Multivariate Laurent Orthogonal Polynomials and Integrable
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Parametrizations of K-Nonnegative Matrices
Parametrizations of k-Nonnegative Matrices Anna Brosowsky, Neeraja Kulkarni, Alex Mason, Joe Suk, Ewin Tang∗ October 2, 2017 Abstract Totally nonnegative (positive) matrices are matrices whose minors are all nonnegative (positive). We generalize the notion of total nonnegativity, as follows. A k-nonnegative (resp. k-positive) matrix has all minors of size k or less nonnegative (resp. positive). We give a generating set for the semigroup of k-nonnegative matrices, as well as relations for certain special cases, i.e. the k = n − 1 and k = n − 2 unitriangular cases. In the above two cases, we find that the set of k-nonnegative matrices can be partitioned into cells, analogous to the Bruhat cells of totally nonnegative matrices, based on their factorizations into generators. We will show that these cells, like the Bruhat cells, are homeomorphic to open balls, and we prove some results about the topological structure of the closure of these cells, and in fact, in the latter case, the cells form a Bruhat-like CW complex. We also give a family of minimal k-positivity tests which form sub-cluster algebras of the total positivity test cluster algebra. We describe ways to jump between these tests, and give an alternate description of some tests as double wiring diagrams. 1 Introduction A totally nonnegative (respectively totally positive) matrix is a matrix whose minors are all nonnegative (respectively positive). Total positivity and nonnegativity are well-studied phenomena and arise in areas such as planar networks, combinatorics, dynamics, statistics and probability. The study of total positivity and total nonnegativity admit many varied applications, some of which are explored in “Totally Nonnegative Matrices” by Fallat and Johnson [5]. -
1111: Linear Algebra I
1111: Linear Algebra I Dr. Vladimir Dotsenko (Vlad) Lecture 11 Dr. Vladimir Dotsenko (Vlad) 1111: Linear Algebra I Lecture 11 1 / 13 Previously on. Theorem. Let A be an n × n-matrix, and b a vector with n entries. The following statements are equivalent: (a) the homogeneous system Ax = 0 has only the trivial solution x = 0; (b) the reduced row echelon form of A is In; (c) det(A) 6= 0; (d) the matrix A is invertible; (e) the system Ax = b has exactly one solution. A very important consequence (finite dimensional Fredholm alternative): For an n × n-matrix A, the system Ax = b either has exactly one solution for every b, or has infinitely many solutions for some choices of b and no solutions for some other choices. In particular, to prove that Ax = b has solutions for every b, it is enough to prove that Ax = 0 has only the trivial solution. Dr. Vladimir Dotsenko (Vlad) 1111: Linear Algebra I Lecture 11 2 / 13 An example for the Fredholm alternative Let us consider the following question: Given some numbers in the first row, the last row, the first column, and the last column of an n × n-matrix, is it possible to fill the numbers in all the remaining slots in a way that each of them is the average of its 4 neighbours? This is the \discrete Dirichlet problem", a finite grid approximation to many foundational questions of mathematical physics. Dr. Vladimir Dotsenko (Vlad) 1111: Linear Algebra I Lecture 11 3 / 13 An example for the Fredholm alternative For instance, for n = 4 we may face the following problem: find a; b; c; d to put in the matrix 0 4 3 0 1:51 B 1 a b -1C B C @0:5 c d 2 A 2:1 4 2 1 so that 1 a = 4 (3 + 1 + b + c); 1 8b = 4 (a + 0 - 1 + d); >c = 1 (a + 0:5 + d + 4); > 4 < 1 d = 4 (b + c + 2 + 2): > > This is a system with 4:> equations and 4 unknowns. -
Section 2.4–2.5 Partitioned Matrices and LU Factorization
Section 2.4{2.5 Partitioned Matrices and LU Factorization Gexin Yu [email protected] College of William and Mary Gexin Yu [email protected] Section 2.4{2.5 Partitioned Matrices and LU Factorization One approach to simplify the computation is to partition a matrix into blocks. 2 3 0 −1 5 9 −2 3 Ex: A = 4 −5 2 4 0 −3 1 5. −8 −6 3 1 7 −4 This partition can also be written as the following 2 × 3 block matrix: A A A A = 11 12 13 A21 A22 A23 3 0 −1 In the block form, we have blocks A = and so on. 11 −5 2 4 partition matrices into blocks In real world problems, systems can have huge numbers of equations and un-knowns. Standard computation techniques are inefficient in such cases, so we need to develop techniques which exploit the internal structure of the matrices. In most cases, the matrices of interest have lots of zeros. Gexin Yu [email protected] Section 2.4{2.5 Partitioned Matrices and LU Factorization 2 3 0 −1 5 9 −2 3 Ex: A = 4 −5 2 4 0 −3 1 5. −8 −6 3 1 7 −4 This partition can also be written as the following 2 × 3 block matrix: A A A A = 11 12 13 A21 A22 A23 3 0 −1 In the block form, we have blocks A = and so on. 11 −5 2 4 partition matrices into blocks In real world problems, systems can have huge numbers of equations and un-knowns. -
Math 217: True False Practice Professor Karen Smith 1. a Square Matrix Is Invertible If and Only If Zero Is Not an Eigenvalue. Solution Note: True
(c)2015 UM Math Dept licensed under a Creative Commons By-NC-SA 4.0 International License. Math 217: True False Practice Professor Karen Smith 1. A square matrix is invertible if and only if zero is not an eigenvalue. Solution note: True. Zero is an eigenvalue means that there is a non-zero element in the kernel. For a square matrix, being invertible is the same as having kernel zero. 2. If A and B are 2 × 2 matrices, both with eigenvalue 5, then AB also has eigenvalue 5. Solution note: False. This is silly. Let A = B = 5I2. Then the eigenvalues of AB are 25. 3. If A and B are 2 × 2 matrices, both with eigenvalue 5, then A + B also has eigenvalue 5. Solution note: False. This is silly. Let A = B = 5I2. Then the eigenvalues of A + B are 10. 4. A square matrix has determinant zero if and only if zero is an eigenvalue. Solution note: True. Both conditions are the same as the kernel being non-zero. 5. If B is the B-matrix of some linear transformation V !T V . Then for all ~v 2 V , we have B[~v]B = [T (~v)]B. Solution note: True. This is the definition of B-matrix. 21 2 33 T 6. Suppose 40 2 05 is the matrix of a transformation V ! V with respect to some basis 0 0 1 B = (f1; f2; f3). Then f1 is an eigenvector. Solution note: True. It has eigenvalue 1. The first column of the B-matrix is telling us that T (f1) = f1. -
Determinants
12 PREFACECHAPTER I DETERMINANTS The notion of a determinant appeared at the end of 17th century in works of Leibniz (1646–1716) and a Japanese mathematician, Seki Kova, also known as Takakazu (1642–1708). Leibniz did not publish the results of his studies related with determinants. The best known is his letter to l’Hospital (1693) in which Leibniz writes down the determinant condition of compatibility for a system of three linear equations in two unknowns. Leibniz particularly emphasized the usefulness of two indices when expressing the coefficients of the equations. In modern terms he actually wrote about the indices i, j in the expression xi = j aijyj. Seki arrived at the notion of a determinant while solving the problem of finding common roots of algebraic equations. In Europe, the search for common roots of algebraic equations soon also became the main trend associated with determinants. Newton, Bezout, and Euler studied this problem. Seki did not have the general notion of the derivative at his disposal, but he actually got an algebraic expression equivalent to the derivative of a polynomial. He searched for multiple roots of a polynomial f(x) as common roots of f(x) and f (x). To find common roots of polynomials f(x) and g(x) (for f and g of small degrees) Seki got determinant expressions. The main treatise by Seki was published in 1674; there applications of the method are published, rather than the method itself. He kept the main method in secret confiding only in his closest pupils. In Europe, the first publication related to determinants, due to Cramer, ap- peared in 1750. -
Moments of Logarithmic Derivative SO
Alvarez, E. , & Snaith, N. C. (2020). Moments of the logarithmic derivative of characteristic polynomials from SO(2N) and USp(2N). Journal of Mathematical Physics, 61, [103506 (2020)]. https://doi.org/10.1063/5.0008423, https://doi.org/10.1063/5.0008423 Peer reviewed version Link to published version (if available): 10.1063/5.0008423 10.1063/5.0008423 Link to publication record in Explore Bristol Research PDF-document This is the author accepted manuscript (AAM). The final published version (version of record) is available online via American Institute of Physics at https://aip.scitation.org/doi/10.1063/5.0008423 . Please refer to any applicable terms of use of the publisher. University of Bristol - Explore Bristol Research General rights This document is made available in accordance with publisher policies. Please cite only the published version using the reference above. Full terms of use are available: http://www.bristol.ac.uk/red/research-policy/pure/user-guides/ebr-terms/ Moments of the logarithmic derivative of characteristic polynomials from SO(N) and USp(2N) E. Alvarez1, a) and N.C. Snaith1, b) School of Mathematics, University of Bristol, UK We study moments of the logarithmic derivative of characteristic polynomials of orthogonal and symplectic random matrices. In particular, we compute the asymptotics for large matrix size, N, of these moments evaluated at points which are approaching 1. This follows work of Bailey, Bettin, Blower, Conrey, Prokhorov, Rubinstein and Snaith where they compute these asymptotics in the case of unitary random matrices. a)Electronic mail: [email protected] b)Electronic mail (Corresponding author): [email protected] 1 I. -
Homogeneous Systems (1.5) Linear Independence and Dependence (1.7)
Math 20F, 2015SS1 / TA: Jor-el Briones / Sec: A01 / Handout Page 1 of3 Homogeneous systems (1.5) Homogenous systems are linear systems in the form Ax = 0, where 0 is the 0 vector. Given a system Ax = b, suppose x = α + t1α1 + t2α2 + ::: + tkαk is a solution (in parametric form) to the system, for any values of t1; t2; :::; tk. Then α is a solution to the system Ax = b (seen by seeting t1 = ::: = tk = 0), and α1; α2; :::; αk are solutions to the homoegeneous system Ax = 0. Terms you should know The zero solution (trivial solution): The zero solution is the 0 vector (a vector with all entries being 0), which is always a solution to the homogeneous system Particular solution: Given a system Ax = b, suppose x = α + t1α1 + t2α2 + ::: + tkαk is a solution (in parametric form) to the system, α is the particular solution to the system. The other terms constitute the solution to the associated homogeneous system Ax = 0. Properties of a Homegeneous System 1. A homogeneous system is ALWAYS consistent, since the zero solution, aka the trivial solution, is always a solution to that system. 2. A homogeneous system with at least one free variable has infinitely many solutions. 3. A homogeneous system with more unknowns than equations has infinitely many solu- tions 4. If α1; α2; :::; αk are solutions to a homogeneous system, then ANY linear combination of α1; α2; :::; αk is also a solution to the homogeneous system Important theorems to know Theorem. (Chapter 1, Theorem 6) Given a consistent system Ax = b, suppose α is a solution to the system. -
1 1.1. the DFT Matrix
FFT January 20, 2016 1 1.1. The DFT matrix. The DFT matrix. By definition, the sequence f(τ)(τ = 0; 1; 2;:::;N − 1), posesses a discrete Fourier transform F (ν)(ν = 0; 1; 2;:::;N − 1), given by − 1 NX1 F (ν) = f(τ)e−i2π(ν=N)τ : (1.1) N τ=0 Of course, this definition can be immediately rewritten in the matrix form as follows 2 3 2 3 F (1) f(1) 6 7 6 7 6 7 6 7 6 F (2) 7 1 6 f(2) 7 6 . 7 = p F 6 . 7 ; (1.2) 4 . 5 N 4 . 5 F (N − 1) f(N − 1) where the DFT (i.e., the discrete Fourier transform) matrix is defined by 2 3 1 1 1 1 · 1 6 − 7 6 1 w w2 w3 ··· wN 1 7 6 7 h i 6 1 w2 w4 w6 ··· w2(N−1) 7 1 − − 1 6 7 F = p w(k 1)(j 1) = p 6 3 6 9 ··· 3(N−1) 7 N 1≤k;j≤N N 6 1 w w w w 7 6 . 7 4 . 5 1 wN−1 w2(N−1) w3(N−1) ··· w(N−1)(N−1) (1.3) 2πi with w = e N being the primitive N-th root of unity. 1.2. The IDFT matrix. To recover N values of the function from its discrete Fourier transform we simply have to invert the DFT matrix to obtain 2 3 2 3 f(1) F (1) 6 7 6 7 6 f(2) 7 p 6 F (2) 7 6 7 −1 6 7 6 . -
Me Me Ft-Uiowa-Math2550 Assignment Hw6fall14 Due 10/09/2014 at 11:59Pm CDT
me me ft-uiowa-math2550 Assignment HW6fall14 due 10/09/2014 at 11:59pm CDT The columns of A could be either linearly dependent or lin- 1. (1 pt) Library/WHFreeman/Holt linear algebra/Chaps 1-4- early independent depending on the case. /2.3.40.pg Correct Answers: • B n Let S be a set of m vectors in R with m > n. Select the best statement. 3. (1 pt) Library/WHFreeman/Holt linear algebra/Chaps 1-4- /2.3.42.pg • A. The set S is linearly independent, as long as it does Let A be a matrix with more columns than rows. not include the zero vector. Select the best statement. • B. The set S is linearly dependent. • C. The set S is linearly independent, as long as no vec- • A. The columns of A are linearly independent, as long tor in S is a scalar multiple of another vector in the set. as they does not include the zero vector. • D. The set S is linearly independent. • B. The columns of A could be either linearly dependent • E. The set S could be either linearly dependent or lin- or linearly independent depending on the case. early independent, depending on the case. • C. The columns of A must be linearly dependent. • F. none of the above • D. The columns of A are linearly independent, as long Solution: (Instructor solution preview: show the student so- as no column is a scalar multiple of another column in lution after due date. ) A • E. none of the above SOLUTION Solution: (Instructor solution preview: show the student so- By theorem 2.13, a linearly independent set in n can contain R lution after due date. -
Review Questions for Exam 1
EXAM 1 - REVIEW QUESTIONS LINEAR ALGEBRA Questions (answers are below) Examples. For each of the following, either provide a specific example which satisfies the given description, or if no such example exists, briefly explain why not. (1) (JW) A skew-symmetric matrix A such that the trace of A is 1 (2) (HD) A nonzero singular matrix A 2 M2×2. (3) (LL) A non-zero 2 x 2 matrix, A, whose determinant is equal to the determinant of the A−1 Note: Chap. 3 is not on the exam. But this is a great one for next time. (4) (MS) A 3 × 3 augmented matrice such that it has infinitely many solutions (5) (LS) A nonsingular skew-symmetric matrix B. (6) (BP) A symmetric matrix A, such that A = A−1. (7) (CB) A 2 × 2 singular matrix with all nonzero entries. (8) (LB) A lower triangular matrix A in RREF. 3 (9) (AP) A homogeneous system Ax = 0, where A 2 M3×3 and x 2 R , which has only the trivial solution. (10) (EA) Two n × n matrices A and B such that AB 6= BA (11) (EB) A nonsingular matrix A such that AT is singular. (12) (EF) Three 2 × 2 matrices A, B, and C such that AB = AC and A 6= C. (13) (LD) A and B are n × n matrices with no zero entries such that AB = 0. (14) (OH) A matrix A 2 M2×2 where Ax = 0 has only the trivial solution. (15) (AL) An elementary matrix such that E = E−1. -
Numerically Stable Coded Matrix Computations Via Circulant and Rotation Matrix Embeddings
1 Numerically stable coded matrix computations via circulant and rotation matrix embeddings Aditya Ramamoorthy and Li Tang Department of Electrical and Computer Engineering Iowa State University, Ames, IA 50011 fadityar,[email protected] Abstract Polynomial based methods have recently been used in several works for mitigating the effect of stragglers (slow or failed nodes) in distributed matrix computations. For a system with n worker nodes where s can be stragglers, these approaches allow for an optimal recovery threshold, whereby the intended result can be decoded as long as any (n − s) worker nodes complete their tasks. However, they suffer from serious numerical issues owing to the condition number of the corresponding real Vandermonde-structured recovery matrices; this condition number grows exponentially in n. We present a novel approach that leverages the properties of circulant permutation matrices and rotation matrices for coded matrix computation. In addition to having an optimal recovery threshold, we demonstrate an upper bound on the worst-case condition number of our recovery matrices which grows as ≈ O(ns+5:5); in the practical scenario where s is a constant, this grows polynomially in n. Our schemes leverage the well-behaved conditioning of complex Vandermonde matrices with parameters on the complex unit circle, while still working with computation over the reals. Exhaustive experimental results demonstrate that our proposed method has condition numbers that are orders of magnitude lower than prior work. I. INTRODUCTION Present day computing needs necessitate the usage of large computation clusters that regularly process huge amounts of data on a regular basis. In several of the relevant application domains such as machine learning, arXiv:1910.06515v4 [cs.IT] 9 Jun 2021 datasets are often so large that they cannot even be stored in the disk of a single server. -
Solution to Section 3.2, Problem 9. (A) False. If the Matrix Is the Zero Matrix, Then All of the Variables Are Free (There Are No Pivots)
Solutions to problem set #3 ad problem 1: Solution to Section 3.2, problem 9. (a) False. If the matrix is the zero matrix, then all of the variables are free (there are no pivots). (b) True. Page 138 says that “if A is invertible, its reduced row echelon form is the identity matrix R = I”. Thus, every column has a pivot, so there are no free variables. (c) True. There are n variables altogether (free and pivot variables). (d) True. The pivot variables are in a 1-to-1 correspondence with the pivots in the RREF of the matrix. But there is at most one pivot per row (every pivot has its own row), so the number of pivots is no larger than the number of rows, which is m. ad problem 2: Solution to Section 3.2, problem 12. (a) The maximum number of 1’s is 11: 0 1 1 0 1 1 1 0 0 1 B 0 0 1 1 1 1 0 0 C B C @ 0 0 0 0 0 0 1 0 A 0 0 0 0 0 0 0 1 (b) The maximum number of 1’s is 13: 0 0 1 1 0 0 1 1 1 1 B 0 0 0 1 0 1 1 1 C B C @ 0 0 0 0 1 1 1 1 A 0 0 0 0 0 0 0 0 ad problem 3: Solution to Section 3.2, problem 27. Assume there exists a 3 × 3-matrix A whose nullspace equals its column space.