Show that $D \cong {\rm End}_A(D^n)$ where $D$ is a division algebra and $A\cong M_n(D)$ Define $A$ to be a finite dimensional simple algebra over a field $k$. $D$ is a $k$-division algebra (not necessarily commutative) such that $A\cong M_n(D)$ for some integer $n$. Let $L$ be a minimal left ideal of $A$. Then $L\cong D^n$. 
I want to show that $D\cong {\rm End}_A(D^n)$. Do I do this by mapping the matrix of $f\in {\rm End}_A(D^n)$ to its determinant? 
Many thanks! 
 A: There is a natural ring homomorphism $\theta:D\to End(_RV)$ given by right multiplication by elements of $D$. That is:
$\theta(d)=\text{right multiplication map by $d$ on $V$}$, 
i.e. $(\theta(d))(v)=v\cdot d$
This is immediately injective since $D$ is a division ring.  So it remains to show that it is surjective. This involves little more than some elementary linear algebra.
Let $f\in End(_RV)$ be some element, and (writing it as acting on the right of column vectors) suppose
$$\begin{bmatrix}1 \\ 0 \\ \vdots  \\ 0\end{bmatrix}f=\begin{bmatrix}d \\ \ast \\ \vdots  \\ \ast\end{bmatrix}$$
for some element $d\in D$.
Then 
$$
\begin{bmatrix}a_1 \\ a_2 \\ \vdots  \\ a_n\end{bmatrix}f=\begin{bmatrix}a_1 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots  \\ a_n & 0 & \cdots & 0\end{bmatrix}\begin{bmatrix}1 \\ 0 \\ \vdots  \\ 0\end{bmatrix}f=\begin{bmatrix}a_1 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots  \\ a_n & 0 & \cdots & 0\end{bmatrix}\begin{bmatrix}d \\ \ast \\ \vdots  \\ \ast\end{bmatrix}=\begin{bmatrix}a_1d \\ a_2d \\ \vdots  \\ a_nd\end{bmatrix}=\begin{bmatrix}a_1 \\ a_2 \\ \vdots  \\ a_n\end{bmatrix}\theta(d)
$$
This shows that $f=\theta(d)$, proving surjectivity.

This is a special case of the double centralizer theorem for Artinian rings.  You can find many versions of the proof online and in textbooks.
A: EDIT: The below is wrong if $D$ is not commutative. Not sure what I was thinking 5.5 years ago, so not sure what the thought process was/whether it's salvagable.

Note that 
$$\text{End}_D(D^n)=\text{Mat}_n(D)=A$$
Taking the center on both sides yields
$$Z(\text{End}_D(D^n))\cong D$$
But, the center of $\text{End}_D(D^n)$ is just the matrices which are $\text{Mat}_n(D)$-linear. So, by tracking this through the isomorphism $A\cong \text{Mat}_n(D)$, you see this turns into the statement $Z(\text{End}_D(D^n))\cong \text{End}_A(D^n)$.
