Uniqueness of Unitary operator i saw the post "Polar decomposition normal operator" (Polar decomposition normal operator). There was that such a $U$ is unique iff the image of $T$ is dense. Some lines later by the comments there is that we also can say that $||T||>\delta||x||$, but this means that $T$ is invertible. Can we thus also say:
$U$ in unique iff $T$ is invertible.
I thought about that: One direction is obvious. But suppose $U$ is unitary how can we see that $T$ is not invertible? My idea was: Suppose we have that $T^*=UT$ and $T^*=U'T$, $U\neq U'$, then $(U-U')T=0$. May we say that then $T=0$ and therefore not invertible? Or is this not correct?
Thank you.

For a bounded operator $A$ with polar decomposition $A = UP$, $U$ is the (canonical, if you'd like) partial isometry
$$
 (A^*A)^{\frac{1}{2}}h \stackrel{U}{\mapsto} Ah
$$ 
from $Ran(A^*A) = Ran(A^*A)^{\frac{1}{2}}$ to $Ran(A)$. If $A$ is invertible, so is $A^*A$. This make $U$ unique.
Assuming $U$ admits unitary extensions and has more than one extension, $Ran(A^*A)$ is not dense. So $A$ is not invertible.
