Exercise 7.D.9

Proof. Consider the proof of 7.45. $T$ being invertible implies $\dim <!--\; FUNCTION\; APPLICATION\; -->{(\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T)}^{\text{⊥}}=0$, thus $S_2=0$ and so $S=S_1$. Clearly $S_1$ is unique, so $S$ must also be unique. Conversely, if $S$ is unique then $S_2=0$, otherwise we could set $S=S_1-S_2$ and it would still be an isometry satisfying $T=S\sqrt{T^{\ast }T}$. This implies that $m=0$, because from the defintion we see that $S_2$ must be invertible for any positive integer $m$. Thus $\dim <!--\; FUNCTION\; APPLICATION\; -->{(\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T)}^{\text{⊥}}=0$ and so $T$ is invertible. □