Operators on Hilbert Spaces

1. Sep 29, 2005

Oxymoron

Let $\mathcal{H}$ be a Hilbert space over $\mathbb{C}$ and let $T \in \mathcal{B(H)}$.

I want to prove that $\|Tx\| = \|x\| \, \Leftrightarrow \, T^{\ast}T = I$ for all $x \in \mathbb{H}$ and where $I$ is the identity operator in the Hilbert space.

Since this is an if and only if statement I began with the reverse inclusion:

$$(\Leftarrow)$$

Suppose $T^{\ast}T = I$, then for every $x \in \mathcal{H}$ we have

$$\|Tx\|^2 = \langle Tx\,|\,Tx \rangle = \langle x\,|\,T^{\ast}Tx \rangle = \langle x\,|\, x \rangle = \|x\|^2$$

And therefore $T$ is isometric (after taking square roots of both sides).

But now with the forward inclusion I am having some difficulties.

$$(\Rightarrow)$$

Suppose that $T$ is isometric. Then for every $x \in \mathcal{H}$ we have

$$\|Tx\|^2 = \|x\|^2$$
$$\Rightarrow \, \langle Tx\,|\,Tx \rangle = \langle x\,|\,x \rangle$$
$$\Rightarrow \, \langle T^{\ast}Tx \,|\, x \rangle = \langle x\,|\,x \rangle$$
$$\Rightarrow \, \langle (I-T^{\ast}T)x\,|\,x \rangle = 0$$

Now since $I-T^{\ast}T$ is self-adjoint and we know that for self-adjoint operators

$$\|T\|_{op} = \sup \{\langle Tx\,|\,Tx \rangle \,:\, \|x\| = 1\}$$

then $\|I - T^{\ast}T\|_{op} = 0$ and therefore $T^{\ast}T = I$.

This last bit in red I dont understand. Does anyone know how to explain it?

Last edited: Sep 29, 2005
2. Sep 29, 2005

Oxymoron

In other words I want to prove that $\|Tx\| = \|x\| \, \Leftrightarrow \, T^{\ast}T = I$ and Im not sure how to do the $\Rightarrow$ bit.

3. Sep 29, 2005

looth

Hi,
After
$$\Rightarrow \, \langle (I-T^{\ast}T)x\,|\,x \rangle = 0$$
Just express $I-T^{\ast}T$ as its spectral representation and then
take $x$ as eigen vector. You can get $I-T^{\ast}T=0$.

Actualy, the proof can be found in most of the relevent textbooks, for example
Hirvensalo's Quantum computing provides another method to proof it.

regards
looth

4. Sep 29, 2005

Oxymoron

Ah yes of course - I remember that. Thanks.

5. Sep 29, 2005

Oxymoron

Actually I dont remember

6. Sep 29, 2005

Oxymoron

Oooh...I came up with something else!

From the Polarization identity we know that a bounded linear operator $T$ is isometric if and only if it preserves the inner product. Now, since I want to prove that $T$ is isometric if and only if $T^{\ast}T = I$. Therefore, $T$ is isometric if and only if $\langle Tx\,|\,Ty \rangle = \langle x\,|\,y \rangle$ (preserving the inner product), and continuing on...

$$T \mbox{ is isometric } \Leftrightarrow \langle Tx\,|\,Ty \rangle = \langle x\,|\,y \rangle$$
$$T \mbox{ is isometric } \Leftrightarrow \langle x\,|\,T^{\ast}Ty \rangle= \langle x\,|\,y \rangle$$
$$T \mbox{ is isometric } \Leftrightarrow T^{\ast}T = I$$

Therefore $T$ is isometric if and only if $T^{\ast}T = I$. So I have proven both if statements in one go. How does this look?

Last edited: Sep 29, 2005
7. Sep 29, 2005

HallsofIvy

Staff Emeritus
If $\|I - T^{\ast}T\|_{op} = 0$ then
$$\sup \{\langle (I-T)x\,|\,(I-T)x \rangle:\|x\|= 1 \}= 0$$
Since that can never be negative that says
$$\langle (I-T)x\,(I-T)x\rangle= 0$$
for all x which means that (I-T)x= 0 for all x so I-T= 0 and T= I.

Last edited: Sep 29, 2005
8. Sep 29, 2005

Oxymoron

Yes! Straight from the definition of the operator norm. Thanks Halls.

9. Sep 29, 2005

Oxymoron

Ok so I have

$$\|Tx\|^2 = \|x\|^2 \Rightarrow \langle (I-T^{\ast}T)x\,|\,x \rangle = 0$$

And from here I want to use that fact that

$$\|I-T^{\ast}T\|_{op} = 0$$

To do what Halls said in his post.

BUT, I am missing the step where I go from

$$\langle (I-T^{\ast}T)x\,|\,x \rangle = 0$$

to

$$\|I-T^{\ast}T\|_{op} = 0$$

Im not sure how to jusify this step.

10. Sep 29, 2005

Oxymoron

Forget I wrote that.

Im using the operator norm of $(I-T^{\ast}T)$ to discover that $(I-T)x = 0$.

11. Sep 29, 2005

Hurkyl

Staff Emeritus
Try computing

$$\langle (1 - T^*T)x \, | \, (1 - T^*T)x \rangle$$

. :tongue2:

12. Sep 30, 2005

Oxymoron

$$\langle (I-T^{\ast}T)x \,|\, (I-T^{\ast}T)x \rangle = \langle (I-T^{\ast}T)(I-TT^{\ast}x \,|\, x \rangle$$

$$= \langle (I^2 - TT^{\ast} - T^{\ast}T + (T^{\ast}T)(TT^{\ast}))x\,|\, x\rangle$$ Note $I^2 = I$.

$$= \langle (I - TT^{\ast} - T^{\ast}T + T^2T^{\ast 2})x\,|\,x \rangle$$ Since $T^2 = T$ and $T^{\ast 2} = T^{\ast}$

$$= \langle (I - TT^{\ast} - T^{\ast}T + TT^{\ast})x\,|\,x \rangle$$

$$= \langle (I - T^{\ast}T)x\,|\,x \rangle$$

Does this look right?

This only works if $T = T^2$. Can I assume this?

Last edited: Sep 30, 2005
13. Sep 30, 2005

Oxymoron

Hold on, why am I computing this?

14. Sep 30, 2005

djeipa

Yes, of course.

15. Sep 30, 2005

djeipa

i think these kinds of problems are just another face of vector, matrix computations. you are trying to prove if one vector in H is considered to be perpendicular to itself.

Last edited: Sep 30, 2005
16. Sep 30, 2005

Oxymoron

Im still struggling to understand this.

I am fine up to

$$\Rightarrow \langle (I - T^{\ast}T)x \,|\, x\rangle = 0$$

I know that $I - T^{\ast}T$ is self-adjoint.

And I know that it has a norm:

$$\|I-T^{\ast}T\|_{op} = \sup\{\langle(I-T^{\ast}T)x\,|\,x \rangle \, : \, \|x\| = 1\}$$

by the definition of the operator norm.

Now, from the exercise that Halls made me do I know this norm is the same as

$$\|I-T^{\ast}T\|_{op} = \sup\{\langle (I-T^{\ast}T)x \,|\, (I-T^{\ast}T)x \rangle$$

Now where do I go?

I don't see how Halls wrote in his post:
Quote:
$$\sup\{(I-T)x\,|\,(I-T)x\rangle \,:\, \|x\| = 1\} = 0$$

I dont see how we have gone from $(I-T^{\ast}T)$ tp $(I-T)$.

Last edited: Sep 30, 2005
17. Sep 30, 2005

Hurkyl

Staff Emeritus
No: $(I - T^*T)^* = (I - T^*T)$, and generally $T^*TTT^* \neq T^2 (T^*)^2$, and why do you think $T^2 = T$?

But you're approaching the problem in the wrong way, and haven't used most of the properties of an inner product!

Because you wanted to know its supremum over all unit vectors x. :tongue2:

Last edited: Sep 30, 2005
18. Sep 30, 2005

Oxymoron

Hi Hurkyl,

The operator norm

$$\|S\|_{op} = \sup\{\langle Sx\,|\,x \rangle \,:\, \|x\| = 1\}$$

works for any continuous operator between normed vector spaces right? And in particular will work for any self-adjoint operator. I know that $(I-T^{\ast}T)$ is self adjoint because $(I-T^{\ast}T) = (I-T^{\ast}T)^{\ast}$. So

$$\|(I-T^{\ast}T)\|_{op} = \sup\{\langle(I-T^{\ast}T)x\,|\,x \rangle \,|\, \|x\| = 1\}$$

But we know that $\langle (I-T^{\ast}T)x\,|\,x \rangle = 0$. Therefore, from the operator norm and the supremum we deduce that

$$\|(I-T^{\ast}T)\|_{op} = 0$$

Therefore $(I-T^{\ast}T) = 0 \Rightarrow T^{\ast}T = I$.

This looks fine to me. But that doesn't mean anything. Please tell me if I have made a mistake.

19. Oct 1, 2005

Hurkyl

Staff Emeritus
No. I know it works when S has a complete set of eigenvectors, because this definition (correctly) picks off the eigenvalue with the largest magnitude, but this does not work for everything.

For example, let S be rotation by pi/2 radians, and your normed (real) vector space be R². You have that <Sx|x> = 0 for all x!

I don't have a handy example of a nondiagonalizable matrix over C³ to test.

I guess $1 - T^*T$ has a complete spectrum over C, since it is self-adjoint, so the analog of my comment about having a complete set of eigenvectors would apply, making the proof valid.

By the way, you can directly prove that $\langle (1 - T^*T)x \, | \, (1 - T^*T)x \rangle = 0$ given just the assumption that T is isometric.

20. Oct 1, 2005

Oxymoron

If $T$ is unitary, then

$$T^{\ast}T = TT^{\ast} = I$$.

I want to prove that $T$ is unitary if and only if $T$ is a surjective isometry.

Now I haven't proved it yet, but I just want to ask a question before we go on. This statement is similar to the one we have just been looking at, except now $T^{\ast}T = I$ AND $TT^{\ast} = I$, plus we have that $T$ needs to be surjective.

My question is: Will proving both $T^{\ast}T = I$ AND $TT^{\ast} = I$ require the surjective hypothesis? Is that the difference here?