How to compute a dual of a module? Let $A=M_2(K)$ be the algebra of all $2\times 2$ matrices over $K$. Let $e_1=\left( \begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix} \right)$ and $e_2=\left( \begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix} \right)$. Then $P=Ae_1=\{ \left( \begin{matrix} a & 0 \\ b & 0 \end{matrix} \right) \mid a, b \in K\}$.
Let $D(P)=Hom_K(P, K)$. How to compute $D(P)$? I can verify that $f: P \to K, \left( \begin{matrix} a & 0 \\ b & 0 \end{matrix} \right) \mapsto a$ is in $Hom_K(P, K)$. Are there other non-zero homomorphisms in $Hom_K(P, K)$? Thank you very much.
 A: A priori, $D(P) = \operatorname{Hom}_K(P,K)$ admits the structure of a right $A$-module via
$$
 (f \cdot a)(p) := f(a \cdot p), \quad f \in D(P), \; a \in A, \; p \in P.
$$
Since $A$ admits the $K$-algebra anti-automorphism $A \mapsto A^T$, one can moreover realise $D(P)$ as a left $A$-module via
$$
 (a \cdot f)(p) := f(a^T \cdot p), \quad f \in D(P), \; a \in A, \; p \in P;
$$
I claim that ${}_A D(P) \cong {}_A P$.
To prove this, first identify ${}_A D(P)$ with ${}_A (e_1 A)$, where $e_1 A$ is endowed with the left $A$-module structure
$$
  a \cdot (e_1 b) := e_1 b a^T, \quad a,b \in A,
$$
and where the dual pairing between $e_1 A \cong D(P)$ and $P = A e_1$ is given by
$$
 \left\langle e_1 a, b e_1 \right\rangle := e_1 a b e_1 \in e_1 A e_1 = K e_1 \cong K, \quad a,b\in A.
$$
Then, the transposition $e_1 a \mapsto (e_1 a)^T = a^T e_1$ yields the necessary isomorphism ${}_A e_1 A \cong {}_A P$. 
If you prefer, you can also observe that ${}_A P \cong {}_A K^2$, and hence ${}_A D(P) \cong {}_A (K^2)^\ast \cong {}_A K^2$, where the isomorphism ${}_A (K^2)^\ast \cong {}_A K^2$ is again just transposition, taking row vectors in $(K^2)^\ast$ to column vectors in $K^2$.
A: It might be worth mentioning that $A$ has a unique simple module, up to isomorphism, and so (identifying $A$ with $A^{op}$ as in Branimir Cacic's answer), since $P$ is a simple $A$-module, so is $D(P)$, and hence they are necessarily isomorphic.  
