Show that an isometric linear operator $T:H\to H$ satisfies $T^* T=I$, where $H$ is an inner product space.
I've been stuck on this for a while and don't really know where to start.
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Yiorgos S. Smyrlis
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Desperate Fluffy
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Is $H$ a Hilbert space? Also, presumably $T$ is defined to be surjective, yes? – Ben Grossmann Oct 28 '14 at 05:20
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@Omnomnomnom Yes, $H$ is a Hilbert space, but $T$ isn't defined to be surjective, just isometric. Should it? – Desperate Fluffy Oct 28 '14 at 05:48
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actually, my mistake; surjectivity is not necessary. It would be if we wanted to show $TT^* = I$. – Ben Grossmann Oct 28 '14 at 05:55
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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 $T^*T=I$.
Yiorgos S. Smyrlis
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This only works for real Hilbert spaces, as otherwise the expansion of $||(x+y)||^2$ will have $2\text{Re}(\langle x, y\rangle )$ instead. – beeclu Feb 25 '25 at 20:26
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In the complex case, we can just say $\langle Ix, x \rangle = \langle x, x \rangle = \langle Tx, Tx\rangle = \langle T^*Tx, x, \rangle$ and in complex Hilbert spaces we have $\langle Ax, x\rangle = \langle Bx, x \rangle$ for all $x$ iff $A = B$. – beeclu Feb 25 '25 at 20:28
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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 $$ \|Tx\|^{2}=(Tx,Tx)=(T^{\star}Tx,x)=(x,x)=\|x\|^{2},\;\;\; x \in X. $$
Disintegrating By Parts
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Assuming $H$ is a Hiblert space:
Let $\{e_i\}$ be an orthonormal Schauder basis for the Hilbert space. We note that $$ T^*T = I \iff \forall x,y: \langle x,T^*Ty\rangle = \langle x,y\rangle\\ \iff \forall x,y: \langle Tx, Ty \rangle = \langle x,y \rangle $$ Try to deduce this last statement using the fact that $$ \forall x,y: \langle Tx, Tx \rangle = \langle x,x \rangle $$
Ben Grossmann
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