Does T^*T = I imply that T is an isometry on a Hilbert space?

[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 don't understand. Does anyone know how to explain it?

 

[Oxymoron]

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

 

[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 relevant textbooks, for example
Hirvensalo's Quantum computing provides another method to proof it.

Hopefully my comment could help you.
regards
looth

 

[Oxymoron]

Ah yes of course - I remember that. Thanks.

 

[Oxymoron]

Actually I don't remember

 

[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?

 

[HallsofIvy]

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.

 

[Oxymoron]

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

 

[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.

 

[Oxymoron]

Forget I wrote that.

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

 

[Hurkyl]

Try computing

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

. :tongue2:

 

[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?

 

[Oxymoron]

Hold on, why am I computing this?

 

[djeipa]

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

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

Yes, of course.

 

[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.

 

[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 don't see how we have gone from $(I-T^{\ast}T)$ tp $(I-T)$.

 

[Hurkyl]

Does this look right?

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!

Hold on, why am I computing this?

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

 

[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.

 

[Hurkyl]

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

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.

 

[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?

 

[Hurkyl]

Exercise: find an isometry of [0, &infin;) that is not invertible.

 

[Oxymoron]

Not sure I understand that exercise Hurkyl, sorry.

An isometry is an operator which preserves distance (preserves the norm). If an isometry is not invertible, then it can't be written as $T^{-1}$, therefore it is not a bijection(?).

 

[Oxymoron]

For the forward inclusion

$$(\Rightarrow)$$

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

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

Then take square roots.

Am I correct to assume that I do not need to use surjectiveness in this part of the proof?

 

[Hurkyl]

You are correct, if an isometry is not invertible, then it is not a bijection. (Because you can prove, as functions on sets, that a function is bijective iff it is invertible)

The exercise is not a direct answer to anything you asked -- it is more to remind you of a common technique good for generating counterexamples.

Am I correct to assume that I do not need to use surjectiveness in this part of the proof?

Yes, but you didn't know it was surjective to begin with! Remember that surjectiveness is half of what you were trying to prove in the (=>) direction!

 

[Oxymoron]

With the $(\Leftarrow)$ proof I found that it is almost exactly the same as the proof for $\|Tx\| = \|x\| \Rightarrow T^{\ast}T = I$.

Does the difference occur when

$$\|Tx\|^2 = \|x\|^2 \Rightarrow \langle Tx\,|\,Tx \rangle = \langle x\,|\,x \rangle$$

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

$$\Rightarrow \langle x\,|\,TT^{\ast}x \rangle = \langle x\,|\,x \rangle$$

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

and solve as usual (like we did before)? If not, is this a good start anyway?

 

[Hurkyl]

Maybe I'm just having a memory blank, but how did you go from

$$\langle x \, | \, T^*T x \rangle = \langle x \, | \, x \rangle$$

to

$$\langle x \, | \, TT^* x \rangle = \langle x \, | \, x\rangle$$

?

The thing you want to prove is that $I - T^*T = 0$, so I don't understand why you're involving $TT^*$ in the first place...

 

[Oxymoron]

!

This is what is meant to be:

$$\|Tx\|^2 = \|x\|^2 \Rightarrow \langle Tx\,|\,Tx \rangle = \langle x\,|\,x \rangle$$

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

$$\Rightarrow \langle x\,|\,TT^{\ast}x \rangle = \langle x\,|\,x \rangle$$

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

Actually I am not even sure THIS is right

 

[Hurkyl]

Exercise: compute $(T^*T)^*$, showing and justifying every step.

 

[Oxymoron]

Ok. But I am not too good under pressure

If we take $T:\mathcal{H}_1 \rightarrow \mathcal{H}_2$, that is, a bounded linear operator between two Hilbert spaces. Then $T^{\ast}:\mathcal{H}_2 \rightarrow \mathcal{H}_1$. So now we take any element $x \in \mathcal{H}_2$ and apply $T^{\ast}$ and then $T$, ie the composition, we have

$$(T \circ T^{\ast})(x) = x \in \mathcal{H}_2$$

That is, after operating on x by the two operators in the correct order we should end up back where we started.

Now $(T^{\ast}T)^{\ast}$ means we do the same process in reverse order. And we get

$$(T^{\ast}T)^{\ast} = (T)^{\ast}(T^{\ast})^{\ast} = T^{\ast}T$$

How does this look?

 

[Hurkyl]

Yes, that looks right. (Actually, all I was looking for was the last LaTeX image! )

I'm not sure anymore I guessed correctly what mistake you were making...

I had been assuming that when you went from

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

to

$$\langle x\,|\,TT^{\ast}x \rangle = \langle x\,|\,x \rangle$$

that you went through the intermediate step

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

and had simplified incorrectly... if this isn't what you did, then what did you do?

 

[Oxymoron]

Yes, that is the mistake I made.

Is the following step correct?

$$\quad \quad \langle Tx\,|\,Tx \rangle = \langle x\,|\,x \rangle$$

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

 

[Hurkyl]

Yes it is.

 

[Oxymoron]

Just re-stating the question

Prove that $T^{\ast}T = TT^{\ast} = I$ $\Leftrightarrow$ $T$ is surjective and $\|Tx\| = \|x\|\quad \forall \, x \in \mathcal{H}$.

$$(\Rightarrow)$$

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

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

and the result is obtained after taking the square root.

Now as Hurkyl said, this is only half of the $\Rightarrow$ proof. So now we have to prove that $T$ is surjective.

Well (im not sure if this proves surjectiveness?)

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

 

[Hurkyl]

I don't see what that has to do with surjectiveness. You've almost said in this thread what is necessary to prove surjectiveness... I'll give you a hint: you only need to consider T as a function on sets.


By the way, here is an example of why you need the surjective hypothesis, if you haven't come up with it yet:

One answer to my exercise is to consider the isometry on $[0, \infty)$ defined by translating to the right. This is one of the basic tricks for generating counterexamples: to have some sort of infinite thing and shift towards the open end.

Here's an actual example of an linear operator on a Hilbert space that is an isometry:

Consider the space l², the space of square summable sequences -- right shifting is an isometry of this space, but is clearly not a surjection, and not invertible.

 

[Oxymoron]

I don't see what that has to do with surjectiveness. You've almost said in this thread what is necessary to prove surjectiveness... I'll give you a hint: you only need to consider T as a function on sets.

Well, if I treat T as a function on sets. And we know that $T^{\ast}T = TT^{\ast} = I$ then operating on a set by T and then T* does nothing to that set. The same occurs by doing it backwards (T* and then T). If $T^{\ast}T$ does the same thing to a set as $TT^{\ast}$ does, then $T^{\ast}T$ is invertible, no?

Consider the space $\ell^2$, the space of square summable sequences -- right shifting is an isometry of this space, but is clearly not a surjection, and not invertible.

Is that because right-shifting on this space preserves the norm of elements in $\ell^2$, but it is not a surjection because 0 does not have an inverse(?) - a surjective isometry means that every element in $[0,\infty)$ has to have an pre-image. But if the operation is right-shifting then 0 is not the image of any element under the operation. So the operation is not surjective.

Excuse me for thinking out aloud. So there does exist an isometry which is not surjective(?) hence not invertible. BUT if there was an isometry which WAS surjective then it WOULD be invertible (?).

When you say invertible, do you by any chance $T^{\ast}T = TT^{\ast} = I$.