Generalized polar decomposition

Question

Let $x\in B(H)$. We say $(x,v,y)$ is a polar decomposition for $x$ if,

$\bullet$ $y$ is positive.

$\bullet$ $v$ is a partial isometry with $x=vy$.

$\bullet$ Ker$(x)$=Ker$(y)$=Ker($v$)

The polar decomposition $x=u|x|$ induces a polar decomposition for $x$. These natural problems give rise:

1. Does there exists any (non-trivial) polar decomposition for $x$?
2. Any characterization for all polar decomposition for $x$?

Answer 1

This is exactly the uniqueness in the polar decomposition. You have, since $v $ is a partial isometry, $$\tag {2}{\text {ran}\,v^*v}= {\text {ran}\,v^*}=(\ker v)^\perp=(\ker y)^\perp=\overline {\text {ran}\,y}. $$ Suppose that $w,z $ gives another such decomposition of $x $. Let $p=v^*v=w^*w $. Then, since $py=y$, we have $$ y^2=y^*y=y^*py=y^*v^*vy=x^*x=|x|^2. $$ It follows that $y=|x|$, and repeating the argument we get $z=y=|x|$. Now $$ wy=vy, $$ so $w=v$ on $\overline{\text{ran}\,y}$; we also have $\ker w=\ker v=\ker y$, so $w=v$.