Absolute value of an element in a C*-algebra

Question

Is absolute value of a partial isometry a partial isometry itself?

Answer 1

We have, as square roots are self-adjoint $\def\abs#1{\left|#1\right|}$ \begin{align*} \abs v^*\abs v &= [(v^*v)^{1/2}]^*(v^*v)^{1/2}\\ &= (v^*v)^{1/2}(v^*v)^{1/2}\\ &= v^*v \end{align*}