= Solution
By the <Artin–Wedderburn theorem>, a finite-dimensional complex <simple algebra> is $M_d(\mathbb C)$, with a unique <simple module> up to isomorphism, namely the column <module> $S=\mathbb C^d$. Its <left regular module> is $S^{\oplus d}$. The <Schur lemma> gives $\operatorname{End}_A(S)=\mathbb C$, so
$$
\boxed{\dim_{\mathbb C}V_M=d=\dim_{\mathbb C}M=\sqrt{\dim_{\mathbb C}A}.}
$$
The right action can be seen explicitly. For a row vector $r$, the map $f_r(m)=mr$ is $A$-linear, and all such maps are obtained this way, since the row space and $V_M$ both have <dimension> $d$. Moreover $f_r\cdot x=f_{rx}$. Any nonzero row generates the whole row space under right multiplication by matrices: if $r_j\ne0$, suitable <matrix units> send it to each coordinate row. Thus \b[$V_M$ is a simple right $A$-<module>].
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