# Linear operator decomposition

1. Oct 6, 2008

### maze

In finite dimensions, a matrix can be decomposed into the sum of rank-1 matrices. This got me thinking - in what situations can a bounded linear operator mapping between infinite dimensional spaces be written as an (infinite) sum of rank-1 operators?

eg, let A be a bounded linear operator from banach spaces X to Y, then perhaps we might try

$$A = \sum_{i=1}^\infty y_i \phi_i[/itex] for some functionals $\phi_i$ in X', and elements y_i in Y. Is there anything to this idea? 2. Oct 6, 2008 ### morphism Presumably we want the series $\sum_{i=1}^\infty y_i \phi_i$ to converge in the operator norm. In this case, notice that for each N, the operator $\sum_{i=1}^N y_i \phi_i$ is finite rank. So if we let N tend to infinity, we get a compact operator. On general Banach spaces things get a little hairy. But if we're working on one Hilbert space H (so we're in B(H)), things are a little nicer, and the converse is true: every compact operator is the uniform limit of a sequence of finite ranks. And in fact we get a sort of singular value decomposition: if K is a compact operator on H, then there exists a sequence {s_i} of complex numbers and orthonormal bases {e_i} and {f_i} for H such that [tex]K = \sum_{i=1}^\infty s_i (f_i \otimes e_i^*).$$

This is easy to prove. It basically follows from the fact that if K is compact, then K*K is compact and positive. So we can apply the spectral theorem to its square root (K*K)1/2. The "singular values" s_i of K are then nothing but the square roots of the eigenvalues of K*K.

Last edited: Oct 6, 2008
3. Oct 6, 2008

### HallsofIvy

If your vector space has uncountable basis (which, I think, can't happen for Banach or Hilbert spaces), you would have to use an integral rather than a sum.

4. Oct 6, 2008

### maze

Err, doesnt $L^\infty$ have an uncountable basis?

5. Oct 6, 2008

### morphism

Yes, the algebraic dimension of a Banach space is either finite or uncountable.

I don't really know what Halls meant by his comment. Maybe he was talking about Schauder bases? If your Banach spaces X and Y have (countable) Schauder bases, then a statement similar to the one in post #2 can be made. By the way, in that post I implicitly assumed that H was separable, so that we get countable orthonormal bases. There is really no loss in making this assumption, because if K is a compact operator between Hilbert spaces, then $(\ker K)^\perp$ and $\overline{\mathrm{ran} K}$ are separable Hilbert spaces - so compact operators essentially live in separable Hilbert spaces.

6. Oct 7, 2008

### HallsofIvy

Hey, I said "I think"! It's been a long time!

7. Oct 28, 2008

### maze

On topological vector spaces with no norm, is it true that the limit of finite rank operators is compact?

8. Oct 29, 2008

### morphism

Limit in what sense? And how are you defining compact operators?

9. Oct 30, 2008

### maze

Limit in the topological sense. In the sense that there exists an operator, the limit, such that every neighborhood of the limit contains a tail of the sequence of finite rank operators. We can make the space hausdorff too so that the limit is unique.

Then a compact operator would be defined as one where the image of the unit ball is precompact. All of this requires only topology and a vector space, so we ought to be able to say something about it.

10. Oct 30, 2008

### morphism

You need a norm to define the unit ball.

And still, what topology are you talking about? We're taking a sequence of operators - so what is the topology on the space of operators between two your vector spaces? The topology of pointwise/uniform convergence?

11. Oct 30, 2008

### Office_Shredder

Staff Emeritus

There's a standard norm on bounded linear operators. The definition of a bounded linear operator is there exists M s.t. |Tx| <= M |x| for all x in your domain. Then taking the infimum of all such M gives |T|

12. Oct 30, 2008

### morphism

That only makes sense if there are norms on the underlying spaces. In any case, it's already been mentioned that the uniform limit (i.e. the limit in the topology generated by the operator norm) of finite ranks is compact.

13. Oct 30, 2008

### maze

Ahh right, cant use the unit ball. No matter, a compact operator can still be defined as an operator that maps a neighborhood of 0 into a subset of a compact set. As for the operator topology, I agree the topology of pointwise convergence would be a reasonable choice.

This should be true, but I can't see an easy proof.

14. Oct 30, 2008

### morphism

I don't think it's true. Take for instance a tvs X with a countably infinite (Hamel) basis {x_1, x_2, ...}. Then the identity I is not compact - otherwise X would be locally compact and hence finite-dimensional (a standard result). Nevertheless, I is the pointwise limit of finite rank operators. For instance if we define F_n(x_k)=x_k if k<=n and 0 otherwise, and extend by linearity, then F_n(x) -> x for all x, and F_n is a finite rank operator for each n.

15. Oct 30, 2008

### maze

Nice one. I should have seen that