Then, 
\begin{align*}V\otimes_K A&= K^n \otimes_K K[X] = (K\oplus\cdots\oplus K)\otimes K[X]
 = (K\otimes K[X])\oplus\cdots\oplus(K\otimes K[X]) 
\\ &= K[X]\oplus\cdots\oplus K[X] = K[X]^n
\end{align*}
Now, the $K[X]$-endomorphisms of $K[X]^n$ are just what you think they are, they are $n\times n$ matrices with polynomials as entries. 
And this is quite generally the case. Since $V\cong K^n$ for some $n$, you will always get an induced isomorphism $V\otimes_K A\cong K^n\otimes_K A\cong A^n$. With this isomorphism, $\operatorname{End}_A(A^n)\cong A^{n\times n}$. 
But similarly, you have $\operatorname{End}_K(V)\cong K^{n\times n}$ and therefore, $\operatorname{End}_K(V)\otimes_K A \cong A^{n\times n}$, so the two objects are isomorphic as $K$-vector spaces.
Now, you claim that there is a canonical isomorphism $\operatorname{End}(V)\otimes_K A\cong \operatorname{End}_A(V\otimes_K A)$. This usually means that there is an isomorphism that can be expressed without choosing bases. It also suggests that the way to write it should be somewhat straightforward. I suggest
\begin{align*}
\phi:\operatorname{End}(V)\otimes_K A &\longrightarrow \operatorname{End}_A(V\otimes_K A) \\
\varphi\otimes a &\longmapsto \varphi_a,
\end{align*}
where
\begin{align*}
\varphi_a: V\otimes_K A &\longrightarrow V\otimes_K A\\
 v\otimes b &\longmapsto \varphi(v)\otimes ab.
\end{align*}
We now have to prove that this is actually an isomorphism. However, we can prove this by choosing bases, as long as we did not use any in the definition. You should be able to show that $\phi$ maps a basis bijectively onto a basis. I am currently too lazy to do it, but if you need more help leave a comment.
