Subalgebra Generated by Self-Adjoint Set is Self-Adjoint

Theorem

Let $\tuple {A, \ast}$ be a $\ast$-algebra over $\C$.

Let $S \subseteq A$ be self-adjoint.

Let $K \sqbrk S$ be the subalgebra generated by $S$.

Then $\C \sqbrk S$ is a $\ast$-subalgebra of $A$.

Proof

From Explicit Form for Generated Subalgebra, we have:

$\C \sqbrk S = \map \span \AA$

where:

$\AA = \set {x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} : x_1, \ldots, x_n \in S, \, k_1, \ldots, k_n \ge 1}$

From Star of Product of Elements in *-Algebra we have:

$\paren {x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} }^\ast = x_n^{k_n} \ldots x_2^{k_2} x_1^{k_1} \in S$

Hence $\AA$ is self-adjoint.

Then $\C \sqbrk S$ is self-adjoint from Linear Span of Self-Adjoint Subset of *-Algebra is Self-Adjoint.

Hence $\C \sqbrk S$ is a $\ast$-subalgebra of $A$.

$\blacksquare$