Corollary

Let A be a star algebra and let B be a star subalgebra of A. Then one has Sp$'_B(x) =$ Sp$'_A(x)$ for every $x$ in B.

By adjoining a unit element (prop. 3), this follows from prop. 5.

Proposition 6 (Fuglede–Putnam Theorem)

Let A be a unital star algebra. Let $a$ and $b$ be normal elements of A. If $c\in A$ satisfies $ac=cb$, then $a^*c=cb^*$.

The hypothesis implies $(wa)^kc=c(wb)^k$ for every integer $k\geqslant 0$ and every $w\in \mathbf{C}$, hence $e^{wa}c=ce^{wb}$ (formula (9) of I, p. 78). Consider the function $f$ from $\mathbf{C}$ into A defined by $z\mapsto e^{-za^*}ce^{zb^*}$. It is a holomorphic function on $\mathbf{C}$, whose derivative satisfies

for every $z\in \mathbf{C}$. Since $c=e^{-za}ce^{zb}$, one can write

Since $a$ and $b$ are normal, the elements $e^{-za^*}e^{za}=e^{-za^*+za}$ and $e^{-zb}e^{zb^*}=e^{-zb+zb^*}$ of A are unitary for every $z\in \mathbf{C}$ (lemma 5 of I, p. 102), hence of norm 1. Consequently, one has $\|f(z)\|\leqslant \|c\|$ for every $z\in \mathbf{C}$. The function $f$ is therefore constant (VAR, R1, 3.3.6, p. 29), that is to say that $f(z) =f(0) =c$ for every $z\in \mathbf{C}$. But then $-a^*c+cb^*=f'(0) = 0$.