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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: 3,892
Registered: 12/13/04
Re: Invariant subspace problem solved?
Posted: Feb 28, 2013 7:56 PM
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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
>>>> the problem. Did anyone around here read something about this subject?

>>> [...]
>>> 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?

>> 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
> of the invariant subspace problem had been found. They had
> 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 and now they have withdrawn their submission.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:
> ---
> 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.:
> and/or
> 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

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

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