Finding $n\dim B$ linearly independent invertible matrices in $M(n,B)$? I'm trying to solve the following exercise:

If $A$ is a finite-dimensional $k$-algebra, then a minimal representation of $A$ is a representation of minimal $k$-dimension. Show that if $B$ is a finite-dimensional division algebra over $k$ and $A=M(n,B)$, then the action of $A$ on $B^n$ is a minimal representation.

To show this, we need to show that if we have any representation $A\to\text{End}(V)$ for a $k$-space $V$, then $\dim V\ge n\dim B$. I will write this map as $E\mapsto\varphi_E$ for $E\in A$. Now, if I suppose that $A$ has $nm$ linearly independent invertible matrices $E_1,\dots,E_{nm}$, where $m=\dim B$ then I will be done. By taking some nonzero $v\in V$, I let $v_i=\varphi_{E_i}(v)$ then this is nonzero since $\varphi_{E_i}$ is invertible (since $E_i$ is), and I can show that $\{v_1,\dots,v_{nm}\}$ is linearly independent using the linear independence of $E_1,\dots,E_{nm}$ in $A$.
So my problem just lies now in finding $nm$ linearly independent, invertible elements of $A$ (the latter of which is the same as having $\det E\neq0$ since $B$ is a division algebra). Does anybody have a suggestion on how to obtain these?
 A: Your strategy here is not correct.  To conclude that $\{v_1,\dots,v_{nm}\}$ are linearly independent you need to know not just that the $E_{ij}$ are invertible and linearly independent but that every nonzero linear combination of them is invertible.  This amounts to finding a division algebra of dimension $mn$ in $A$, which is not always possible (consider the case $k=B=\mathbb{R}$ and $n=3$, for instance).
Here is what I would suggest instead.  Note that $B$ embeds in $A$ as the multiples of the identity matrix, so any representation of $A$ is a $B$-module.  Now note that if $e_{ij}$ is the matrix with $ij$ entry $1$ and all other entries $0$, then if $M$ is any $A$-module, then as a $B$-module $M$ splits as the direct sum of the $B$-submodules $e_{ii}M$ for $i=1,\dots,n$ (since the $e_{ii}$ form a family of orthogonal idempotents in $A$, which commute with the elements of $B$).  Moreover, these $B$-modules $e_{ii}M$ are all isomorphic , since multiplication by $e_{ij}$ gives isomorphisms between them.  So this means that as a $B$-module, $M$ is a direct sum of $n$ isomorphic submodules.  So if $M$ is nontrivial, the dimension of $M$ over $B$ must be at least $n$.
