Introduction to Chemical Engineering Mathematics
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Math 4571 (Advanced Linear Algebra) Lecture #27
Math 4571 (Advanced Linear Algebra) Lecture #27 Applications of Diagonalization and the Jordan Canonical Form (Part 1): Spectral Mapping and the Cayley-Hamilton Theorem Transition Matrices and Markov Chains The Spectral Theorem for Hermitian Operators This material represents x4.4.1 + x4.4.4 +x4.4.5 from the course notes. Overview In this lecture and the next, we discuss a variety of applications of diagonalization and the Jordan canonical form. This lecture will discuss three essentially unrelated topics: A proof of the Cayley-Hamilton theorem for general matrices Transition matrices and Markov chains, used for modeling iterated changes in systems over time The spectral theorem for Hermitian operators, in which we establish that Hermitian operators (i.e., operators with T ∗ = T ) are diagonalizable In the next lecture, we will discuss another fundamental application: solving systems of linear differential equations. Cayley-Hamilton, I First, we establish the Cayley-Hamilton theorem for arbitrary matrices: Theorem (Cayley-Hamilton) If p(x) is the characteristic polynomial of a matrix A, then p(A) is the zero matrix 0. The same result holds for the characteristic polynomial of a linear operator T : V ! V on a finite-dimensional vector space. Cayley-Hamilton, II Proof: Since the characteristic polynomial of a matrix does not depend on the underlying field of coefficients, we may assume that the characteristic polynomial factors completely over the field (i.e., that all of the eigenvalues of A lie in the field) by replacing the field with its algebraic closure. Then by our results, the Jordan canonical form of A exists. -
Lecture Notes: Qubit Representations and Rotations
Phys 711 Topics in Particles & Fields | Spring 2013 | Lecture 1 | v0.3 Lecture notes: Qubit representations and rotations Jeffrey Yepez Department of Physics and Astronomy University of Hawai`i at Manoa Watanabe Hall, 2505 Correa Road Honolulu, Hawai`i 96822 E-mail: [email protected] www.phys.hawaii.edu/∼yepez (Dated: January 9, 2013) Contents mathematical object (an abstraction of a two-state quan- tum object) with a \one" state and a \zero" state: I. What is a qubit? 1 1 0 II. Time-dependent qubits states 2 jqi = αj0i + βj1i = α + β ; (1) 0 1 III. Qubit representations 2 A. Hilbert space representation 2 where α and β are complex numbers. These complex B. SU(2) and O(3) representations 2 numbers are called amplitudes. The basis states are or- IV. Rotation by similarity transformation 3 thonormal V. Rotation transformation in exponential form 5 h0j0i = h1j1i = 1 (2a) VI. Composition of qubit rotations 7 h0j1i = h1j0i = 0: (2b) A. Special case of equal angles 7 In general, the qubit jqi in (1) is said to be in a superpo- VII. Example composite rotation 7 sition state of the two logical basis states j0i and j1i. If References 9 α and β are complex, it would seem that a qubit should have four free real-valued parameters (two magnitudes and two phases): I. WHAT IS A QUBIT? iθ0 α φ0 e jqi = = iθ1 : (3) Let us begin by introducing some notation: β φ1 e 1 state (called \minus" on the Bloch sphere) Yet, for a qubit to contain only one classical bit of infor- 0 mation, the qubit need only be unimodular (normalized j1i = the alternate symbol is |−i 1 to unity) α∗α + β∗β = 1: (4) 0 state (called \plus" on the Bloch sphere) 1 Hence it lives on the complex unit circle, depicted on the j0i = the alternate symbol is j+i: 0 top of Figure 1. -
MATH 2370, Practice Problems
MATH 2370, Practice Problems Kiumars Kaveh Problem: Prove that an n × n complex matrix A is diagonalizable if and only if there is a basis consisting of eigenvectors of A. Problem: Let A : V ! W be a one-to-one linear map between two finite dimensional vector spaces V and W . Show that the dual map A0 : W 0 ! V 0 is surjective. Problem: Determine if the curve 2 2 2 f(x; y) 2 R j x + y + xy = 10g is an ellipse or hyperbola or union of two lines. Problem: Show that if a nilpotent matrix is diagonalizable then it is the zero matrix. Problem: Let P be a permutation matrix. Show that P is diagonalizable. Show that if λ is an eigenvalue of P then for some integer m > 0 we have λm = 1 (i.e. λ is an m-th root of unity). Hint: Note that P m = I for some integer m > 0. Problem: Show that if λ is an eigenvector of an orthogonal matrix A then jλj = 1. n Problem: Take a vector v 2 R and let H be the hyperplane orthogonal n n to v. Let R : R ! R be the reflection with respect to a hyperplane H. Prove that R is a diagonalizable linear map. Problem: Prove that if λ1; λ2 are distinct eigenvalues of a complex matrix A then the intersection of the generalized eigenspaces Eλ1 and Eλ2 is zero (this is part of the Spectral Theorem). 1 Problem: Let H = (hij) be a 2 × 2 Hermitian matrix. Use the Min- imax Principle to show that if λ1 ≤ λ2 are the eigenvalues of H then λ1 ≤ h11 ≤ λ2. -
Orthogonal Reduction 1 the Row Echelon Form -.: Mathematical
MATH 5330: Computational Methods of Linear Algebra Lecture 9: Orthogonal Reduction Xianyi Zeng Department of Mathematical Sciences, UTEP 1 The Row Echelon Form Our target is to solve the normal equation: AtAx = Atb ; (1.1) m×n where A 2 R is arbitrary; we have shown previously that this is equivalent to the least squares problem: min jjAx−bjj : (1.2) x2Rn t n×n As A A2R is symmetric positive semi-definite, we can try to compute the Cholesky decom- t t n×n position such that A A = L L for some lower-triangular matrix L 2 R . One problem with this approach is that we're not fully exploring our information, particularly in Cholesky decomposition we treat AtA as a single entity in ignorance of the information about A itself. t m×m In particular, the structure A A motivates us to study a factorization A=QE, where Q2R m×n is orthogonal and E 2 R is to be determined. Then we may transform the normal equation to: EtEx = EtQtb ; (1.3) t m×m where the identity Q Q = Im (the identity matrix in R ) is used. This normal equation is equivalent to the least squares problem with E: t min Ex−Q b : (1.4) x2Rn Because orthogonal transformation preserves the L2-norm, (1.2) and (1.4) are equivalent to each n other. Indeed, for any x 2 R : jjAx−bjj2 = (b−Ax)t(b−Ax) = (b−QEx)t(b−QEx) = [Q(Qtb−Ex)]t[Q(Qtb−Ex)] t t t t t t t t 2 = (Q b−Ex) Q Q(Q b−Ex) = (Q b−Ex) (Q b−Ex) = Ex−Q b : Hence the target is to find an E such that (1.3) is easier to solve. -
Solutions to Math 53 Practice Second Midterm
Solutions to Math 53 Practice Second Midterm 1. (20 points) dx dy (a) (10 points) Write down the general solution to the system of equations dt = x+y; dt = −13x−3y in terms of real-valued functions. (b) (5 points) The trajectories of this equation in the (x; y)-plane rotate around the origin. Is this dx rotation clockwise or counterclockwise? If we changed the first equation to dt = −x + y, would it change the direction of rotation? (c) (5 points) Suppose (x1(t); y1(t)) and (x2(t); y2(t)) are two solutions to this equation. Define the Wronskian determinant W (t) of these two solutions. If W (0) = 1, what is W (10)? Note: the material for this part will be covered in the May 8 and May 10 lectures. 1 1 (a) The corresponding matrix is A = , with characteristic polynomial (1 − λ)(−3 − −13 −3 λ) + 13 = 0, so that λ2 + 2λ + 10 = 0, so that λ = −1 ± 3i. The corresponding eigenvector v to λ1 = −1 + 3i must be a solution to 2 − 3i 1 −13 −2 − 3i 1 so we can take v = Now we just need to compute the real and imaginary parts of 3i − 2 e3it cos(3t) + i sin(3t) veλ1t = e−t = e−t : (3i − 2)e3it −2 cos(3t) − 3 sin(3t) + i(3 cos(3t) − 2 sin(3t)) The general solution is thus expressed as cos(3t) sin(3t) c e−t + c e−t : 1 −2 cos(3t) − 3 sin(3t) 2 (3 cos(3t) − 2 sin(3t)) dx (b) We can examine the direction field. -
Graph Equivalence Classes for Spectral Projector-Based Graph Fourier Transforms Joya A
1 Graph Equivalence Classes for Spectral Projector-Based Graph Fourier Transforms Joya A. Deri, Member, IEEE, and José M. F. Moura, Fellow, IEEE Abstract—We define and discuss the utility of two equiv- Consider a graph G = G(A) with adjacency matrix alence graph classes over which a spectral projector-based A 2 CN×N with k ≤ N distinct eigenvalues and Jordan graph Fourier transform is equivalent: isomorphic equiv- decomposition A = VJV −1. The associated Jordan alence classes and Jordan equivalence classes. Isomorphic equivalence classes show that the transform is equivalent subspaces of A are Jij, i = 1; : : : k, j = 1; : : : ; gi, up to a permutation on the node labels. Jordan equivalence where gi is the geometric multiplicity of eigenvalue 휆i, classes permit identical transforms over graphs of noniden- or the dimension of the kernel of A − 휆iI. The signal tical topologies and allow a basis-invariant characterization space S can be uniquely decomposed by the Jordan of total variation orderings of the spectral components. subspaces (see [13], [14] and Section II). For a graph Methods to exploit these classes to reduce computation time of the transform as well as limitations are discussed. signal s 2 S, the graph Fourier transform (GFT) of [12] is defined as Index Terms—Jordan decomposition, generalized k gi eigenspaces, directed graphs, graph equivalence classes, M M graph isomorphism, signal processing on graphs, networks F : S! Jij i=1 j=1 s ! (s ;:::; s ;:::; s ;:::; s ) ; (1) b11 b1g1 bk1 bkgk I. INTRODUCTION where sij is the (oblique) projection of s onto the Jordan subspace Jij parallel to SnJij. -
Parametrization of 3×3 Unitary Matrices Based on Polarization
Parametrization of 33 unitary matrices based on polarization algebra (May, 2018) José J. Gil Parametrization of 33 unitary matrices based on polarization algebra José J. Gil Universidad de Zaragoza. Pedro Cerbuna 12, 50009 Zaragoza Spain [email protected] Abstract A parametrization of 33 unitary matrices is presented. This mathematical approach is inspired by polarization algebra and is formulated through the identification of a set of three orthonormal three-dimensional Jones vectors representing respective pure polarization states. This approach leads to the representation of a 33 unitary matrix as an orthogonal similarity transformation of a particular type of unitary matrix that depends on six independent parameters, while the remaining three parameters correspond to the orthogonal matrix of the said transformation. The results obtained are applied to determine the structure of the second component of the characteristic decomposition of a 33 positive semidefinite Hermitian matrix. 1 Introduction In many branches of Mathematics, Physics and Engineering, 33 unitary matrices appear as key elements for solving a great variety of problems, and therefore, appropriate parameterizations in terms of minimum sets of nine independent parameters are required for the corresponding mathematical treatment. In this way, some interesting parametrizations have been obtained [1-8]. In particular, the Cabibbo-Kobayashi-Maskawa matrix (CKM matrix) [6,7], which represents information on the strength of flavour-changing weak decays and depends on four parameters, constitutes the core of a family of parametrizations of a 33 unitary matrix [8]. In this paper, a new general parametrization is presented, which is inspired by polarization algebra [9] through the structure of orthonormal sets of three-dimensional Jones vectors [10]. -
On the Eigenvalues of Euclidean Distance Matrices
“main” — 2008/10/13 — 23:12 — page 237 — #1 Volume 27, N. 3, pp. 237–250, 2008 Copyright © 2008 SBMAC ISSN 0101-8205 www.scielo.br/cam On the eigenvalues of Euclidean distance matrices A.Y. ALFAKIH∗ Department of Mathematics and Statistics University of Windsor, Windsor, Ontario N9B 3P4, Canada E-mail: [email protected] Abstract. In this paper, the notion of equitable partitions (EP) is used to study the eigenvalues of Euclidean distance matrices (EDMs). In particular, EP is used to obtain the characteristic poly- nomials of regular EDMs and non-spherical centrally symmetric EDMs. The paper also presents methods for constructing cospectral EDMs and EDMs with exactly three distinct eigenvalues. Mathematical subject classification: 51K05, 15A18, 05C50. Key words: Euclidean distance matrices, eigenvalues, equitable partitions, characteristic poly- nomial. 1 Introduction ( ) An n ×n nonzero matrix D = di j is called a Euclidean distance matrix (EDM) 1, 2,..., n r if there exist points p p p in some Euclidean space < such that i j 2 , ,..., , di j = ||p − p || for all i j = 1 n where || || denotes the Euclidean norm. i , ,..., Let p , i ∈ N = {1 2 n}, be the set of points that generate an EDM π π ( , ,..., ) D. An m-partition of D is an ordered sequence = N1 N2 Nm of ,..., nonempty disjoint subsets of N whose union is N. The subsets N1 Nm are called the cells of the partition. The n-partition of D where each cell consists #760/08. Received: 07/IV/08. Accepted: 17/VI/08. ∗Research supported by the Natural Sciences and Engineering Research Council of Canada and MITACS. -
Linear Independence, the Wronskian, and Variation of Parameters
LINEAR INDEPENDENCE, THE WRONSKIAN, AND VARIATION OF PARAMETERS JAMES KEESLING In this post we determine when a set of solutions of a linear differential equation are linearly independent. We first discuss the linear space of solutions for a homogeneous differential equation. 1. Homogeneous Linear Differential Equations We start with homogeneous linear nth-order ordinary differential equations with general coefficients. The form for the nth-order type of equation is the following. dnx dn−1x (1) a (t) + a (t) + ··· + a (t)x = 0 n dtn n−1 dtn−1 0 It is straightforward to solve such an equation if the functions ai(t) are all constants. However, for general functions as above, it may not be so easy. However, we do have a principle that is useful. Because the equation is linear and homogeneous, if we have a set of solutions fx1(t); : : : ; xn(t)g, then any linear combination of the solutions is also a solution. That is (2) x(t) = C1x1(t) + C2x2(t) + ··· + Cnxn(t) is also a solution for any choice of constants fC1;C2;:::;Cng. Now if the solutions fx1(t); : : : ; xn(t)g are linearly independent, then (2) is the general solution of the differential equation. We will explain why later. What does it mean for the functions, fx1(t); : : : ; xn(t)g, to be linearly independent? The simple straightforward answer is that (3) C1x1(t) + C2x2(t) + ··· + Cnxn(t) = 0 implies that C1 = 0, C2 = 0, ::: , and Cn = 0 where the Ci's are arbitrary constants. This is the definition, but it is not so easy to determine from it just when the condition holds to show that a given set of functions, fx1(t); x2(t); : : : ; xng, is linearly independent. -
Polynomial Sequences Generated by Linear Recurrences
Innocent Ndikubwayo Polynomial Sequences Generated by Linear Recurrences: Location and Reality of Zeros Polynomial Sequences Generated by Linear Recurrences: Location and Reality of Zeros Linear Recurrences: Location by Sequences Generated Polynomial Innocent Ndikubwayo ISBN 978-91-7911-462-6 Department of Mathematics Doctoral Thesis in Mathematics at Stockholm University, Sweden 2021 Polynomial Sequences Generated by Linear Recurrences: Location and Reality of Zeros Innocent Ndikubwayo Academic dissertation for the Degree of Doctor of Philosophy in Mathematics at Stockholm University to be publicly defended on Friday 14 May 2021 at 15.00 in sal 14 (Gradängsalen), hus 5, Kräftriket, Roslagsvägen 101 and online via Zoom, public link is available at the department website. Abstract In this thesis, we study the problem of location of the zeros of individual polynomials in sequences of polynomials generated by linear recurrence relations. In paper I, we establish the necessary and sufficient conditions that guarantee hyperbolicity of all the polynomials generated by a three-term recurrence of length 2, whose coefficients are arbitrary real polynomials. These zeros are dense on the real intervals of an explicitly defined real semialgebraic curve. Paper II extends Paper I to three-term recurrences of length greater than 2. We prove that there always exist non- hyperbolic polynomial(s) in the generated sequence. We further show that with at most finitely many known exceptions, all the zeros of all the polynomials generated by the recurrence lie and are dense on an explicitly defined real semialgebraic curve which consists of real intervals and non-real segments. The boundary points of this curve form a subset of zero locus of the discriminant of the characteristic polynomial of the recurrence. -
4.1 RANK of a MATRIX Rank List Given Matrix M, the Following Are Equal
page 1 of Section 4.1 CHAPTER 4 MATRICES CONTINUED SECTION 4.1 RANK OF A MATRIX rank list Given matrix M, the following are equal: (1) maximal number of ind cols (i.e., dim of the col space of M) (2) maximal number of ind rows (i.e., dim of the row space of M) (3) number of cols with pivots in the echelon form of M (4) number of nonzero rows in the echelon form of M You know that (1) = (3) and (2) = (4) from Section 3.1. To see that (3) = (4) just stare at some echelon forms. This one number (that all four things equal) is called the rank of M. As a special case, a zero matrix is said to have rank 0. how row ops affect rank Row ops don't change the rank because they don't change the max number of ind cols or rows. example 1 12-10 24-20 has rank 1 (maximally one ind col, by inspection) 48-40 000 []000 has rank 0 example 2 2540 0001 LetM= -2 1 -1 0 21170 To find the rank of M, use row ops R3=R1+R3 R4=-R1+R4 R2ØR3 R4=-R2+R4 2540 0630 to get the unreduced echelon form 0001 0000 Cols 1,2,4 have pivots. So the rank of M is 3. how the rank is limited by the size of the matrix IfAis7≈4then its rank is either 0 (if it's the zero matrix), 1, 2, 3 or 4. The rank can't be 5 or larger because there can't be 5 ind cols when there are only 4 cols to begin with. -
An Introduction to the HESSIAN
An introduction to the HESSIAN 2nd Edition Photograph of Ludwig Otto Hesse (1811-1874), circa 1860 https://upload.wikimedia.org/wikipedia/commons/6/65/Ludwig_Otto_Hesse.jpg See page for author [Public domain], via Wikimedia Commons I. As the students of the first two years of mathematical analysis well know, the Wronskian, Jacobian and Hessian are the names of three determinants and matrixes, which were invented in the nineteenth century, to make life easy to the mathematicians and to provide nightmares to the students of mathematics. II. I don’t know how the Hessian came to the mind of Ludwig Otto Hesse (1811-1874), a quiet professor father of nine children, who was born in Koenigsberg (I think that Koenigsberg gave birth to a disproportionate 1 number of famous men). It is possible that he was studying the problem of finding maxima, minima and other anomalous points on a bi-dimensional surface. (An alternative hypothesis will be presented in section X. ) While pursuing such study, in one variable, one first looks for the points where the first derivative is zero (if it exists at all), and then examines the second derivative at each of those points, to find out its “quality”, whether it is a maximum, a minimum, or an inflection point. The variety of anomalies on a bi-dimensional surface is larger than for a one dimensional line, and one needs to employ more powerful mathematical instruments. Still, also in two dimensions one starts by looking for points where the two first partial derivatives, with respect to x and with respect to y respectively, exist and are both zero.