mm-219 == === Subject: Re: Non-separable lbert spaces > I was wondering if anyone knows some good references to techniques in> non-separable lbert spaces.For the best studied concrete model, search under almost periodic functions.When it comes to generalities, non-separable lbert spaces aretoo big to be interesting, as they are locally separable in at leasttwo senses:* with a specific orthonormal basis (uncountable, of course), the Fourierexpansion of every vector involves only a countable subset of the basis;the rest of the coefficients are zeros;* every bounded linear operator T has a non-trivial separable reducingsubspace (invariant under T and also under its adjoint); the same can besaid about every countable family of bounded linear operators.(Then, using Zorn's Lemma, you can cut up the space into mutuallyorthogonal separable reducing pieces.)In contrast, the invariant subspace problem for infinite dimensionalseparable lbert spaces is still open, === bergv@math.uiuc.edu (Maarten Bergvelt)> Two independent questions.1) Does there exist an infinite, finitely presented, simple group G,> and a two-transitive action of G?I tnk that R.Thompson's group T may be an example. See Cannon, J.W.; Floyd, W. J.; Parry, W. R. Introductory notes on RichardThompson's groups. Enseign. Math. (2) 42 (1996), no. 3-4, 215--256.That group T is finitely presented and simple. It has a subgroupdenoted by F (another R. Thompson's group). F is torsion-free andevery element of T is a product fgf' where f, f' are in F, g istorsion. I did not check, but it might be true that T=FyF for everyelement y from T F. Ts would imply that T acts 2-transitively onthe set of left cosets of F.Mark === undecided-unixOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)> let G = (V,E) be a graph such that to each vertex and to each edge there > is assigned a positive weight. The weight of a clique sums the the > weight of all its vertices and edges. Is there some work about the > maximum weighted clique problem? I'm only aware of numerous algorithms > dealing with the maximum vertex weighted clique problem? The following suggests a way to convert such a problem to finding a max clique in an edge-weighted graph (i.e. removes the vertex weights.) (Definition: A graph H is 'substituted' for a vertex v (in graph G) byreplacing v with the graph H and vertex u of G is adjacent to all verticesof H if and only if u was adjacent to v) If, in your vertex and edge weighted graph, you substitute a vertex v of weight w with a clique of size w, ts transformation will maintain the maximum cliques (since if v was in a maximum clique, then the w-clique is contained in the maximum clique of the new graph.) The edges from vertices a(1), a(2), ..., a(k) to the substituted vertexv have weights b(1), b(2), ..., b(k) (in the original graph.) Of all theedges created from any a(i) vertex to the substituted graph H (for v), leave one of the edges with weight b(i) and make the rest 0. (Or if you insist that all edge weights are positive, make each edge weight b(i)/|H|,since if one of these a(i)-to-H edges exist in the max clique, then all the a(i)-to-H edges will.) Ts would create a graph where all vertices are equally weighted (weight 1) and only the edges are weighted. Of course, ts only works if your vertex weights are whole integers (or rational numbers, but the problem size blowup could be large in that case.) One might tnk that taking the line graph of ts resultant graph (i.e. mapping each edge to a vertex in a new graph) might create a vertex-weighted version of the problem, but ts is not true as a claw (one vertex adjacent to three independent vertices) would map to a 3-clique (so cliques are not perserved in the inverse mapping.) Perhaps someone sees a clever way to transform ts to === bergv@math.uiuc.edu (Maarten Bergvelt) all,Could someone give me an example of a complete uniform space wchis not a Baire space? I've tried to construct such a space, but invain. Furthermore, there's no exemple of such a space in Steen &Seebach's Counterexamples in Topology.Of course, since I'm not aware of the existence of a completeuniform space wch is not a Baire space, perhaps the the questionshould be Is it true or false that... ?, but I would be verymuch surprised if it turned out the that every complete uniformspace is a Baire === bergv@math.uiuc.edu (Maarten Bergvelt),I hope you can help me with the following problem.If you have a rand() function that generates independent uniform values from [0,1], can you simulate any variable with polynomial distribution, using max, min and filtering only?Equivalently: suppose you have a set of distribution functions on [0,1], wch contains F(x)=x, and for any 0 all,I have a dataset with large number of features. I would like to reduce> the dimensionality of the dataset first. In order to do that, I am> tnking of clustering the features first, then replace each cluster> with a meta-feature wch best describe the features in that cluster.> The features from the original dataset are not independent to each> other. I would like to have the clusters to be as pure, and as tight> as possible -- All features in a cluster are similar. Not all to say much without more details. Principal component analysis may help define uncorrelated new features wch are linear combinations of the ones that you have. If the ultimate problem is classification, feature selection can be aceved by taking the features wch have the ghest mutual information with === bergv@math.uiuc.edu (Maarten Bergvelt)I am looking for information on the properties of the Laplacian ona bounded domain of R^n.In particular, if one consider a bounded open set O with a regularboundary, I would like to have a complete study of the properties ofthe eigenvectors {v1,v2,...}of the laplacian with Neumann boundary conditions.In fact, I am looking forward some approximation results.For example, given a C^1 regular function f on O,wch is the convergence speed of the approximationsum{i vi} when N-->infinity ?Ts is in fact the study of the spectrum {}of a regular function f.At last (but not least), I would like to have some insightabout discrete approximation results.For example, when I have a grid that approximates the domainO, and the laplacian is replaced by a finite differencescheme.Best === wemhexskermOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)If S is selfadjoint (not symmetric) ts is very easy: first make thematrices act on C^n, so that you can take scalar products of(eigen)vectors. Then let x be a (non zero) eigenvector of S+iD witheigenvalue lambda. Then Sx+iDx=lambda xTake the scalar product with x of both sides of ts equation(x|S|x)+i(x|D|x)=||x||^2 lambda.D is positive and therefore (x|D|x) is positive. S is selfadjoint andso (x|S|x) is real. The left hand side of the second equation is thena complex number with positive imaginary part. The imaginary part ofthe right hand side is ||x||^2 times Im lambda and so it must bepositive. If S is symmetric, but not selfadjoint, I don't tnk you have thedesired conclusion. Unless my calculations fail me (ts may be thecase, since I do not have a piece of paper handy) takingS = antidiagonal 2x2 matrix with sqrt(2) times i on the antidiagonalD = the 2x2 identity matrixwe get | i i sqrt(2)|S+iD = | | |i sqrt(2) i |wch has eigenvalues i(1+sqrt(2)) and i(1-sqrt(2)). Ts last oneclearly === undecided-unixEpigone-thread: wemhexskermOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)If S is selfadjoint (not symmetric) ts is very easy: first make thematrices act on C^n, so that you can take scalar products of(eigen)vectors. Then let x be a (non zero) eigenvector of S+iD witheigenvalue lambda. Then Sx+iDx=lambda xTake the scalar product with x of both sides of ts equation(x|S|x)+i(x|D|x)=||x||^2 lambda.D is positive and therefore (x|D|x) is positive. S is selfadjoint andso (x|S|x) is real. The left hand side of the second equation is thena complex number with positive imaginary part. The imaginary part ofthe right hand side is ||x||^2 times Im lambda and so it must bepositive. If S is symmetric, but not selfadjoint, I don't tnk you have thedesired conclusion. Unless my calculations fail me (ts may be thecase, since I do not have a piece of paper handy) takingS = antidiagonal 2x2 matrix with sqrt(2) times i on the antidiagonalD = the 2x2 identity matrixwe get | i i sqrt(2)|S+iD = | | |i sqrt(2) i |wch has eigenvalues i(1+sqrt(2)) and i(1-sqrt(2)). Ts last === undecided-unixOriginator: bergv@math.uiuc.edu (Maarten Bergvelt) steveare you sure, that ts is true?the eigenvalues are the solution of the equation det(xI-(S+iD))=0let a+ib be a solution for the equation, then a-ib is also a solution.so you have -b and b as imaginary parts.if b is not equal to 0 then you have one positive and one negative imaginarypart.danielsteve fisk schrieb im Newsbeitrag> Does anyone know a reference/proof for the following?> S is a symmetric matrix.> D is a diagonal matrix with positive values on the diagonal.> i is sqrt(-1)> then the eigenvalues of S + i D all have positive imaginary === bergv@math.uiuc.edu (Maarten Bergvelt)> Does anyone know a reference/proof for the following?S is a symmetric matrix.> D is a diagonal matrix with positive values on the diagonal.> i is sqrt(-1)then the eigenvalues of S + i D all have positive imaginary part.Counterexample: D the identity and S is -1000i*D.Probably you want S to be Hermitian. Then the statement follows bylooking at the Rayleigh quotient x^*(S+iD)x/x^*xwhere x is an eigenvector.Arnold === bergv@math.uiuc.edu (Maarten Bergvelt)> Does anyone know a reference/proof for the following?S is a symmetric matrix.> D is a diagonal matrix with positive values on the diagonal.> i is sqrt(-1)then the eigenvalues of S + i D all have positive imaginary part., Steve,is ts some homework problem?What do you know about the eigenvalues of symmetric matrices?Look at the eigensystem equation:(S+iD - lambda*UnitMatrix)*x=0 and try to rewrite it as(S - sigma*UnitMatrix)*x=0.What do you know about === undecided-unixOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)>Does anyone know a reference/proof for the following?>S is a symmetric matrix.Real, I presume.>D is a diagonal matrix with positive values on the diagonal.>i is sqrt(-1)>then the eigenvalues of S + i D all have positive imaginary part.Suppose S + i D has an eigenvalue s - i t where t >= 0, s real.Replacing S by S - sI, we may assume s = 0. Say an eigenvectoris a + i b where a and b are real. Then by expanding (S + iD)(a + ib) = -i t (a + ib) and taking real and imaginaryparts, we have S a - D b = t bS b + D a = -t aand thus(D+tI) b = S a(D+tI) a = -S bNoting that D+tI is positive definite and thus invertible, let(D+tI)^(-1) = Rwhere R is again positive definite. Now b = R S a and a = -R S b so (R S)^2 a = -a, and(R S)^2 has eigenvalue -1. But (R S)^2 = R^(1/2) R^(1/2) S R S has the same eigenvalues as R^(1/2) S R S R^(1/2), wch is positivesemidefinite and therefore can't have a negative eigenvalue,contradiction.