= Hom into the regular algebra as a right module
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{title2=$\operatorname{Hom}_A(M,A)_A$}
For a left <module> $M$ over a unital <associative algebra> $A$, right multiplication in $A$ gives a <right module> structure on $\operatorname{Hom}_A(M,A)$ 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>.
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