= Solution
The <commutant> is $A'=\{T\in\operatorname{End}V:Ta=aT\text{ for every }a\in A\}$. Clearly $A\subseteq A''$. For the reverse inclusion, first note that $A\xi$ is an <invariant subspace> for the unital star-algebra $A$, and its <orthogonal complement> is invariant too. Its projection belongs to $A'$. Thus if $B\in A''$, $B\xi\in A\xi$.
To make the same algebra element work for all <vectors>, amplify to $V^d$, where $d=\dim V$. The diagonal action $a^{(d)}$ has <commutant> consisting of block <matrices> whose entries belong to $A'$. Therefore $B^{(d)}$ commutes with this amplified <commutant>. For a basis $\xi_1,\ldots,\xi_d$, the subspace $\{(a\xi_1,\ldots,a\xi_d):a\in A\}$ is invariant and reducing for the amplified algebra. Its projection lies in that <commutant>. The preceding argument gives some $a\in A$ with $(B\xi_1,\ldots,B\xi_d)=(a\xi_1,\ldots,a\xi_d)$. Since these <vectors> form a basis, $B=a$. Hence
$$
\boxed{A''=A.}
$$
This finite-dimensional <double commutant theorem for star-algebras> uses the star hypothesis to make the orbit subspaces reducing.
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