Dualizing simple module which is given by primitive idempotent Let $e$ be a primitive idempotent in an associative finite-dimensional $k$-algebra $A$. Then the two modules $Ae/\text{rad}(Ae)$ and $D(eA/\text{rad}(eA))$ are both simple, where $D: \text{mod}(A^{\text{op}}) \to \text{mod}(A)$ is the standard dualization.
Question: Is there an easy proof of the fact (Or is it even true) that those two simple modules are isomorphic to each other as $A$-modules?

Denote by $k$ the ground field of the algebra $A$. Since both modules are simple, every nonzero homomorphism of $A$-modules $\varphi: Ae/\text{rad}(Ae) \to D(eA/\text{rad}(eA))$ will be an isomorphism. In order to find this, we first construct some $\tilde{\varphi}: Ae \to D(eA/\text{rad}(eA))$:
Let $g \in D(eA/\text{rad}(eA))$ be an arbitrary linear map $eA/\text{rad}(eA) \to k$ such that $g(\overline{e}) \neq 0$. This exists since $e \notin \text{rad}(eA)$, i.e. $\overline{e} \neq 0$. We claim $eg \neq 0$: Indeed, we have
$$(eg)(\overline{e}) = g(\overline{e}e) = g(\overline{e}) \neq 0$$
by definition of $g$. This gives rise to a map $\tilde{\varphi}: Ae \to D(eA/\text{rad}(eA)), \ ae \mapsto aeg$, which is clearly $A$-linear. We claim $\text{rad}(Ae) \subseteq \ker(\tilde{\varphi})$: For $ae \in \text{rad}(Ae)$ we get for all $a' \in A$
$$\tilde{\varphi}(ae)\left(\overline{ea'}\right) = (aeg)\left( \overline{ea'}\right) = g\left(\overline{ea'ae} \right) = g(0) = 0$$,
where the second to last equality follows from $ea'ae \in eA\text{rad}(Ae) = e\text{rad}(A)e = \text{rad}(eA)e \subseteq \text{rad}(eA)$, i.e. $\overline{ea'ae} = 0$. Therefore, $\tilde{\varphi}(ae) = 0$ and thus $\text{rad}(Ae) \subseteq \ker(\tilde{\varphi})$. Hence we get an $A$-linear map on the quotient
$$\varphi: Ae/\text{rad}(Ae) \to D(eA/\text{rad}(eA)), \ \overline{ae} \mapsto \tilde{\varphi}(ae) = aeg.$$ 
We are done if $\varphi \neq 0$. This is in fact the case, since $\varphi(\overline{e})(\overline{e}) = (eg)(\overline{e}) \neq 0$.
Follow-up Question: Is there a more natural isomorphism?
