# Is it true that $|x^2|=|x|^2$ for a normal $x$ in a $C^*$-algebra?

I know that for a selfadjoint $x$ in a $C^*$-algebra $A$, it holds that $$|x^2|=|x|^2,$$ where $|\cdot |$ is the norm in $A$ as well as $$r(x)=|x|$$ (by spectral radius formula). I'm wondering if this identity also holds for a normal element. Couldn't think of a proof yet.

• You should refrain from using $\lvert x \rvert$ for the norm of an element $x$ in a C*-algebra, since this is usually used to denote the element $\sqrt{x^*x}$.
– user42761
Commented Mar 19, 2018 at 20:14

Firstly, the spectral radius formula is irrelevant. If $x\in A$ is self-adjoint, $$\|x^2\|=\|x^*x\|=\|x\|^2.$$ If now $x\in A$ is normal, we have $$\|x^2\|^2=\|(x^*)^2x^2\|=\|(x^*x)^2\|=\|x^*x\|^2=\|x\|^4$$ and taking square roots gives the result.

Note: I wasn't paying much attention to what I was proving. The below is a proof that $\|x\|$ is the spectral radius of $x$ if $x$ is normal. It does lead to the desired conclusion, but it's overkill in this context.

Begin by noting that the spectral radius of $x$ is given by $$\rho = \lim_{n \to \infty}\|x^n\|^{1/n}$$ With that in mind, we note that $\|x\|^2 = \|x^*x\|$, and that $x^*x$ is self-adjoint. With that in mind, we see that for $k \geq 1$, we have $$\left\|x^{2^{k}}\right\|^2 = \left\|\left(x^{2^k}\right)^*\left(x^{2^k}\right)\right\| = \|(x^*x)^{2^k}\| = \|x\|^{2^{k+1}}.$$ That is, we prove by induction that $\left\|x^{2^{k}}\right\| = \|x\|^{2^k}$ for all $k \geq 1$. Thus, we can conclude that $$\rho = \lim_{n \to \infty}\|x^n\|^{1/n} = \lim_{k \to \infty}\left\| x^{2^k}\right\|^{1/2^k} = \lim_{k \to \infty}\left( \|x\|^{2^k}\right)^{1/2^k} = \|x\|$$ as desired.