# Tag Info

9

This is an extended version of my observations in the comments. The upshot is that there exist pointed convex cones without a convex base, but every cone has a base. Hence what the OP is trying to do is bound not to work. (1) There are pointed convex cones that do not have a convex base. To see this, take $V=\mathbb{R}^2$ as a simple example, with $C$ ...

9

Take $\mathcal X = L^1(\Omega,\mu)$ where $\Omega = \{1,2,3,\dots\}$ and measure $\mu(\{k\}) = p_k$ with $p_k > 0$, $\sum p_k = 1$. The norm in $\mathcal X$ is $$\|f\|_{\mathcal X} = \sum_k |f(k)|p_k .$$ Let $\mathcal Y = \mathbb R^2$ with norm $$\|(x,y)\|_{\mathcal Y} = \frac{1}{2}|x|+\frac{1}{2}|y|.$$ Thus $\mathcal Y$ is also $L^1$ of a ...

7

This is the problem of finding spherical codes. Putatively optimal solutions can be found at Neil Sloane's website. For an upper bound, there's $d\leq\sqrt{4-csc^2[\frac{πn}{6(n-2)}]}$, where $d$ is the 3d distance between some two points.

7

The answer to the second part is also no. That is, $aa^*+bb^*$ could be zero for nonzero $a$ and $b$. Look at the $*$-algebra $\mathbb{C}^2$ with component-wise addition and multiplcation and $*$ given by $(x,y)^* = (\overline{y},\overline{x})$. It's straightforward to calculate that anything of the form $aa^*$ will look like $(z,\overline{z})$ for some $z\... 7 This is really just rephrasing Simon Henry's comment, but there is a natural transformation$F\to\text{Hom}(C,-)$given by$(f,s)\mapsto f$, and so if$\text{Cone}(C)$represents$F$then by Yoneda's lemma this natural transformation is induced by a map$C\to\text{Cone}(C)$. You can also see that this map is a monomorphism without using the construction of ... 5 The answer to the edited question is yes. For any non-empty, closed, convex cone$C\subseteq V$for any locally convex and hausdorff topological vector space$V$the equation$C^{\ast\ast}=C$holds. One inclusion follows immediately from the definition, the other follows easily by aiming for a contradiction and applying the Hahn-Banach theorem about ... 5 Presumably you are working in$\mathbb{R}^d$for$d > 3$. There are specialized algorithms in$d=2,3$. Here is one route. You have, essentially, what is known as the V-representation of your polytopal cones, V=vertex. An alternate representation is the H-representation, H=halfspace, the intersection of halfspaces. If you convert each cone's V-... 5 The decomposition of polyhedral cones as mentioned in Part 1 is treated in my article Completions of fans, J. Geom. 100 (2011), 147--169. Following Igor's wish I will give a short version here. (1) Consider a real vector space$V$and a family$(A_i)_{i\in I}$of polyhedral cones (called just cones in the following). The sum of vector spaces$\sum_{i\in I}\...

4

The answer to the first part of the question, in general, is no. Consider the following counter-example: let $\mathscr{D}(\mathbb{R}^{kd})$ be the complex vector space of (complex-valued) smooth functions with compact support on $\mathbb{R}^{kd}$ (seen as functions $f(x_1,\ldots,x_k)$ with $k>0$ arguments in $\mathbb{R}^d$), and define $$\mathfrak{A}=\... 4 The second fundamental form II of \partial V vanishes in the radial direction (along any line that goes through the origin), so one eigenvalue of II is zero. There are only two eigenvalues since \dim(\partial V)=2 and the mean curvature is (up to a factor of 2) the trace of II. Therefore the nonzero eigenvalue of II has the same sign as the mean ... 4 If the column of AU belongs (strictly) inside the space generated by the columns of U then you have a dominant eigenvector inside U (ie an eigenvector whose eigenvalue is real positive and of maximum modulus). It is Perron-Frobenius theorem. EDIT: Now given a matrix A, a necessary condition is to check that it admits a real positive dominant ... 4 The definition of an "interval" given in the current edit trivially implies that every interval has empty interior in the cone of PSD matrices. The same holds for any other convex cone of dimension at least 2. Therefore, by the Baire category theorem, a countable union of such intervals cannot cover the cone. The alternative definition given in the edit, ... 4 The answer to your question is "No" (but the first set, obviously, always contains the second). Example showing that the second set can be strictly smaller: Denote by \{e_n\} the unit vector basis in \ell_1 and consider the following set P:=\{e_1+\frac1ne_n\}_{n=2}^\infty in \ell_1. It is clear that e_1 is in the first set, but not in the second. 4 A good reference for volumetric arguments for the maximum number of 'cones' or spherical 'caps' that one can fit, is a series of papers by Jon Hamkins. The density of a packing of these caps can be at most \frac{\pi}{2\sqrt{3}}, (this being known as the Fejes Tóth bound) and in the minimal distance between centers of a packing, d, the density is ... 4 A few hours after having posted the question, a counterexample arose (this is the magic of MO). It can now be found in the question. Note However the converse is true in finite dimensions. Let V a real vector space of dimension n, then our maximal cone is contained in a (unique) closed half space (call H its hyperplan), all elements of the ... 4 To prove that (iii) implies (i), assume w.l.o.g. that \tau\neq\sigma. We first need to show that \tau\subset \partial \sigma. If this is not the case then either there exists a hyperplane H containing \tau such that H is not a supporting hyperplane for \sigma, or \tau is full-dimensional. If H exists then there are x,y\in\sigma on ... 3 Yes for dimension 2 (pick the cone with larger angle; one of its sides fits). No for larger dimensions. For a counterexample in dimension 3, let C_1 be a positive orthant, and let C_2 be a negative orthant rotated by \pi/3 around x=y=z. (Note that you need to check just the faces of C_1, since the cones can be swapped by an orthogonal transform.)... 3 @TobiasFritz's answer is very excellent. Here is another nice example of a Banach lattice with order unit. Consider a measurable set (S,\Sigma) and a \sigma-ring N\subseteq \Sigma, elements of which we call null. Now consider L_\infty(\Sigma|N) of all equivalence classes of bounded measurable functionals such that f\sim g if they differ on a null ... 3 Hall and Newman (1963) cite this work as Motzkin, T., Copositive quadratic forms. National Bureau of Standards Report 1818 (1952), pp. 11–12. This cited part of the NBS report is available online at this link on pages 259-260. Please notice however that it's not a standalone work, but rather a review of other (possibly unpublished) manuscripts. 2 Take some element in the dual cone g\in L^1([0,1],Y_+)^*\subset L^\infty_{w^*}([0,1],Y^*). Then, by definition, for every f\in L^1([0,1],Y_+)$$\int_0^1 \langle f(t),g(t)\rangle\mathbb{d}t\geq0$$In particular, for any y\in Y_+ and [a,b]\subset [0,1], we can take f \equiv y\cdot \chi_{[a,b]} so that$$\int_a^b \langle y,g(t)\rangle \mathbb{d}t\...

2

Another code you may wish to try is 4ti2's rays-function. It can also be called via polymake.

2

The prescribed group action allows to start from it and build the cones in question. It is well-known how the permutation representation of $S_n$ on 3-subsets decomposes into irreducibles - there will be just 4 of these, one of them trivial 1-dimensional one. With the other ones (more precisely, subsets of these) you have a graph embedding, affording the ...

2

In general, your problem can become infeasible under arbitrarily small perturbations $\Delta b$. For example, consider the problem $\min X_{11}+X_{22}$ subject to $X_{11}=0$ $X_{22}=0$ $X \succeq 0$. The only feasible solution $X=0$, and the problem becomes infeasible if $\Delta b_{1}$ or $\Delta b_{2}$ is negative. The book "Perturbation ...

2

A positive matrix is primitive and so it has only one nonzero eigenvalue of maximum modulus. An irreducible which is not primitive has (by definition) strictly more than one eigenvalue with maximum modulus. As basis change leave eigenvalues invariant, this shows that such change of basis might not exists if we only assume irreducibility.

2

A geometric interpretation of the envelope of a one-parameter family of surfaces is the following (slightly adapted from Wikipedia): a point on the envelope is a point in the intersection of two "adjacent" surfaces. If you don't mind infinitesimals this could be rephrased as: $(x_0,y_0,z_0)$ is a point on the envelope, if $(x_0,y_0,z_0)\in S_{a_0}\cap S_{a_0+... 2 Assume$C$and$\mathbb{R}_{\ge 0}^n$can be (non-strictly) separated by a subspace of dimension$n-1$. Then a normal vector$x$to that subspace lies in$\mathbb{R}_{\ge 0}^n$; see e.g. here for a proof. But then by your assumption, there exists$y \in C$on the same side of this subspace as$\mathbb{R}_{\ge 0}^n$, contradiction. 1 For$R=\mathbb{C}$, this is exercise 3.2.10 in Cox-Little-Schenck (most likely for the proof outlined there this assumption is not really necessary). 1 The proof can be found in H.H. Bauschke's 1996 doctoral dissertation: Projection Algorithms and Monotone Operators (p. 40, Theorem 3.3.6). P.S. I wonder what the downvote is for. 1 All such vectors$x\geq 0$form a polyhedron in$\mathbb{R}^n$, which may be empty. The latter can be certified using Farkas Lemma, as David points out in the comment. On the other hand, a nonempty polyhedron is a Minkowski sum of a subspace, a polyhedral cone, and a polytope. (see Wikipedia for terms I used). Your extra conditions on$M$and$y\$ imply ...

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