The operators $u$ and $|u^*|$ have the same range.

Question

I'm reading the proof of the following theorem in Murphy's book "$C^*$-algebras and operator theory" (only the relevant parts have been shown)

Why do $u$ and $|u^*|:= \sqrt{uu^ *}$ have the same image? The author mentions polar decomposition of $u^*$, so I tried writing $$u^* = w|u^*| $$ with $w$ a partial isometry with $\ker (w) = \ker(u^*)$

We also have $$\operatorname{im}(|u^*|)^\perp=\ker(|u^*|) =\ker(u^*) = \operatorname{im}(u)^\perp$$ but I don't think this implies $\operatorname{im}(u)= \operatorname{im}(|u^*|)$.

Why is this true?

Answer 1

Taking the adjoint of the polar decomposition equation you get $$u=|u^*|w^*$$ from where it follows that the range of $u$ is contained in the range of $|u^*|$. Can you get the other inclusion by yourself?