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

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

 

[Hurkyl]

If S and T are operators, and ST and TS do the same thing, that just means they commute. Surely you can find examples of commuting operators whose product is not invertible?

And even if ST is invertible, that does not mean S and T are invertible.


In my example, if we right-shift 0 (=(0, 0, 0, ...)), we get zero, so it does have a pre-image. In fact, 0 has at least one pre-image under any linear operator. (the proof is easy)


What condition is necessary for an arbitrary surjection to be invertible? Does an isometry satisfy that condition?


When I say T is invertible, I mean that there exists some S such that ST = TS = I. (In other words, exactly what it means for T to be an invertible function on sets, and also exactly what it means for T to be an invertible element of a ring)

 

[Oxymoron]

Hurkyl, I see your point.

For the forward implication, if I suppose that $TT^{\ast} = T^{\ast}T = I$ then $T$ is isometric! (by the very definition - and it was also proved earlier in this thread). By hypothesis now, $T$ obviously has an inverse, namely $T^{\ast}$ is the inverse of $T$ (because by acting on $T$ by $T^{\ast}$ we get the identity and this is the very definition of inverse from group theory, and other places).

From this we can conclude that $T$ is an isomorphism, which means that it is certainly an surjective isometry.


For the reverse implication, suppose that $T$ is a surjective isometry. Then $T$ has an inverse which is also isometric (because since $T$ is an isometry, $T^{\ast}T = I$ by definition).
From this we can deduce that $T^{-1} = IT^{-1} = T^{\ast}TT^{-1} = T^{\ast}I = T^{\ast}$.

How does this look?

 

[Hurkyl]

$T^*T = I$ by definition

By what definition? I can't think of any that would let you conclude that.

 

[Oxymoron]

Definition:

T is an isometry if and only if T*T = I

i.e.

$$\|Tx\| = \|x\| \Leftrightarrow T^{\ast}T = I \quad \quad \forall\, x\in \mathcal{H}$$

 

[Hurkyl]

This must be the source of your confusion -- that's not the definition of an isometry.

An isometry is a transformation that preserves the metric. I.E. if T is a map from X to Y, then T is an isometry if and only if $d_X(a, b) = d_Y(Ta, Tb)$ (where $d_X$ is the metric on X)

In a normed linear space, this is equivalent to the statement: a linear map T is an isometry if and only if $||a||_X = ||Ta||_Y$. (Where $|| \cdot ||_X$ is the norm on X)

 

[Oxymoron]

Wait a minute! Is isometry and isometric the same thing?

Because T is isometric if and only if T*T = I

 

[Oxymoron]

Remember, we are in Hilbert spaces. Does that make a difference?

 

[Hurkyl]

Wait a minute! Is isometry and isometric the same thing?

Ack, maybe it's the source of my confusion. For some reason, I was thinking an "isometry" just had to be distance preserving, not that it also had to be an isomorphism.

(Incidentally, isometric is an adjective, and isometry is a noun. )

The right shift operator is merely distance-preserving, and is not an isomorphism.

But $T^*T = I$ is still not the definition of isometry.

Furthermore, if we let E be the right-shift operator on l², then if my algebra is right, $E^*$ is the left-shift operator, and clearly $E^*E = I$ is true, so your if and only if is incorrect.

(The definition of $T^*$ is that it's the map such that $\langle Tv|w\rangle = \langle v | T^*w\rangle$ for all v, w, right?)