For a hermitian element $a$ in a $C^*$-algebra, show that $\|a^{2n}\| = \|a\|^{2n}$
In this post, I saw a comment stating that "More generally, if $a$ is normal then $∥a^n∥=∥a∥^n$ for each positive integer $n$. "
Can some one explain why this is true?
Thanks!
Normal means $a^*a=aa^*$.
In a C$^*$-algebra, the norm is algebraic: $$\tag{1}\|a\|=\text{spr}(a^*a)^{1/2}.$$ When $a$ is normal, we know that $C^*(a)\simeq C(\sigma(a))$ via the Gelfand transform, where $a$ is mapped to the identity function. Also $a^*a$ is mapped to the function $f:t\longmapsto |t|^2$, and it follows that $$\tag{2}\sigma(a^*a)=\sigma(f)=f(\sigma(a))=\{|\lambda|^2:\ \lambda\in\sigma(a)\}.$$ From $(1)$ and $(2)$, $$ \|a\|=\sup\{|\lambda|:\ \lambda\in\sigma(a)\} $$ As $a$ normal implies that $a^n$ is normal, now we have $$ \|a^n\|=\sup\{|\lambda|^n:\ \lambda\in\sigma(a)\}=\sup\{|\lambda|:\ \lambda\in\sigma(a)\}^n=\|a\|^n. $$