Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 10 / 1 / b

The commutant is $A^{\prime }=\{T\in \mathrm{End}V:Ta=aT\text{ for every }a\in A\}$. Clearly $A\subseteq A^{\prime \prime }$. 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 $A^{\prime }$. Thus if $B\in A^{\prime \prime }$, $B\xi \in A\xi$.

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 $A^{\prime }$. Therefore $B^{(d)}$ commutes with this amplified commutant. For a basis ${\xi }_1,\dots ,{\xi }_d$, the subspace $\{(a{\xi }_1,\dots ,a{\xi }_d):a\in A\}$ is invariant and reducing for the amplified algebra. Its projection lies in that commutant. The preceding argument gives some $a\in A$ with $(B{\xi }_1,\dots ,B{\xi }_d)=(a{\xi }_1,\dots ,a{\xi }_d)$. Since these vectors form a basis, $B=a$. HenceThis finite-dimensional double commutant theorem for star-algebras uses the star hypothesis to make the orbit subspaces reducing.