Endomorphisms of a semisimple module Is there an easy way to see the following: 
Given a $k$-algebra $A$, with $k$ a field, and a finite dimensional semisimple $A$-module $M$. Look at the natural map $A \to \mathrm{End}_k(M)$ that sends an $a \in A$ to 
$$
M \to M: m \mapsto a \cdot m.
$$
Then the image of $A$ is a finite-dimensional semisimple algebra. 
 A: Here's one way to look at it: Notice that the kernel of the map is exactly $ann(M)$, which necessarily contains the Jacobson radical $J(A)$ of $A$. Since $A/J(A)$ and all of its quotients are semiprimitive, it follows that $A/ann(M)$ is semiprimitive. 
Now $M$ as a faithful $A/ann(M)$ module. Since the simple submodules of $M$ remain the same during this passage, $M$ is also still semisimple over this new ring. You can see in this question why a ring with a faithful module of finite length must be Artinian. Now we have that $A/ann(M)$ is Artinian and semiprimitive: so it is semisimple.

I see I overlooked a simple way for concluding that the image is finite dimensional. Of course our image ring is a subalgebra of $End(M_k)$ which is finite dimensional... so the subring is finite dimensional as well. The argument I gave before essentially proves a more general case: "If $M$ is a semisimple $R$ module of finite length, the image of the natural map is a semisimple ring."
A: Here's another slightly different approach. Let $B$ be the image algebra, which is a finite dimensional algebra over $k.$ Note that $M$ is a $B$-module and its $B$-submodules are precisely the $A$-submodules of $M.$ Hence $M$ is also a semisimple $B$-module, and the Jacobson radical $J(B)$ annihilates $M.$ Thus $J(B) = \{0 \} $ and $B$ is semisimple.
($B$ is a genuine subalgebra of ${\rm End}_{k}(M),$ so is faithfully embedded there, and the elements of $J(B)$ are annihilating $M$, so only the zero endomorphism can be in $J(B) ).$
