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

[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?)