= Solution
Let $V$ be a finite-dimensional real or complex <inner product space>, and let $A\subseteq\operatorname{End}(V)$ be a unital <subalgebra> stable under the <adjoint operator>. Define its <commutant of an operator algebra> by $A'=\{T:Ta=aT\text{ for all }a\in A\}$. The <double commutant theorem for star-algebras> asserts $\boxed{A''=A}$. The inclusion $A\subseteq A''$ follows immediately from the definition; we prove the reverse.
Write $N=\dim V$, fix a <basis> $v_1,\ldots,v_N$, and act diagonally on $V^N$. The <linear subspace>
$$
W=\{(av_1,\ldots,av_N):a\in A\}
$$
is invariant under every diagonal $a$. Since $A$ is closed under adjoints, its <orthogonal complement> $W^\perp$ is invariant too: $\langle aw,z\rangle=\langle w,a^*z\rangle=0$ for $w\in W^\perp$, $z\in W$. Hence the <orthogonal projection> $P$ onto $W$ commutes with every diagonal $a$.
Writing $P$ as an $N$ by $N$ block <matrix> $(P_{ij})$, this says $P_{ij}a=aP_{ij}$, so each $P_{ij}\in A'$. If $b\in A''$, it commutes with every $P_{ij}$, and its diagonal action therefore commutes with $P$ and preserves $W$. Because $1\in A$, $(v_1,\ldots,v_N)\in W$. Therefore there is one $a\in A$ such that
$$
(bv_1,\ldots,bv_N)=(av_1,\ldots,av_N).
$$
Equality on the <basis> gives $b=a$, proving the theorem over either scalar field.
The identity hypothesis is necessary. For a possibly nonunital adjoint-stable <subalgebra>, adjoining the ambient identity does not change its <commutant of an operator algebra>, so the exact statement is $A''=A+\mathbb F I$. For example $A=\{0\}$ has $A''=\mathbb F I$ on a nonzero $V$. Thus the usual exam formulation takes the algebra to contain the identity.
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