# Show that an isometric linear operator $T:H\to H$ satisfies $T^* T=I$, where $I$ is the identity operator on $H$.

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## Solution 1

Isometric means $$\|Tx\|=\|x\|,$$ for all $x\in H$. Equivalently, for all $x,y\in H$ $$\|T(x+y)\|^2=\|(x+y)\|^2.$$ But $$\|(x+y)\|^2=\langle x+y,x+y\rangle=\|x\|^2+2\langle x,y\rangle+\|y\|^2,$$ while $$\|T(x+y)\|^2=\langle T(x+y),T(x+y)\rangle=\|Tx\|^2+2\langle Tx,Ty\rangle+\|Ty\|^2.$$ Thus, for all $x,y\in H$ $$\langle Tx,Ty\rangle=\langle x,y\rangle.$$ Note that $\langle x,Ty\rangle=\langle T^*,y\rangle$, and the above becomes $$\langle T^*Tx,y\rangle=\langle x,y\rangle\quad\text{or}\quad \langle (T^*T-I)x,y\rangle=0$$ for all $x,y\in H$. In particular, for $y=(T^*T-I)x$, it becomes $$0=\langle (T^*T-I)x,(T^*T-I)x\rangle=\|(T^*T-I)x\|^2,$$ for all $x\in H$, and thus $T^*T=I$.

## Solution 2

The polarization identity for a complex Hilbert space $H$ is $$(x,y) = \frac{1}{4}\sum_{n=0}^{3}i^{n}\|x+i^{n}y\|^{2},\;\;\; x,y\in H.$$ If $T$ is a linear isometry, then \begin{align} (T^{\star}Tx,y) & =(Tx,Ty) \\ & = \frac{1}{4}\sum_{n=0}^{3}i^{n}\|Tx+i^{n}Ty\|^{2} \\ & = \frac{1}{4}\sum_{n=0}^{3}i^{n}\|T(x+i^{n}y)\|^{2} \\ & = \frac{1}{4}\sum_{n=0}^{3}i^{n}\|x+i^{n}y\|^{2} = (x,y). \end{align} Therefore, $((T^{\star}T-I)x,y)=0$ for all $x,y$. By a judicious choice of $y$, it follows that $(T^{\star}T-I)x=0$ for all $x$ and, hence, $T^{\star}T-I=0$.

Conversely, if $T$ is a bounded linear operator for which $T^{\star}T=I$, then $T$ is isometric because $$\|Tx\|^{2}=(Tx,Tx)=(T^{\star}Tx,x)=(x,x)=\|x\|^{2},\;\;\; x \in X.$$

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### Desperate Fluffy

Updated on October 28, 2020

Show that an isometric linear operator $T:H\to H$ satisfies $T^* T=I$, where $I$ is the identity operator on $H$.