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For a left module $M$ over a unital associative algebra $A$, right multiplication in $A$ gives a right module structure on ${\mathrm{Hom}}_A(M,A)$ by $(f\cdot a)(m)=f(m)a$. This remains a module homomorphism because $f(bm)a=bf(m)a$; associativity and the unit verify the action axioms. When $A=M_d(\mathbb{C})$ and $M={\mathbb{C}}^d$ is its simple module of columns, every such map is $m\mapsto mr$ for a unique row $r$, making the Hom space the simple row right module. This illustrates how left and right simple modules correspond for a semisimple algebra.

For $a\in A$, $m\in M$,Additivity and complex linearity follow from those of $f$ and multiplication in $A$, so $f\cdot x$ is an module homomorphism. In particular the action stays inside $V$. Associativity givesBoth distributive laws hold pointwise. These are exactly the axioms for a right module, proving ${\mathrm{Hom}}_A(M,A)$ is a right $A$-module.

The Jacobson radical is $J=\mathbb{C}{e}_{12}$. Indeed $J^2=0$, so $J$ is contained in the Jacobson radical, and $A/J\cong \mathbb{C}\times \mathbb{C}$ is a semisimple algebra, giving the reverse inclusion. The radical of a module of the left regular module is thereforeIts socle is the sum of its simple modules. The displayed left action shows that its $S_1$-submodules span $\mathbb{C}{e}_{11}\oplus \mathbb{C}{e}_{12}$. There is no $S_2$-submodule: if $av=tv$ for every $a$, applying $e_{11}$ forces the first row of $v$ to vanish, and applying $e_{12}$ then forces the remaining $e_{22}$ coefficient to vanish. HenceA homomorphism from $S_1$ is determined by any image $\alpha e_{11}+\beta e_{12}$ of $s_1$. Thus $V_{S_1}$ is the two-dimensional row right module with actionIts line $\mathbb{C}{e}_{12}$ is a nonzero proper submodule, so $V_{S_1}$ is not simple. Since $S_2$ has no copy in the socle, any homomorphism $S_2\to A$ is zero. Thus $V_{S_2}=0$ is not simple either, since a simple module is required to be nonzero.