# Hilbert Space Isomorphisms

1. Jan 18, 2006

### benorin

If H1 and H2 are two Hilbert spaces, prove that one of them is isomorphic to a subspace of the other. (Note that every closed subspace of a Hilbert space is a Hilbert Space.)

What I'm thinking is that every separable Hilbert space is isomorphic to L2. If I recall, a Hilbert space is separable iff it has a countable basis, so what of uncountable bases? Do I then consider a subspace of that space spanned by a countable subset of the uncountable basis?

2. Jan 18, 2006

### matt grime

What has separability got to do with it?

Given two vector spaces V and W one is isomorphic to a subspace of the other.

3. Jan 18, 2006

### benorin

Duh! Thanks Matt.

4. Jan 18, 2006

### Hurkyl

Staff Emeritus
(Don't forget you have slightly more to prove here -- a Hilbert space has a little more structure than a vector space!)

5. Jan 18, 2006

### HallsofIvy

Staff Emeritus
Is that really true? Certainly if V and W are finite dimensional then the one with smaller dimension is isomorphic to a subspace of the other. But I thought "If two spaces have the same dimension then they are isomorphic" was only true for finite dimensional spaces.

6. Jan 19, 2006

### matt grime

A vector space is determined up to isomorphism by the cardinality of its basis, and since the cardinals are well ordered (modulo the axiom of choice) I think we're okay.

I am leaving a lot unanswered (this is just my thought as to how to start the proof), since there is more structure to verify than is 'a vector space' but as long as the image is closed, then we're ok.

7. Jan 19, 2006

### benorin

The definition of a Hilbert space isomorphism is:

Two Hilbert spaces, say H1 and H2, are isomorphic iff there exists a one-to-one linear mapping $$\Lambda$$ of H1 onto H2 such that $$\left( \Lambda x, \Lambda y\right) = (x,y), \\ \forall x,y\in H_1$$.

8. Dec 21, 2009

### LinnsburghF

I agree with HallsOfIvy. The polynomials with domain [0,1] form a vector space with countable basis, as does the l2 space of the natural numbers. These vector spaces are not isomorphic, since the first is not complete. A Hilbert space is determined up to isomorphism by the cardinality of its basis.

My suggestion for the proof is as follows: If H is an arbitrary Hilbert space over the field F with orthonormal basis B, show that H is isomorphic to the l2 space of B. The isomorphism is exactly what you'd expect. To define a linear map, you need only define it on the basis B. If v is a basis element, let T take v to the function fv, where
fv:B -> F is just the characteristic function of {v}, so fv(u)=1 if u=v, and fv=0 otherwise. This is a linear map. Surjectivity is pretty easy from the definition of an l2 space, injectivity comes from the fact that B is a basis, and the isometry <Tu,Tv>=<u,v> is immediate once you write as a combination of basis vectors.

Once you've done this, you've shown that a Hilbert space is determined up to isomorphism by the card of its basis. Then for two Hilbert spaces H,K, assume H has the lesser dimension. Take an orthonormal set S in K with cardinality equal to the cardinality of an ONB for H. Once you've established that span S is closed, it's a Hilbert space, and you're done.

9. Dec 21, 2009

### Hurkyl

Staff Emeritus
Welcome to PF, LinnsburghF!

While we do appreciate having more people to give homework help, we strongly discourage giving complete (or near-complete) solutions here -- the student will learn much more if they did the exercise themselves (with our help) rather than watch someone else do the exercise (which is why it was given as an exercise in the first place!)... and giving answers amounts to cheating anyways.

But the question is nearly four years old, so no harm done!