Remark

Let A be a star-algebra and let $x\in A$. The element $x^*x$ of A is positive; one then sets $|x|= (x^*x)^{1/2}$. We have $\|x\|^2=\|x^*x\|=$ $\||x|^2\|=\||x|\|^2$, hence $\|x\|=\||x|\|$.

When $x$ is normal, we have also $|x|=f(x)$, where $f$ is the mapping of $\mathbf{C}$ into $\mathbf{C}$, zero at 0, given by $f(z) =|z|$. In particular, when $x$ is hermitian, $|x|$ coincides with the element defined by formula (3).

Suppose moreover that A is unital and that $x$ is invertible. Then $|x|$ is likewise invertible, the element $u=x|x|^{-1}$ is unitary and one has $x=u|x|$ ("polar decomposition"; see also I, p. 139, n$^o8$ for the case of endomorphisms of hilbert spaces).