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Topic: Invariant subspace problem solved?
Replies: 9   Last Post: Mar 6, 2013 7:34 AM

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 David Bernier Posts: 3,755 Registered: 12/13/04
Re: Invariant subspace problem solved?
Posted: Feb 28, 2013 8:12 PM

On 02/28/2013 07:56 PM, David Bernier wrote:
> On 02/28/2013 01:24 AM, David Bernier wrote:
>> On 01/28/2013 11:19 AM, José Carlos Santos wrote:
>>> On 28-01-2013 15:35, David Bernier wrote:
>>>

>>>>> I found several texts across the Internet stating that the Invariant
>>>>> Subspace problem for Hilbert spaces was solved by Carl Cowen (USA) and
>>>>> Eva Gallardo (Spain). However, each of these texts is written
>>>>> either in
>>>>> Spanish or in Portuguese. This is rather odd, given the importance of

>>>> [...]
>>>>
>>>> I found something similar to what you found.
>>>>
>>>> From the "Real Sociedad Matemática Española" web-site, a 2013 meeting:
>>>>
>>>> "del 21 al 25 de Enero" 21-25 January?
>>>>
>>>> http://www.rsme.es/content/view/1199/1/

>>>
>>> Interesting. This site is the website of the Royal Spanish Mathematical
>>> Society. It looks serious.
>>>

>>>> I think they mention John von Neumann and the 1930s ...
>> [...]
>>
>> This development seems to be of some interest, and me included,
>> to some degree.
>>
>> The authors announced in early February that a gap in their proof
>> submitted a paper to a journal containing that claimed proof,
>> and now they have withdrawn their submission.
>>
>> They are working to bridge the gap, and there are results
>> independent of the existence of the gap. Their statement
>> was posted at cafe metematico blog here:
>>
>> http://cafematematico.com/2013/02/05/statement-from-cowen-and-gallardo/
>>
>> ---
>>
>> For complex Hilbert spaces H of dimension n, n finite and n>=2,
>> the related question is if an linear operator T: H-> H
>> that is continuous has a Hilbert sub-space W
>> such that T(W) is contained within W, and
>> W is non-trivial, meaning both
>> dim(W) >= 1 and dim(W) < dim(H) = n.
>>
>> So the sought-for W has 1<= dim(W) <= n-1, if it exists.
>>
>> All n-dimensional Hilbert spaces are isomorphic in the strongest
>> sense, essentially they are all like C^n with
>> (z1, ... z_n).(w1, ... w_n) = sum_{i=1, .. n} z_i w_i^{bar}.
>>
>> Operators on H of finite dimension n>=2 can be thought of
>> as nxn square matrices with entries in C.
>>
>> If A is an nxn matrix, the equation for eigenvalues is:
>> det(A - lambda I) = 0. 'I' is the nxn identity matrix.
>> lambda is the unknown scalar in C.
>>
>> det(A - lambda I) = 0 is equivalent (possibly after mutiplying by +/-1)
>> to lambda^n + a_{n-1} lambda^(n-1) + ... + a_0 = 0,
>> a degree 'n' polynomial equation with indeterminate
>> lambda obtained by fully expanding the determinant:
>> det(A - lambda I)
>> by using the coefficients of A as things that don't
>> depend on lambda (i.e. parameters, coefficients).
>>
>> Then, we apply the fundamental theorem of algebra for
>> C to this degree n polynomial equation.
>>
>> Suppose lambda = r0 is one root of the polynomial.
>> Then r0 is an eigenvalue of the matrix A, and
>> there exists a non-zero vector v in C^n such that
>> Av = r0 v. If k in C is any scalar,
>> A (kv) = k (Av) = k (r0 v) = r0 (kv),
>> so (A - r0 I) (kv) = 0.
>>
>> Or, one could say that for any scalar k in C,
>> A maps kv to (r0 k) v, an element of the 1-dim.
>> space C.v . So W = {kv, k in C the complex numbers},
>> v is non-zero, dim(W) = 1 exactly , W is a non-trivial
>> subspace of C^n, and W is mapped by A to W (although
>> not always onto; this is shown in case A is the zero
>> matrix).
>>
>> In countably infinite dimension, there's essentially one
>> complex Hilbert space, say l^2, the Banach space
>> of square-summable sequences of complex numbers.
>>
>> When the topic is Hilbert spaces, by a subspace one
>> always means/assumes a complete (vector) subspace, unless
>> it's stated otherwise. That's always been my working
>> assumption.
>>
>> This also applies, if I remember correctly, to Banach spaces
>> and subspaces, where the assumption would be that a
>> subspace must be norm-complete, a complete metric space.
>>
>> So now if H is l^2 or any other essentially equivalent
>> separable infinite dimensional complex Hilbert space,
>> one supposes given a continuous complex-linear mapping
>> T: H -> H .
>>
>> That's what an operator from H to H is. (However,
>> in quantum mechanics they need mappings that aren't
>> necessarily continuous, so-called unbounded operators).
>>
>> Just like there is a ring of nxn matrices
>> in finite dimension, there is a ring with identity
>> of all the continuous complex-linear mappings
>> T: H -> H that can be. That huge ring is denoted
>> B(H).
>>
>> B(H) is deceptively large.
>> For example, if sigma: N^* -> N^* is a permutation
>> on N^*, and e_i is the sequence
>> (0, 0, ... 1, ... 0, ....)
>> ^ i'th component ,
>>
>> then one can arrange provisionally to map
>> each e_i to e_{sigma(i)} , for i = 1, 2, 3, ad infinitum
>>
>> This can be extended by linearity to all linear combinations
>> of the e_i, i in N^*. Then, by continuity, the mapping
>> can be extended to the closure in l^2 of these linear
>> combinations.
>>
>> This works and will give a bounded operator U_sigma on
>> l^2. This U_sigma has norm 1 in the operator norm.
>> There are 2^\omega permutations on N^*, resulting in
>> 2^\omega = continuum distinct operators U_sigma.
>> If sigma and sigma' are distinct permutations, then
>> for the operators U_sigma and U_sigma' , I get:
>> || U_sigma - U_sigma' || >= 1 (hopefully ... ).
>>
>> { U_sigma, all permutations sigma: N^*->N^*} forms
>> a set of the power of the continuum of points in
>> B(H) any two of which are at an operator-norm distance
>> of at least 1 from each other.
>> It follows that B(H) has no countable dense subset, i.e.
>> B(H) is a non-separable metric space.
>>
>> For any A: H-> H, there's something called the
>> polar decomposition. One lets P = sqrt(A A^*) (I think)
>> and then A = O P , where O is an isometry or a partial
>> isometry and P is a so-called "positive" operator.
>> This might seem like a promising divide-and-conquer
>> approach to the invariant sub-space problem, but
>> it doesn't seem to say much about what happens when,
>> starting with a non-zero x in H, one iterates
>> A, constructing Ax, AAx, A^3 x, etc or using
>> the polar decomposition,
>> (OPx), (OPOP)x, (OPOPOP)x, etc [i.e. which way are things
>> going? we still don't know any better after applying
>> the polar decomposition to A].
>>
>> Per Enflo first constructed an infinite-dim. Banach space
>> and an operator on it that had no non-trivial closed
>> sub-space. See e.g.:
>> archive.numdam.org/article/SAF_1975-1976____A11_0.pdf
>>
>> and/or
>>
>> http://en.wikipedia.org/wiki/Per_Enflo
>>
>>
>> For separable Hilbert spaces, the best known case is when
>> the operator T is self-adjoint, where the so-called
>> spectral theorem in functional analysis applies
>> (Rudin: Functional Analysis).

> [...]
>
> I was reading on problems considered "modulo a compact
> operator K".
>
> If T is a bounded operator on a separable, infinite
> dimensional Hilbert space H,
>
> can one show that for some compact operator K
> on H, T+K has a non-trivial invariant
> subspace?

[...]

Ok. I think we can find a finite-rank K such that
T+K has some non-zero eigenvector, i.e.
one can construct K such that for some v in H,

T+K maps v to a scalar multiple of v.

[Now assuming K may even be compact]

Still, one hasn't found an infinite-dimensional
invariant subspace of T+K, or alternatively
infinitely many "linearly independent"
invariant subspaces
where dim(W_i) >=1 in each case.

--
dracut:/# lvm vgcfgrestore
File descriptor 9 (/.console_lock) leaked on lvm invocation. Parent PID
993: sh
Please specify a *single* volume group to restore.

Date Subject Author
1/28/13 David Bernier
1/28/13 David Bernier
1/30/13 Miguel Lacruz
1/28/13 Jose Carlos Santos
1/29/13 David Bernier
2/28/13 David Bernier
2/28/13 David Bernier
2/28/13 David Bernier
3/6/13 David Bernier