Show that $v|a|=|a^*|v$, when $aa^*=va^*av^*$

Question

Suppose that $A$ is a $C^*$-algebra. Let $a=v|a|$ be the polar decomposition of $a$ in $A^{**}$. Show that $v|a|=|a^*|v$, when $aa^*=va^*av^*$.

I have $aa^*=va^*av^*$ which implies that $aa^*v=va^*a$ by multiplying $v$ on both sides. This further gives that $|a^*|^2v=v|a|^2$.

I don't know how to conclude from here.

Any hint will be greatly appreciated.

Thank you!!

Answer 1

Let $b=|a^*|^2$, $c=|a|^2$. You know that $bv=vc$. Then $$ b^2v=bvc=vc^2. $$ Inductively, show that $b^nv=vc^n$. Then, that $p(b)v=vp(c)$ for all polynomials $p$. Then take limits and show that $f(b)v=vf(c)$ for all continuous functions $f$ with support containing both the spectra of $b$ and $c$.