
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 28012013 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" website, a 2013 meeting: >>>> >>>> "del 21 al 25 de Enero" 2125 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 >> 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 and now they have withdrawn their submission.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: >> >> http://cafematematico.com/2013/02/05/statementfromcowenandgallardo/ >> >>  >> >> 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 subspace W >> such that T(W) is contained within W, and >> W is nontrivial, meaning both >> dim(W) >= 1 and dim(W) < dim(H) = n. >> >> So the soughtfor W has 1<= dim(W) <= n1, if it exists. >> >> All ndimensional 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_{n1} lambda^(n1) + ... + 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 nonzero 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 1dim. >> space C.v . So W = {kv, k in C the complex numbers}, >> v is nonzero, dim(W) = 1 exactly , W is a nontrivial >> 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 squaresummable 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 normcomplete, 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 complexlinear 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, socalled unbounded operators). >> >> Just like there is a ring of nxn matrices >> in finite dimension, there is a ring with identity >> of all the continuous complexlinear 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 operatornorm distance >> of at least 1 from each other. >> It follows that B(H) has no countable dense subset, i.e. >> B(H) is a nonseparable 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 socalled "positive" operator. >> This might seem like a promising divideandconquer >> approach to the invariant subspace problem, but >> it doesn't seem to say much about what happens when, >> starting with a nonzero 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 infinitedim. Banach space >> and an operator on it that had no nontrivial closed >> subspace. See e.g.: >> archive.numdam.org/article/SAF_19751976____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 selfadjoint, where the socalled >> 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 nontrivial invariant > subspace? [...]
Ok. I think we can find a finiterank K such that T+K has some nonzero 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 infinitedimensional invariant subspace of T+K, or alternatively infinitely many "linearly independent" invariant subspaces W_1, W_2, W_3, ad infinitum 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.

