Question about polar decomposition inside a von Neumann algebra.

Question

Let $M \subseteq B(H)$ be a von Neumann algebra.

Let $x = v|x|$ be the polar decomposition of $x$. It is well-known that $v,|x| \in M$. Is it true that the element $v^*v$ is an element of the von Neumann algebra generated by $x^*x$?

Attempt: Maybe it is useful to look at the polar decomposition of $x^*x$. We also know that $v^*v$ is the projection on $ker(x)^\perp$.

Answer 1

Yes, because $v^*v$ is the projection onto the closure of the range of $x^*x$ (as this is equal to the closure of the range of $x^*$). Von Neumann algebras contain range projections.