Dual and simple modules

Question

Let $A$ be a $k$ finite dimensional algebra and let $M$ be a simple finite dimensional right $A$-module. Why is the dual of $M$, i.e $\operatorname{Hom}_{k}(M,k)$ a simple left $A$-module?

Answer 1

Let $M$ be any finite dimensional $A$-module.

Suppose $N$ is a submodule of $\hom_k(M,k)$. Then $N^\perp=\{m\in M:\forall\phi\in N,\phi(m)=0\}$ is a submodule of $M$, and this establishes a bijection $$N\in\operatorname{Sub(hom_k(M,k))}\mapsto N^\perp\in\operatorname{Sub} M$$ between the set of submodules of $M$ and those of $\hom_k(M,k)$.

Now, a module is simple iff it has exactly two submodules, so this bijection tells us that the dual of a finite dimensional simple module is simple.