{"responseHeader":{"status":0,"QTime":7,"params":{"q":"{!q.op=AND}id:\"196852\"","hl":"true","hl.simple.post":"","hl.fragsize":"5000","fq":"!embargo_tdt:[NOW TO *]","hl.fl":"ocr_t","hl.method":"unified","wt":"json","hl.simple.pre":""}},"response":{"numFound":1,"start":0,"docs":[{"ark_t":"ark:/87278/s6186fqn","doi_t":"doi:10.26053/0H-PCBD-ETG0","setname_s":"ir_etd","subject_t":"Algebraic geometry; Tropical geometry","restricted_i":0,"department_t":"Mathematics","format_medium_t":"application/pdf","identifier_t":"etd3/id/3287","date_t":"2014-12","mass_i":1515012026,"publisher_t":"University of Utah","description_t":"Tropical geometry connects the fields of algebraic and polyhedral geometry. This connection has been used to discover much simpler proofs of fundamental theorems in algebraic geometry, including the Brill-Noether theorem. Tropical geometry has also found applications outside of pure mathematics, in areas as diverse as phylogenetic models and auction theory. This dissertation seeks to answer the question of when the minors of a symmetric matrix form a tropical basis. The first chapter introduces the relevant ideas and concepts from tropical geometry and tropical linear algebra. The second chapter introduces different notions of rank for symmetric tropical matrices. The third chapter is devoted to proving all the cases, outside symmetric tropical rank three, where the minors of a symmetric matrix form a tropical basis. The fourth chapter deals with symmetric tropical rank three. We prove that the 4 × 4 minors of an n×n symmetric matrix form a tropical basis if n ≤ 5, but not if n ≥ 13. The question for 5 < n < 13 remains open. The fifth chapter is devoted to when the minors of a symmetric matrix do not form a tropical basis. We prove the r × r minors of an n × n symmetric matrix do not form a tropical basis when 4 < r < n. We also prove that, when the minors of a matrix (general or symmetric) define a tropical variety and tropical prevariety that are different, then, with one exception, the two sets differ in dimension. The exception is the 4 × 4 minors of a symmetric matrix, where the question is still unresolved. The sixth chapter explores tropical conics. A correspondence between a property of the symmetric matrix of a quadric and the dual complex of that quadric is demonstrated for conics, and proposed for all quadrics. The seventh chapter reviews the results and proposes possible questions for further study. The first appendix is devoted to correcting a proof in a paper cited by this dissertation. The second appendix is a transcript of the Maple worksheets used to perform the computer calculations from the fifth chapter.","rights_management_t":"Copyright © Patrick Dylan Zwick 2014","title_t":"Variations on a theme of symmetric tropical matrices","id":196852,"publication_type_t":"dissertation","parent_i":0,"type_t":"Text","dissertation_institution_t":"University of Utah","thumb_s":"/e6/37/e63757c220df84c6066e0a3d5efee54a1577a27c.jpg","oldid_t":"etd3 3287","author_t":"Zwick, Patrick Dylan","metadata_cataloger_t":"CLR","format_t":"application/pdf","modified_tdt":"2018-05-01T14:45:17Z","dissertation_name_t":"Doctor of Philosophy","school_or_college_t":"College of Science","language_t":"eng","file_s":"/b7/49/b74982d1122bc4a7952e219bc9081b8e22d5ec89.pdf","format_extent_t":"5,083,554 bytes","created_tdt":"2015-02-11T00:00:00Z","permissions_reference_url_t":"https://collections.lib.utah.edu/details?id=1318473","_version_":1664094708349009920,"ocr_t":"VARIATIONS ON A THEME OF SYMMETRIC TROPICAL MATRICES by Patrick Dylan Zwick A dissertation submitted to the faculty of The University of Utah in partial fulfillment of the requirements for the degree of Doctor of Philosophy Department of Mathematics The University of Utah December 2014 Copyright c Patrick Dylan Zwick 2014 All Rights Reserved Th e Un i v e r s i t y o f Ut a h Gr a d u a t e Sc h o o l STATEMENT OF DISSERTATION APPROVAL The dissertation of Patrick Dylan Zwick has been approved by the following supervisory committee members: Aaron Bertram , Chair 5/20/2014 Date Approved Tommaso de Fernex , Member 5/20/2014 Date Approved Anurag Singh , Member 5/20/2014 Date Approved Tyler Jarvis , Member 5/20/2014 Date Approved Carleton DeTar , Member 5/20/2014 Date Approved and by Peter Trapa , Chair/Dean of the Department/College/School of Mathematics and by David B. Kieda, Dean of The Graduate School. ABSTRACT Tropical geometry connects the fields of algebraic and polyhedral geometry. This con-nection has been used to discover much simpler proofs of fundamental theorems in algebraic geometry, including the Brill-Noether theorem. Tropical geometry has also found applica-tions outside of pure mathematics, in areas as diverse as phylogenetic models and auction theory. This dissertation seeks to answer the question of when the minors of a symmetric matrix form a tropical basis. The first chapter introduces the relevant ideas and concepts from tropical geometry and tropical linear algebra. The second chapter introduces different notions of rank for symmetric tropical matrices. The third chapter is devoted to proving all the cases, outside symmetric tropical rank three, where the minors of a symmetric matrix form a tropical basis. The fourth chapter deals with symmetric tropical rank three. We prove that the 4 × 4 minors of an n×n symmetric matrix form a tropical basis if n ≤ 5, but not if n ≥ 13. The question for 5 < n < 13 remains open. The fifth chapter is devoted to when the minors of a symmetric matrix do not form a tropical basis. We prove the r × r minors of an n × n symmetric matrix do not form a tropical basis when 4 < r < n. We also prove that, when the minors of a matrix (general or symmetric) define a tropical variety and tropical prevariety that are different, then, with one exception, the two sets differ in dimension. The exception is the 4 × 4 minors of a symmetric matrix, where the question is still unresolved. The sixth chapter explores tropical conics. A correspondence between a property of the symmetric matrix of a quadric and the dual complex of that quadric is demonstrated for conics, and proposed for all quadrics. The seventh chapter reviews the results and proposes possible questions for further study. The first appendix is devoted to correcting a proof in a paper cited by this dissertation. The second appendix is a transcript of the Maple worksheets used to perform the computer calculations from the fifth chapter. To all my friends and family who generously supported me throughout. In particular to my father, Patrick Donovan Zwick, my mother, Rebecca Terry Heal, my stepfather, Gilbert \"Scott\" Heal, and my adviser, Aaron Bertram. \"City,\" he cried, and his voice rolled over the metropolis like thunder, \"I am going to tropicalize you.\" -The Satanic Verses, Salman Rushdie CONTENTS ABSTRACT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii LIST OF FIGURES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ix ACKNOWLEDGEMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x CHAPTERS 1. BASICS OF TROPICAL GEOMETRY AND TROPICAL LINEAR ALGEBRA. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.1 Ranks of Tropical Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Initial Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.3 Tropical Varieties and Prevarieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.4 Tropical Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 2. RANKS OF SYMMETRIC TROPICAL MATRICES . . . . . . . . . . . . . . . 10 2.1 Symmetric Kapranov Rank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.1.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.1.2 Elementary Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.1.3 Tropical Determinantal Varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.2 Symmetric Tropical Rank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.2.1 Tropical Determinantal Prevarieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 2.2.2 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 2.2.3 Cycle-Similar Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 2.2.4 Tropical and Symmetric Tropical Ranks . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.2.5 Basic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.3 Symmetric Barvinok Rank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 2.3.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 2.3.2 Example of Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 3. WHEN THEMINORS OF A SYMMETRICMATRIX FORMA TROPICAL BASIS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 3.1 Singular Symmetric Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 3.2 Rank One Symmetric Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 3.3 Rank Two Symmetric Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 3.3.1 Matrix Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 3.3.2 Kapranov and Tropical Rank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 3.3.3 Supporting Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 3.3.4 Completed Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 4. SYMMETRIC TROPICAL RANK THREE . . . . . . . . . . . . . . . . . . . . . . . . 45 4.1 Definitions and Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 4.1.1 Symmetric Scaling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 4.1.2 The Form Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 4.2 The Method of Joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 4.2.1 The Definition of Joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 4.2.2 Joints and Kapranov Rank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 4.3 The Exceptional Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 4.4 Searching for Joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 4.4.1 There Must Be a Transposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 4.4.2 Not Two Transpositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 4.4.3 The Case with Five Zeros . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 4.4.4 The Case with Six Zeros . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 4.4.5 The Case with Seven Zeros . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 4.4.6 The Case with Eight Zeros . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 4.4.7 The Case with Nine Zeros . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 4.5 The 4 × 4 Minors of a 5 × 5 Symmetric Matrix . . . . . . . . . . . . . . . . . . . . . . . 69 5. WHEN THEMINORS OF A SYMMETRICMATRIX DO NOT FORM A TROPICAL BASIS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 5.1 The Foundational Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 5.1.1 Rank Three . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 5.1.2 Rank Four . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 5.2 Dimension Growth of Determinantal Prevarieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 5.2.1 The Standard Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 5.2.2 The Symmetric Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 5.2.3 Dimension Growth for Standard Matrices . . . . . . . . . . . . . . . . . . . . . . . 78 5.2.4 Dimension Growth for Symmetric Matrices . . . . . . . . . . . . . . . . . . . . . . 80 5.3 The Base Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 5.3.1 The Standard Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 5.3.2 The Symmetric Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 5.4 The Dimension Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 5.4.1 The Standard Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 5.4.2 The Symmetric Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 6. TROPICAL QUADRICS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 6.1 Determinantal Profiles and Dual Complexes . . . . . . . . . . . . . . . . . . . . . . . . . . 92 6.2 Exploring Tropical Conics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 7. FURTHER QUESTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 7.1 Tropical Bases for Symmetric Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 7.2 The Dimensions of Determinantal Prevarieties . . . . . . . . . . . . . . . . . . . . . . . . 101 7.3 Dual Complexes of Tropical Quadrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 7.4 Shellability of Symmetric Rank Two Matrices . . . . . . . . . . . . . . . . . . . . . . . . 102 7.5 Computing and Comparing Symmetric Ranks . . . . . . . . . . . . . . . . . . . . . . . . 102 7.6 Other Special Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 APPENDICES vii A. CORRECTION TO A PROOF IN \"ON THE RANK OF A TROPICAL MATRIX\" BY DEVELIN, SANTOS, AND STURMFELS . . . . . . . . . . . 104 B. MAPLE CODE USED TO PERFORM THE COMPUTATIONS IN CHAPTER 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 viii LIST OF FIGURES 1.1 This tropical hypersurface contains the point (0, 0), but not the point (0, 1). . . 3 1.2 Two tropical lines intersecting at a ray. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.3 An example of a basis that is not a tropical basis, but in which both the tropical variety and tropical prevariety have the same dimension. . . . . . . . . . . 8 1.4 A connected example of a basis that is not a tropical basis, but in which both the tropical variety and tropical prevariety have the same dimension. . . . . . . . 9 6.1 Combinatorially distinct class of tropical conics 1 . . . . . . . . . . . . . . . . . . . . . . . 94 6.2 Combinatorially distinct class of tropical conics 2 . . . . . . . . . . . . . . . . . . . . . . . 95 6.3 Combinatorially distinct class of tropical conics 3 . . . . . . . . . . . . . . . . . . . . . . . 96 6.4 Combinatorially distinct class of tropical conics 4 . . . . . . . . . . . . . . . . . . . . . . . 97 6.5 Combinatorially distinct class of tropical conics 5 . . . . . . . . . . . . . . . . . . . . . . . 97 6.6 Combinatorially distinct class of tropical conics 6 . . . . . . . . . . . . . . . . . . . . . . . 98 6.7 Combinatorially distinct class of tropical conics 7 . . . . . . . . . . . . . . . . . . . . . . . 98 ACKNOWLEDGEMENTS This dissertation is the culmination of a lengthy, tortuous, and difficult journey. Along the way I've relied upon the help and encouragement of many comrades, and the list that follows is far from complete. Thank you to those who have helped me at home. Thank you to my brother, Baxter, with whom it's honestly been a pleasure to live these past years. I'm glad for the time we've shared, and I look forward to much more in the future. Thank you to my brother, Marek, who during my time in graduate school I've seen grow from a confused, uncertain teenager into a confident, competent, and focused man. Thank you to my dear friend, Brady Barten, who's always been there to listen, or just to relax. And thank you to my cat, Markov, for his humbling and grounding insistence that he is, in fact, the center of the universe, and that I am not. Thank you to my confreres. Thank you to my friends Matt Housley, Peter Marcy, Chris Remien, William Malone, Michael Purcell, Bobby Hanson, Jack Jeffries, and in particular Julian Chan, who have been both professional colleagues and personal friends. Getting to know you all has been one of the greatest pleasures of graduate school. Thank you to Lance Miller, Robert Easton, Drew Johnson, and Tyler Jarvis, for fruitful conversations not just about math. Thank you Lance, in particular, for your support. Thank you to Melody Chan and Brian Osserman for helpful correspondence. Thank you to Bernd Sturmfels who, although we've never met, seems like my second academic father. Thank you to all my math teachers throughout the years. Thank you Tomasso de Fernex, Anurag Singh, Tyler Jarvis, and Carleton DeTar for being on my supervisory committee. Thank you to the staff of the University of Utah Mathematics Department for, among many other things, being so helpful all those times I locked myself out of my office. Thank you to the 7-11 by President's Circle, for many late night coffees. Thank you to Mar´ıa del Mar Gonz´alez for being a caring and wonderful companion these last two years, and for your understanding, even when I was difficult. Thank you very much to my mother, Rebecca Heal, for being there for me from, literally, day one. Thank you to my father, Patrick Zwick, for your generosity, time, encouragement, and love. It's meant so much to me. Thank you to my stepfather, Gilbert \"Scott\" Heal, for being there for me when I needed you. I wouldn't have even begun, let alone finished, this dissertation without you, and I wish you were here to celebrate its completion with me. Not lastly but at last it's my pleasure and honor to thank my adviser and academic father, Aaron Bertram. Thank you for your support, your time, your intelligence, your guidance, and your patience. Thank you also for not settling for substandard work, even when, in fact especially when, I wanted you to do so. You haven't just made me a better mathematician, but a better person. I mentioned at the beginning that a Ph.D. in math, or in any field, is a difficult journey. I must also acknowledge how extremely fortunate I've been in having the opportunity to make the journey at all. If you, dear reader, are yourself on that journey let me just say that it is worth it. It might not always feel that way, or it might seem the journey will never end, but the experience of discovering something, even something small, at the frontier of man's knowledge is one of the greatest pleasures life has to offer. Mathematics is the pinnacle intellectual construction of mankind, and one could do worse than to contribute a stone, or even a pebble, to its elegant and beautiful structure. xi CHAPTER 1 BASICS OF TROPICAL GEOMETRY AND TROPICAL LINEAR ALGEBRA Tropical geometry is a relatively new area of mathematics that incorporates ideas and methods from both algebraic geometry and polyhedral geometry. As such, it is both interesting in its own right, and a source of possible tools and insights for approaching problems in related areas. This chapter introduces the fundamental definitions and concepts from tropical geometry and tropical linear algebra that will motivate the rest of the dissertation. A good general reference for the basics of tropical geometry is the book by Maclagan and Sturmfels [13]. 1.1 Ranks of Tropical Matrices In classical linear algebra there are many equivalent definitions of the rank of a matrix. In particular, the following three are equivalent: • The rank of a matrix A is the smallest integer r for which A can be written as the sum of r rank one matrices. A matrix has rank one if it is the outer product of a column vector and a row vector. • The rank of A is the smallest dimension of any linear space containing the columns of A. • The rank of A is the largest integer r such that A has a nonsingular r × r minor. Develin, Santos, and Sturmfels [8] define analogs of these three definitions for tropical matrices, and call them, respectively, the Barvinok rank, the Kapranov rank, and the tropical rank. These three definitions are not equivalent, and satisfy the inequalities tropical rank(A) ≤ Kapranov rank(A) ≤ Barvinok rank(A). Both inequalities may be strict ([8] Theorem 1.4). 2 In this dissertation we define and investigate analogs of these notions of rank for sym-metric matrices, with particular attention to Kapranov and tropical rank. We work over the tropical semiring (R,⊕,⊙), whose arithmetic operations are a ⊕ b := min(a, b) and a ⊙ b := a + b. Note that here we are working over the real numbers R, and not the extended real numbers R = R ∪ {∞}, and so for all the definitions of rank the minimum possible rank is one. In general, letters and variables will be upper case when working in the tropical semiring, and lower case when not. When denoting the element in row i and column j of a matrix these indices will be separated by a comma, so Ai,j is element (i, j) of the matrix A. The notation Aij will mean the submatrix formed by eliminating row i and column j from the matrix A. 1.2 Initial Definitions We first define the basic objects of tropical algebra and tropical geometry. Definition 1.1. A tropical monomial Xa1 1 · · ·Xam m is a symbol, and represents a function equivalent to the linear form P i aiXi (standard addition and multiplication). Definition 1.2. A tropical polynomial is a tropical sum of tropical monomials F(X1, . . . ,Xm) = M a∈A CaXa1 1 Xa2 2 · · ·Xam m , with A ⊂ Nm, Ca ∈ R (tropical addition and multiplication), and represents a piecewise linear convex function F : Rm → R. Note, unlike with standard polynomials, it is possible for two distinct tropical poly-nomials to represent the same linear convex function. For example, the distinct tropical polynomials F1 = X2 ⊕ Y 2, and X2 ⊕ 2XY ⊕ Y 2 represent the same linear convex function. Definition 1.3. The tropical hypersurface V(F) defined by a tropical polynomial F is the set of all points P ∈ Rm such that at least two monomials in F are minimal at P. This is also called the double-min locus of F. So, for example, the tropical hypersurface defined by the tropical polynomial 3 1X2 ⊕ XY ⊕ X ⊕ 1Y 2 ⊕ Y ⊕ 1 = min{2x + 1, x + y, x, 2y + 1, y, 1} would include the point (0, 0), but not the point (0, 1). This hypersurface is graphed in Figure 1.1. Just as in standard algebraic geometry, there is a tropical notion of projective space. Definition 1.4. The tropical projective space TPn−1 is the quotient of Rn by the equivalence relation (a1, . . . , an) ∼ (c ⊙ a1, . . . , c ⊙ an), where c ∈ R. As in standard algebraic geometry, a homogeneous tropical polynomial defines a projec-tive tropical hypersurface in tropical projective space. We will occasionally be working in tropical projective space, particularly in Chapter 6. When we are it will be made clear. 1.3 Tropical Varieties and Prevarieties Let k be an algebraically closed field. Let f ∈ k[x1, . . . , xm] be a polynomial. The set of points p ∈ km such that f(p) = 0 is a hypersurface, and is denoted V(f). Let I be a prime ideal of k[x1, . . . , xm]. The ideal I defines a algebraic variety, (or variety, for short) V(I), in km, which is the set of points p ∈ km such that f(p) = 0 for all f ∈ I. If I = (f1, . . . , fn) then the set {f1, . . . , fn} is a basis for I, and V(I) is equal to the set of points p ∈ km such that fi(p) = 0 for all fi in the basis. Put succinctly Figure 1.1. This tropical hypersurface contains the point (0, 0), but not the point (0, 1). 4 V(I) = \\ V(fi). So, a variety is an intersection of hypersurfaces. By the Hilbert basis theorem every ideal of k[x1, . . . , xm] is finitely generated, so any variety is a finite intersection of hypersurfaces. In the tropical setting there is an analog of a hypersurface, and we would like an analog of a variety. It might seem natural to define a tropical variety as the intersection of a finite set of tropical hypersurfaces, but these sets do not always have the properties we need in order for them to be useful analogs of algebraic varieties, and we instead call these sets tropical prevarieties. Definition 1.5. A tropical prevariety V(F1, . . . , Fn) is a finite intersection of tropical hypersurfaces: V(F1, . . . , Fn) = \\n i=1 V(Fi). A tropical variety is defined differently. First, one must define the field of Puiseux series. The use of this field goes all the way back to Isaac Newton [15], although the field is named after Puiseux, because he was the first to prove it is algebraicaly closed [17]. Let K = C{{t}} be the set of formal power series a = c1ta1 + c2ta2 + · · · , where a1 < a2 < a3 < · · · are rational numbers that have a common denominator. This set is an algebraically closed field of characteristic zero ([21], Theorem 2.4.3), and for any nonzero element a in this set we define the degree of a to be the value of the leading exponent a1. This gives us a degree map deg : K∗ → Q. For any two elements a, b ∈ K∗ we have deg(ab) = deg(a) + deg(b) = deg(a) ⊙ deg(b). Generally, we also have deg(a + b) = min(deg(a), deg(b)) = deg(a) ⊕ deg(b). The only case when this addition relation is not true is when a and b have the same degree, and the coefficients of the leading terms cancel. We would like to do tropical arithmetic over R, and not just over Q, so we enlarge the field of Puisieux series to allow this. Define the field ˜K by ˜K = ( X α∈A cαtα|A ⊂ R well-ordered, cα ∈ C ) . This field contains the field of Puisieux series, and is also an algebraically closed field of characteristic zero. We will define a tropical variety in terms of a variety over ˜K . 5 Definition 1.6. The degree map on ( ˜K ∗)m is the map T taking points (p1, . . . , pm) ∈ ( ˜K ∗)m to points (deg(p1), deg(p2), . . . , deg(pm)) ∈ Rm. A tropical variety is the image of a variety in ( ˜K ∗)m under the degree map. We call taking this image the tropicalization of a set of points in ( ˜K ∗)m. The tropicalization of a polynomial f ∈ ˜K [x1, . . . , xm] is the tropical polynomial T (f) formed by tropicalizing the coefficients of f, and converting addition and multiplication into their tropical counterparts. For example, the tropicalization of the polynomial f = 3t2xy − 7tx3 is the tropical polynomial T (f) = 2XY ⊕ 1X3. Sturmfels [20] proved that the tropicalization of a d-dimensional variety in ( ˜K∗)m is a pure d-dimensional polyhedral fan. That the dimension of the tropicalization is the dimension of the variety originates with Bieri and Groves [2]. In an unpublished manuscript from the early 1990s, Mikhail Kapranov proved the following useful and fundamental result. Theorem 1.7 (Kapranov's Theorem). For f ∈ ˜K [x1, . . . , xm] the tropical variety T (V(f)) is equal to the tropical hypersurface V(T (f)) determined by the tropical polynomial T (f). Given Kapranov's theorem if I = (f1, . . . , fn), then obviously the tropical prevariety determined by the set of tropical polynomials {T (f1), . . . , T (fn)} contains the tropical variety determined by I: T (V(I)) ⊆ \\n i=1 V(T (fi)). While Kapranov's theorem gives us the two sets are equal if N = 1, in general the containment may be strict. For example, the lines in ( ˜K ∗)2 defined by the linear equations f = 2x + y + 1, and g = tx + ty + 1, intersect at the point (t−1−1,−2t−1+1). The tropicalization of this point is (−1,−1), and so if I = (f, g) then T (V(I)) = (−1,−1). However, is we tropicalize the linear equations we get: 6 T (f) = X ⊕ Y ⊕ 0, and T (g) = 1X ⊕ 1Y ⊕ 0. Each of V(T (f)) and V(T (g)) is a tropical line, and their intersection is the tropical prevariety consisting of all points (a, a) with a ≤ −1. This intersection is graphed in Figure 1.2. This tropical prevariety properly contains the tropical variety (−1,−1), but the preva-riety is clearly much larger than the variety. That the intersection of two distinct tropical lines is not necessarily a point is an example of why we do not want to define a tropical variety to be a finite intersection of tropical hypersurfaces. 1.4 Tropical Bases There are sets of polynomials f1, . . . , fn ∈ ˜K [x1, . . . , xm] generating a prime ideal for which the tropical variety T \\n i=1 V(fi) ! and tropical prevariety Figure 1.2. Two tropical lines intersecting at a ray. 7 \\n i=1 V(T (fi)) are equal, and these sets are given special distinction. Definition 1.8. A basis for a prime ideal I = (f1, . . . , fn) ⊆ ˜K [x1, . . . , xm] is a tropical basis if T (V(I)) = \\n i=1 V(T (fi)). In [3] it is proven that every prime ideal I in ˜K [x1, . . . , xm] has a tropical basis, but not every basis is a tropical basis. A question of central importance to this dissertation, first posed by Chan, Jensen, and Rubei [7], is when the r × r minors of an n × n symmetric matrix form a tropical basis. In Chapters 2 through 5 we will answer this question entirely, apart from the 4 × 4 minors of an n × n symmetric matrix with 5 < n < 13. We saw earlier an example of a tropical prevariety that is not a tropical variety. Namely, two tropical lines intersecting at a ray. In this case the tropical prevariety corresponding to the basis is not just larger than the tropical variety, but is in fact of greater dimension. Generally, if a basis for a prime ideal is not a tropical basis, a natural question to ask is whether the corresponding tropical prevariety has a larger dimension than the corresponding tropical variety. That is to say, if I = (f1, . . . , fn) is a prime ideal, and the containment T (V(I)) ⊂ \\n i=1 V(T (fi)) is proper, is it the case that dim(T (V(I))) < dim \\n i=1 V(T (fi)) ! ? In general, the answer is no [16], as can be seen with the ideal I = ((x − 1)(x − t), (x − 1)(x − 2t)) ⊂ ˜K [x]. The tropical variety T (V(I)) is the point {0}, while the tropical prevariety V(X2 ⊕ X ⊕ 1) ∩ V(X2 ⊕ X ⊕ 1) is the set of points {0, 1}. This variety and prevariety are graphed in Figure 1.3. In this last example the tropical prevariety is disconnected, but this is not always the case. We can modify this example slightly to get an example of a connected tropical 8 Figure 1.3. An example of a basis that is not a tropical basis, but in which both the tropical variety and tropical prevariety have the same dimension. prevariety that is larger than its corresponding tropical variety, but does not have greater dimension. Specifically, the ideal J = ((x − y)(x − t), (x − y)(x − 2t)) ⊂ ˜K [x, y] Here the tropical variety T (V(J)) is the line X = Y , while the tropical prevariety V((X2 ⊕ XY ⊕ 1X ⊕ 1Y )) ∩ V((X2 ⊕ XY ⊕ 1X ⊕ 1Y )) is the union of the two lines X = Y and X = 1. This variety is graphed in Figure 1.4. For determinantal varieties of standard matrices, however, it is true that every time the r ×r minors of an m×n matrix of variables are not a tropical basis the tropical prevariety they define has greater dimension than the tropical variety they define. For determinantal varieties of symmetric matrices the same is true when r > 4. When r = 4 it is unknown whether it is true or not. As will be proven in Chapter 3, when r ≤ 3 the minors are always a tropical basis. These dimension inequalities will all be proven in Chapter 5. 9 Figure 1.4. A connected example of a basis that is not a tropical basis, but in which both the tropical variety and tropical prevariety have the same dimension. CHAPTER 2 RANKS OF SYMMETRIC TROPICAL MATRICES In this chapter we define three notions of rank for symmetric tropical matrices: the symmetric Kapranov rank, the symmetric tropical rank, and the symmetric Barvinok rank. These are the symmetric analogs of the corresponding three notions of rank for tropical matrices from [8]. Like their standard matrix analogs, these ranks are not equivalent, and satisfy the inequalities symmetric tropical rank ≤ symmetric Kapranov rank ≤ symmetric Barvinok rank. Both these inequalities may be strict. In this chapter we will prove all these inequalities, and prove the right inequality may be strict. We will prove in Chapter 5 that the left inequality may also be strict. We will focus mostly on symmetric tropical rank and symmetric Kapranov rank, de-scribing how they differ from their general matrix counterparts, and investigating some of their basic properties. 2.1 Symmetric Kapranov Rank Like the Kapranov rank, the symmetric Kapranov rank relates the rank of a symmetric real matrix to the smallest rank of an appropriate \"lift\" of that matrix. 2.1.1 Definition A lift of a real matrix A = (Ai,j) ∈ Rd×n is a matrix ˜ A = (˜ai,j) ∈ ( ˜K ∗)d×n such that deg(˜ai,j) = Ai,j for all i, j. The Kapranov rank of a matrix, as defined in [8], is the smallest rank of any lift of the matrix. For the symmetric Kapranov rank of a real symmetric matrix B ∈ Rn×n we require this lift ˜B to be symmetric. 11 Definition 2.1. The symmetric Kapranov rank of a real symmetric matrix is the smallest rank of any symmetric lift. Denote the set of m×n real matrices with Kapranov rank less than r by ˜ Tm,n,r. Denote the set of n×n real symmetric matrices with symmetric Kapranov rank less than r by ˜ Sn,r. 2.1.2 Elementary Properties Obviously, for any symmetric matrix A, Kapranov rank(A) ≤ symmetric Kapranov rank(A), and, as demonstrated in the example below, this inequality may be strict. Viewing both ˜ Sn,r and ˜ Tn,n,r as subsets of Rn×n we can write this relation as ˜ Sn,r ⊂ ˜ Tn,n,r. An example of a matrix with different Kapranov and symmetric Kapranov ranks is C3 := 1 0 0 0 1 0 0 0 1 . The reason for the terminology C3 will be explained later in this chapter in the section on Barvinok rank. C3 lifts to the rank two matrix t 1 1 + t 1 t 1 + t 1 + t −1 t , and so has Kapranov rank two. However, C3 does not lift to any symmetric rank two matrix. To prove this, for the sake of contradiction suppose there is a lift to a symmetric rank two matrix ˜ C3 := c1,1t + · · · c1,2 + · · · c1,3 + · · · c1,2 + · · · c2,2t + · · · c2,3 + · · · c1,3 + · · · c2,3 + · · · c3,3t + · · · , where ci,j ∈ C. If the first column of ˜ C3 were a ˜K -multiple of the second, ˜ c1 = ˜k˜c2, 12 then the relation from the first row c1,1t + · · · = ˜k(c1,2 + · · · ) would require deg(˜k) = 1, while the relation from the second row c1,2 + · · · = ˜k(c2,2t + · · · ) would require deg(˜k) = −1. This is a contradiction, and so the second column of ˜ C3 is linearly independent of the first. As the first two columns are linearly independent, if ˜ C3 has rank two there must be a linear combination of the first two columns equal to the third ˜k1˜c1 + ˜k2˜c2 = ˜c3. Explicitly, this equality is the three equalities: ˜k1(c1,1t + · · · ) + ˜k2(c1,2 + · · · ) = c1,3 + · · · ; ˜k1(c1,2 + · · · ) + ˜k2(c2,2t + · · · ) = c2,3 + · · · ; ˜k1(c1,3 + · · · ) + ˜k2(c2,3 + · · · ) = c3,3t + · · · . If deg(˜k2) < deg(˜k1) then from the first equality we must have deg(˜k2) = 0, but this would make the third equality impossible. If deg(˜k1) < deg(˜k2) then from the second equality we must have deg(˜k1) = 0, but this would also make the third equality impossible. If deg(˜k1) = deg(˜k2) then from the first equality (or the second) we must have deg(˜k1) = deg(˜k2) = 0. Suppose deg(˜k1) = deg(˜k2) = 0, and denote the leading terms of ˜k1 and ˜k2 by, respec-tively, k1 and k2. Then the first, second, and third equalities above, respectively, require: k2c1,2 = c1,3; k1c1,2 = c2,3; k1c1,3 = −k2c2,3. Substituting the first of these equalities into the left side of the third, and the second into the right side of the third, we derive the equality k1k2c1,2 = −k1k2c1,2. This cannot be as neither k1, k2, nor c1,2 is 0. So, C3 has no rank two symmetric lift, and its symmetric Kapranov rank is three. 13 2.1.3 Tropical Determinantal Varieties An equivalent definition of the Kapranov and symmetric Kapranov rank of a matrix can be given in terms of tropical varieties. A basic result in algebraic geometry is that the r × r minors of an m × n matrix of variables are a basis for a prime ideal, and the variety of this ideal corresponds with the set of m × n matrices of rank less than r. Similarly, the r × r minors of an n × n symmetric matrix of variables are a basis for a prime ideal, and the variety of this ideal corresponds with the set of n × n symmetric matrices of rank less than r. See Harris [11], for example, as a general reference for these results. Proposition 2.2. The Kapranov rank of an m×n real matrix is the smallest r ≤ min(m, n) such that the matrix is not in the set T (V(Ir)), where Ir is the ideal formed by the r × r minors of an m × n matrix of variables. Similarly, the symmetric Kapranov rank of an n×n real symmetric matrix is the smallest r ≤ n such that the matrix is not in the set T (V(Jr)), where Jr is the ideal formed by the r × r minors of an n × n symmetric matrix of variables. Proof. If A is an m× n real matrix and r is the smallest r ≤ min(m, n) such that A is not in the set T (V(Ir)), then if r < min(m, n) by definition there does not exist a lift of A to an m × n matrix over ˜K with rank less than r, while there exists a lift of A to an m × n matrix over ˜K with rank less than r + 1, and therefore r is the smallest rank of a lift of A. A generic lift of A has rank min(m, n), and if r = min(m, n) this is the smallest rank lift of A. If A is an n × n symmetric real matrix the corresponding proof is identical. 2.2 Symmetric Tropical Rank Like the tropical rank, the symmetric tropical rank is an intrinsically tropical property of a symmetric real matrix, and does not depend upon any lift to a matrix over a valued field. 2.2.1 Tropical Determinantal Prevarieties The definition of when a square matrix is tropically singular is the analog of the definition from classical linear algebra. Definition 2.3 ([8] Definition 1.3). A square matrix A = (Ai,j) ∈ Rd×d is tropically singular if the minimum 14 tropdet(A) := M σ∈Sd A1,σ(1) ⊙ A2,σ(2) ⊙ · · · ⊙ Ad,σ(d) is attained at least twice. Here Sd denotes the symmetric group on {1, 2, . . . , d}. We call the number tropdet(A) defined above the tropical determinant of A, and any permutation σ such that tropdet(A) = A1,σ(1) ⊙ A2,σ(2) ⊙ · · · ⊙ Ad,σ(d) realizes the tropical determinant. So, equivalently, a square matrix A is tropically singular if more than one permutation realizes the tropical determinant. The tropical rank of a matrix A ∈ Rm×n is the largest integer r such that A has a tropically nonsingular r×r submatrix. An equivalent definition of the tropical rank of a matrix can be given in terms of tropical prevarieties. In particular, the tropical prevariety defined by the minors of an m×n matrix of variables. Proposition 2.4. The tropical rank of an m × n real matrix is the largest r ≤ min(m, n) such that the matrix is not in the tropical prevariety \\ i V(T (mi)), where {mi} is the set of r × r minors of an m × n matrix of variables. Proof. If X′ is an r × r submatrix of the m × n matrix of variables X := x1,1 x1,2 · · · x1,n x2,1 x2,2 · · · x2,n ... ... . . . ... xm,1 xm,2 · · · xm,n , then the determinant of X′ is the polynomial f := X σ (−1)sgn(σ) Y i xi,σ(ρ(i)), where i runs over the row indices of X′, σ runs over all permutations of the column indices, and ρ is the order-preserving bijection from the row indices to the column indices. The tropicalization of this polynomial will be the tropical polynomial F := T (f) = M σ K i Xi,σ(i), where here addition and multiplication are their tropical counterparts. If A is an m × n real matrix then A′, the submatrix of A with the same row and column indices as X′, is by 15 definition tropically singular if and only if A′ is on the tropical hypersurface V(F). As the tropical rank of A is the largest r such that A contains a nonsingular r × r submatrix the proposition is immediate. Suppose A is an m×n real matrix, and {i1, ı2, . . . , ir} and {j1, j2, . . . , jr} are subsets of {1, . . . ,m} and {1, . . . , n}, respectively. These subsets define an r × r submatrix A′ of A, with column indices {i1, . . . , ir} and row indices {j1, . . . , jr}. A tropical monomial of the form Kr k=1 Xik,ρ(ik), where ρ is a bijection from the column indices to the row indices, is a minimizing monomial for the submatrix A′ if, over all monomials defined by bijections from {i1, i2, . . . , ir} to {j1, j2, . . . , jr}, this monomial is minimal under the valuation Xi,j 7→ Ai,j . In particular, an r × r submatrix of A is tropically nonsingular if and only if it has a unique minimizing monomial. For example, the upper-left principal 3 × 3 submatrix of 2 0 3 0 0 0 5 0 2 1 0 7 1 2 4 0 has the unique minimizing monomial X1,2X2,1X3,3, while the upper-right 3 × 3 submatrix has two minimizing monomials, X1,2X2,4X3,3 and X1,4X2,2X3,3. 2.2.2 Definition Along the lines of Proposition 2.4, we will define the symmetric tropical rank of a real symmetric matrix in terms of tropical prevarieties. However, before we do so, let us examine the symmetric matrix 1 0 0 1 1 0 1 0 1 1 0 0 1 0 1 1 1 0 0 0 1 1 1 0 1 . Viewed as a standard tropical matrix, the matrix has two minimizing monomials; X1,2X2,3X3,1X4,5X5,4, and X1,3X2,1X3,2X4,5X5,4. The upper-left 4 × 4 principal submatrix has three minimizing monomials; 16 X1,2X2,3X3,1X4,4, X1,3X2,1X3,2X4,4, and X1,2X2,1X3,4X4,3. The submatrix with columns {1, 2, 3, 5} and rows {1, 2, 3, 4} has two minimizing monomials; X1,2X2,3X3,1X4,5, and X1,3X2,1X3,2X4,5. So, viewed as a standard tropical matrix, both the matrix and these two submatrices would be tropically singular. However, if we view these monomials as coming from determinants of submatrices of a symmetric matrix of variables, then we have the equivalence Xi,j = Xj,i, and the matrix has only one minimizing monomial, namely X1,2X2,3X1,3X2 4,5. The upper-left 4 × 4 principal submatrix has two, and not three, minimizing monomials; X1,2X2,3X3,1X4,4, and X2 1,2X2 3,4. The submatrix with columns {1, 2, 3, 5} and rows {1, 2, 3, 4} has one minimizing monomial, X1,2X2,3X1,3X4,5. So, when viewed specifically as a symmetric matrix, in this example we would like for both the matrix and the given nonprincipal submatrix to be nonsingular, while the given principal submatrix would still be singular. Our definition of symmetric tropical rank is formulated with this in mind. Definition 2.5. The symmetric tropical rank of a symmetric n×n real matrix is the largest r ≤ n such that the matrix is not in the tropical prevariety \\ i V(T (mi)), where {mi} is the set of r × r minors of a symmetric n × n matrix of variables. An n × n symmetric real matrix is symmetrically tropically singular if it is in the tropical hypersurface defined by the tropicalization of the determinant of an n×n symmetric matrix of variables. An r × r submatrix of an n × n symmetric real matrix is symmetrically tropically singular if it is on the tropical hypersurface defined by the tropicalization of the determinant of the corresponding r×r submatrix of a symmetric n×n matrix of variables. So, the symmetric tropical rank of a symmetric real matrix A is the largest r such that A contains an r × r submatrix that is not symmetrically tropically singular. 17 You can view this as saying that a submatrix is symmetrically tropically singular if there are two permutations that realize its tropical determinant, and these permutations are not required to be equal given the symmetry of the matrix. In terms of minimizing monomials, a submatrix of a symmetric matrix is symmetrically tropically singular if it has two minimizing monomials that are distinct given the equivalence Xi,j = Xj,i. Denote the set of m × n real matrices with tropical rank less than r by Tm,n,r. Denote the set of n × n real symmetric matrices with symmetric tropical rank less than r by Sn,r. ˜ Sn,r is the tropical variety defined by the r × r minors of a symmetric n × n matrix of variables, Sn,r is the tropical prevariety defined by these same minors, and as a tropical variety defined by a basis is always contained in the corresponding tropical prevariety defined by that basis, we must have symmetric tropical rank(A) ≤ symmetric Kapranov rank(A). Equivalently, ˜ Sn,r ⊆ Sn,r. This is just a specific case of the tropical variety and tropical prevariety containment relation from Chapter 1. 2.2.3 Cycle-Similar Permutations We can construct an equivalent definition for when a symmetric matrix is symmetrically tropically singular by defining an equivalence class on the elements of Sn. We declare two permutations to be in the same class if they have the same disjoint cycle decomposition up to inversion of the cycles. In other words, if τ is a permutation with disjoint cycle decomposition: τ = σ1σ2 · · · σk, where the σi are disjoint cycles, then the other elements in its equivalence class are of the form: σ± 1 σ± 2 · · · σ± k . Note that as the parity of a permutation is determined completely by the sizes of the cycles in its disjoint cycle decomposition, and as a cycle and its inverse have the same size, every element in a given equivalence class has the same parity. 18 Denote by ˜ Sn this equivalence class of permutations in Sn. If two permutations are in the same equivalence class they are cycle-similar, and if not they are cycle-distinct. Denote the equivalence class containing the permutation τ by ˜τ . Proposition 2.6. A symmetric matrix is symmetrically tropically singular if and only if it has two cycle-distinct permutations that realize the determinant. Proof. Consider the symmetric matrix of variables: X := x1,1 x1,2 x1,3 · · · x1,n x1,2 x2,2 x2,3 · · · x2,n x1,3 x2,3 x3,3 · · · x3,n ... ... ... . . . ... x1,n x2,n x3,n · · · xn,n . For any cycle σ = (k1k2 · · · km) define the monomial xσ = xk1,k2xk2,k3 · · · xkm,k1 , and for any permutation τ ∈ Sn with disjoint cycle decomposition τ = σ1σ2 · · · σk define the monomial xτ = Yn i=1 xi,τ(i) = Yk i=1 xσi . We have xσ = xk1,k2xk2,k3 · · · xkm,k1 , and xσ−1 = xk1,km · · · kk3,k2xk2,k1 . As xi,j = xj,i we see xσ = xσ−1 , and therefore for any two cycle-similar permutations τ1 and τ2 we must have xτ1 = xτ2 . In other words, the permutations τ1 and τ2 produce the same monomial in the determinant of X. Note that as τ1 and τ2 have the same parity the monomials xτ1 and xτ2 have the same sign in the determinant, and there is no concern about identical monomials cancelling. On the other hand, suppose for two distinct permutations τ1 and τ2 that, given xi,j = xj,i, we have xτ1 = xτ2 . The permutation τ1 will have some disjoint cycle decomposition τ1 = σ1σ2 · · · σt. Suppose σ1 = (k1k2 · · · ks). 19 This means the variables xk1,k2xk2,k3 · · · xks,k1 appear in the product of variables defining the monomial xτ1 . If every one of these variables also appear in xτ2 , then the cycle σ1 also appears in the disjoint cycle decomposition of τ2. If this is the case for every cycle in the cycle decomposition of τ1, then τ1 = τ2. So, assume without loss of generality that σ1 is not in the disjoint cycle decomposition of τ2, and the variable xk1,k2 does not appear in xτ2 . As the only relation between the variables is xi,j = xj,i, if xk1,k2 does not appear in xτ2 , then xk2,k1 must. This means xk2,k3 cannot appear in xτ2 , and so xk3,k2 must. Repeating this argument we see that the product of variables xk2,k1xk3,k2 · · · xk1,ks must appear in xτ2 , which means τ2 must contain in its disjoint cycle decomposition the cycle (ksks−1 · · · k1) = (k1k2 · · · ks)−1. So, for every cycle in the disjoint cycle decomposition of τ1 either that cycle or its inverse appears in τ2, and obviously vice-versa. Ergo, τ1 and τ2 are cycle-similar. From this we conclude the distinct monomials occuring in the determinant of X are the cycle-distinct monomials, and therefore a symmetric matrix is symmetrically tropically singular if and only if it has two cycle-distinct permutations that realize the determinant. 2.2.4 Tropical and Symmetric Tropical Ranks If an r×r submatrix of a symmetric n×n matrix has two distinct minimizing monomials given the equivalence Xi,j = Xj,i then a fortiori it has two distinct minimizing monomials without that equivalence, and so tropical rank(A) ≤ symmetric tropical rank(A). Equivalently, viewing both Sn,r and Tn,n,r as subsets of Rn×n, we can write the above inequality as Sn,r ⊂ Tn,n,r. Just like with Kapranov rank and symmetric Kapranov rank, the tropical rank and symmetric tropical rank of a real symmetric matrix can be different. We can see this for 3 × 3 symmetric matrices. The determinant of the symmetric matrix of variables 20 x1,1 x1,2 x1,3 x1,2 x2,2 x2,3 x1,3 x2,3 x3,3 is the polynomial x1,1x2,2x3,3 + 2x1,2x2,3x1,3 − x22 ,3x1,1 − x21 ,2x3,3 − x21 ,3x2,2. In particular note that, because the matrix is symmetric, the monomial corresponding with the permutation (123) is the same as the monomial corresponding with the permutation (132). The tropicalization of this polynomial is the tropical polynomial X1,1X2,2X3,3 ⊕ X1,2X2,3X1,3 ⊕ X2 2,3X1,1 ⊕ X2 1,2X3,3 ⊕ X2 1,3X2,2. The symmetric matrix C3 defined in the last section has tropical rank two, but it is not on the tropical hypersurface defined by the tropical polynomial above, as X1,2X2,3X1,3 is the unique minimizing monomial for the entries in C3. So, C3 has symmetric tropical rank three. A more interesting example is provided by the matrix 1 1 0 1 0 0 0 0 1 1 0 1 0 0 0 0 1 1 0 1 0 0 0 0 1 1 0 1 1 0 0 0 1 1 0 0 1 0 0 0 1 1 1 0 1 0 0 0 1 . Using ideas and techniques from matroid theory, Develin, Santos, and Sturmfels proved ([8], Section 7) that this matrix, the cocircuit matrix of the Fano matroid, has tropical rank three but Kapranov rank four. If we permute the rows of this matrix with the permutation (275)(34), and the columns with the permutation (25364), we get the symmetric matrix: 1 1 0 0 1 0 0 1 0 1 1 0 0 0 0 1 1 0 0 0 1 0 1 0 1 0 1 0 1 0 0 0 0 1 1 0 0 0 1 1 0 1 0 0 1 0 1 1 0 . The upper-left 4×4 principal submatrix is tropically singular, but symmetrically tropically nonsingular. Consequently, the symmetric 7 × 7 matrix above has tropical rank three, but not symmetric tropical rank three. In particular, it is not an example of a symmetric matrix with symmetric tropical rank three and greater symmetric Kapranov rank. 21 2.2.5 Basic Properties One situation where tropical rank and symmetric tropical rank are necessarily equal is when both are one. Proposition 2.7. A real symmetric matrix A has tropical rank one if and only if it has symmetric tropical rank one. Proof. Rank one is the minimum possible rank. As tropical rank cannot be greater than symmetric tropical rank, symmetric tropical rank one implies tropical rank one. The determinant of a 2 × 2 submatrix of a symmetric n × n matrix of variables is the difference of two monomials, the product of the diagonal terms, and the product of the off-diagonal terms, and these monomials cannot be equal. If a matrix has tropical rank one, then for every 2 × 2 submatrix the sum of the diagonal terms equals the sum of the off-diagonal terms. This means every 2 × 2 submatrix is symmetrically tropically singular, and the matrix has symmetric tropical rank one. Corollary 2.8. If a real symmetric matrix has symmetric tropical rank two then it has tropical rank two. Proof. The tropical rank cannot be greater the symmetric tropical rank, and by the above proposition if the tropical rank were one, the symmetric tropical rank would be one as well. So, the tropical rank must be two. We have seen the matrix C3 has tropical rank two but greater symmetric tropical rank. This is a somewhat special situation. Proposition 2.9. A real symmetric matrix of tropical rank two has greater symmetric tropical rank if and only if a principal 3×3 submatrix is not symmetrically tropically singular. Proof. If any 3 × 3 submatrix of a real symmetric matrix is not symmetrically tropically singular, then the matrix has symmetric tropical rank greater than two. So, what must be proven is that for a real symmetric matrix if a 3×3 submatrix is not a principal submatrix then tropically singular implies symmetrically tropically singular. Take any 3 × 3 submatrix from an n × n symmetric matrix of variables xi,p xi,q xi,r xj,p xj,q xj,r xk,p xk,q xk,r , where i < j < k, and p < q < r. The determinant of this submatrix is the polynomial 22 xi,pxj,qxk,r + xi,qxj,rxk,p + xi,rxj,pxk,q − xi,pxj,rxk,q − xi,qxj,pxk,r − xi,rxj,qxk,p. Suppose, given the symmetry of the n×n matrix of variables, that two of these monomials are equal. If i < p then i is not the index of any column in our submatrix, and symmetry provides no duplication of variables from row i. This means if a monomial in the 3 × 3 determinant above is duplicated, the monomials in a 2 × 2 minor are duplicated. This is impossible. Identical logic applies if p < i, and therefore i = p. Applying the same argument we get j = q and k = r. So, the only situation where tropically singular and symmetrically tropically singular can differ for a 3 × 3 submatrix is if that submatrix is principal. In standard linear algebra if one column (or row) of a square matrix is a multiple of another, then that matrix must be singular. The same is true for symmetric tropical matrices. Proposition 2.10. If A is an r ×r submatrix of an n×n symmetric matrix, and one row of A is a tropical multiple of another, then A is symmetrically tropically singular. The same is true if one column of A is a tropical multiple of another. Proof. Suppose A is formed from the row indices i1, . . . , ir and the column indices j1, . . . , jr of the n × n symmetric matrix. Denote the rows of A by ai1 , ai2 , . . . , air . We may assume without loss of generality that air = c ⊙ air−1 , where c ∈ R. Suppose the monomial X1 = Xi1,ρ(i1) ⊙ Xi2,ρ(i2) ⊙ · · · ⊙ Xir−1,ρ(ir−1) ⊙ Xir,ρ(ir), where ρ is a bijection from the column indices of A to the row indices, is a minimizing monomial for A. Given the equivalence of air and c ⊙ air−1 the monomial X2 = Xi1,ρ(i1) ⊙ Xi2,ρ(i2) ⊙ · · · ⊙ Xir−1,ρ(ir) ⊙ Xir,ρ(ir−1) must have the same valuation as X1, and therefore also be a minimizing monomial. If X1 = X2 under the equivalence Xi,j = Xj,i then this would require one of the four equalities below to be true: Xir−1,σ(ir−1) = Xir−1,σ(ir); Xir−1,σ(ir−1) = Xσ(ir),ir−1 ; Xir−1,σ(ir−1) = Xir,σ(ir−1); Xir−1,σ(ir−1) = Xσ(ir−1),ir . Given ir−1 6= ir and ρ is a bijection, none of these equalities is possible. So, even under the equivalence Xi,j = Xj,i the minimizing monomials X1 and X2 are distinct, and therefore A is symmetrically tropically singular. An identical proof applies if one column is a tropical multiple of another. 23 The symmetric tropical rank exhibits some interesting properties that are not encoun-tered with the standard tropical rank. For example, suppose A is the real symmetric matrix 3 0 4 1 0 6 1 9 4 1 7 2 1 9 2 5 . The tropical determinant of A is realized by the permutation (1234), and also by the permutations (12)(34) and (14)(23). These permutations are all cycle-distinct, and therefore A is symmetrically tropically singular. In fact, for any 4 × 4 real symmetric matrix if the tropical determinant is realized by the permutation (1234), then the matrix is symmetrically tropically singular, and this is an example of a general phenomenon. Proposition 2.11. If a permutation has a disjoint cycle decomposition containing an odd-cycle larger than a transposition, then, if this permutation realizes the tropical determinant of a real symmetric matrix, the matrix must be symmetrically tropically singular. Proof. Suppose A is an n × n real symmetric matrix. For a permutation σ ∈ Sn we define the tropical product Aσ = Kn i=1 Ai,σ(i). For a cycle σ′ = (k1k2 · · · km) we define the tropical product Aσ′ = Ak1,k2 ⊙ Ak2,k3 ⊙ · · · ⊙ Akm,k1 . In particular, if σ has the disjoint cycle decomposition σ = σ1σ2 · · · σp, then Aσ = Kp i=1 Aσi . Suppose σ has the disjoint cycle decomposition above, and σj = (k1k2 · · · km) is an odd-cycle larger than a transposition. We define the permutations τ ′ and τ ′′ τ ′ = σ1σ2 · · · σj−1(k1k2)(k3k4) · · · (km−1km)σj+1 · · · σp, τ ′′ = σ1σ2 · · · σj−1(k2k3)(k4k5) · · · (kmk1)σj+1 · · · σp. As σj is an odd-cycle m must be even, and so τ ′ and τ ′′ are well-defined. As σj is larger than a transposition, σ, τ ′, and τ ′′ are cycle-distinct. As A is symmetric we have the following sequence of equalities (standard addition) 24 2Aσj = 2(Ak1,k2+Ak2,k3+· · ·+Akm,k1) = Ak1,k2+Ak2,k1+Ak2,k3+Ak3,k2+· · ·+Akm,k1+Ak1,km = (Ak1,k2 + Ak2,k1 + Ak3,k4 + Ak4,k3 + · · · + Akm−1,km + Akm,km−1) +(Ak2,k3 + Ak3,k2 + Ak4,k5 + Ak5,k4 + · · · + Akm,k1 + Ak1,km) = (A(k1k2) + A(k3k4) + · · ·A(kn−1kn)) + (A(k2k3) + A(k4k5) + · · · + A(kmk1)). From these equalities we get 2Aσ = Aτ′ + Aτ′′ . If the permutation σ realizes the tropical determinant of A, then we must have Aσ ≤ Aτ′ and Aσ ≤ Aτ′′ . These inequalities combined with the above equality give us Aσ = Aτ′ = Aτ′′ . As σ, τ ′, and τ ′′ are all cycle-distinct the matrix A is symmetrically tropically singular. So, if the disjoint cycle decomposition of σ contains an odd-cycle larger than a trans-position it is impossible for σ to realize the determinant of a symmetric matrix that is not symmetrically tropically singular. A natural question to ask, then, is whether this is the only type of permutation for which this is the case. The answer is yes. Proposition 2.12. Suppose that a permutation cannot realize the tropical determinant of any real symmetric matrix that is not symmetrically tropically singular. Then this permuta-tion has a disjoint cycle decomposition containing an odd-cycle larger than a transposition. Proof. We first note that for any permutation σ ∈ Sn we can find an n × n symmetric matrix for which σ realizes the tropical determinant. Using σ we define the matrix A such that Ai,σ(i) = Aσ(i),i = 0 for 1 ≤ i ≤ n, and all other terms in A are 1. Obviously the sum Aσ = Kn i=1 Ai,σ(i) = 0 is minimal over all permutations in Sn, and so σ realizes the tropical determinant. If σ realizing the tropical determinant requires that A is symmetrically tropically singular, then there is a permutation τ ∈ Sn also realizing the tropical determinant, where σ and τ are cycle-distinct. Suppose σ has the disjoint cycle decomposition σ = σ1σ2 · · · σp. 25 As σ and τ are cycle-distinct there must exist a σi = (k1k2 · · · km) such that neither σi nor σ−1 i is in a cycle-decomposition of τ . The only 0 terms on row k1 of A are Ak1,k2 , and Ak1,km. So, we must have either τ (k1) = k2 or τ (k1) = km. If τ (k1) = k2, then as the only 0 terms on row k2 of A are Ak2,k3 and Ak2,k1 , either τ (k2) = k3 or τ (k2) = k1. Suppose τ (k1) = k2 and τ (k2) = k3. The only 0 terms on row k3 of A are Ak3,k4 and Ak3,k2 . As τ (k1) = k2 we cannot have τ (k3) = k2, and so we must have τ (k3) = k4. Repeating this argument we get τ (kj−1) = τ (kj) for all 1 < j ≤ m, and τ (km) = k1. So, τ has a cycle decomposition containing σi, which cannot be. So, if τ (k1) = k2, then τ (k2) = k1. Using the same reasoning used in the paragraph above we get τ (k3) = k4, and either τ (k4) = k3 or τ (k4) = k5. Again, applying the same reasoning as the paragraph above if τ (k4) = k5 we must have τ (kj−1) = τ (kj) for all 3 < j ≤ m, and τ (km) = k1. As τ (k2) = k1 this cannot be, and so τ (k4) = k3. Repeating this argument we get that τ has a cycle decomposition containing the cycles (k1k2)(k3k4) · · · (km−1km). This is only possible if m is even, and as τ does not contain σi in its cycle decomposition we must have m > 2. So, σi is an odd-cycle larger than a transposition. If τ (k1) = km then an essentially identical argument gives us either τ has a cycle decomposition that contains σ−1 i , which cannot be, or τ has a cycle decomposition con-taining (k2k3)(k4k5) · · · (kmk1) with m > 2, implying σi is an odd-cycle larger than a transposition. 2.3 Symmetric Barvinok Rank The symmetric Barvinok rank of a symmetric matrix, along with two additional notions of rank for symmetric matrices (tree rank and star tree rank) have been examined in depth by Cartwright and Chan [4], from whom we lift the definition of symmetric Barvinok rank. We will not explore the symmetric Barvinok rank in much detail, except to prove symmetric Kapranov rank(A) ≤ symmetric Barvinok rank(A) and that this inequality may be strict. 2.3.1 Definition Definition 2.13. The symmetric Barvinok rank of a tropical symmetric matrix A is the smallest integer r for which A can be written as the sum of r rank one symmetric matrices. 26 We note that any rank one symmetric matrix can be written as the outer product X ⊙XT , where X is an n × 1 column vector. 2.3.2 Example of Inequality The proof of the above inequality is a straightforward modification of the proof for the corresponding inequality in Develin, Santos, and Sturmfels ([8], Proposition 3.6). Proposition 2.14. Every symmetric tropical matrix A satisfies symmetric Kapranov rank(A) ≤ symmetric Barvinok rank(A). Proof. If a matrix A has symmetric Barvinok rank one, then A = X ⊙ XT . If we pick a vector ˜X that tropicalizes to X, then the matrix ˜X ⊙ ˜X T will be a rank 1 symmetric matrix that tropicalizes to A. So, A has symmetric Kapranov rank 1 as well. Suppose the matrix A has symmetric Barvinok rank r. Write A = A1 ⊕ A2 ⊕ · · · ⊕ Ar. Each Ai has symmetric Kapranov rank one, so for each i there exists a rank one matrix A˜i that tropicalizes to Ai. By multiplying the matrices A˜i by random complex numbers, we can choose A˜i such that there is no cancallation of leading terms in t when we form the matrix A˜ = A˜1 + · · · + A˜r. This matrix A˜ has rank at most r, and tropicalizes to A. To prove this inequality can be strict we examine the classical identity matrix introduced in [8]: Cn = 1 0 0 · · · 0 0 1 0 · · · 0 0 0 1 · · · 0 ... ... ... . . . ... 0 0 0 · · · 1 . Note this is certainly not the identity matrix in tropical linear algebra, but it is sym-metric. Develin, Santos, and Sturmfels ([8], Examples 3.5 and 4.4) demonstrate the tropical and Kapranov ranks of Cn are both 2 for any n ≥ 2. We now prove the symmetric tropical and symmetric Kapranov ranks of Cn are both 3 for any n ≥ 3. Consider the matrix 27 t 1 1 16 5 + t · · · n2 n+1 + t 1 t 1 1 5 + 4t · · · 1 n+1 + nt 1 1 t 5 − 4 5 t · · · (n + 1) − n n+1 t 16 5 + t 1 5 + 4t 5 − 4 5 t 409 25 t · · · (4−n)2 5(n+1) + 5+21n+20n2 5(n+1) t ... ... ... ... . . . ... n2 n+1 + t 1 n+1 + nt (n + 1) − n n+1 t (n−4)2 5(n+1) + 5+21n+20n2 5(n+1) t · · · 1+2n+n2+2n3+n4 (n+1)2 t . This matrix, which we will denote by ˜ Cn, is defined as follows. The upper-left 3 × 3 submatrix is given. For the rest of the entries in the first three columns we define (i > 3, j ≤ 3) ci,j = c1,j + ic2,j − i 1 + i c3,j , which gives us ci,1 = i2 1 + i + t, ci,2 = 1 1 + i + it, ci,3 = (1 + i) − i 1 + i t. In particular, as i > 3, the constant term for all these elements is never 0. The remaining columns of ˜ Cn are defined in terms of the first three columns, in a matter exactly analogous to how we completed the first three columns above. For j > 3 ci,j = ci,1 + jci,2 − j 1 + j ci,3 = (i − j)2 (1 + i)(1 + j) + 1 + i + j + ij + i2j + ij2 + i2j2 (1 + i)(1 + j) t. The matrix ˜ Cn has rank three by construction, and from these formulas it is obviously symmetric. Also, the constant term is 0 if and only if i = j, and the linear term is never 0. So, ˜ Cn is a lift of Cn, and the symmetric Kapranov rank of Cn is at most three. The matrix Cn contains the matrix C3 as its upper-left 3 × 3 submatrix, and C3 has symmetric tropical rank three. So, the symmetric tropical rank of Cn is at least three. As the symmetric tropical rank cannot be greater than the symmetric Kapranov rank, both must be three. The symmetric Barvinok rank of Cn can be calculated using a proposition from [4], which we cite. Proposition 2.15 ([4], Proposition 3). If M is a symmetric matrix and 2mi,j < mi,i+mj,j for some i and j, then the symmetric Barvinok rank is infinite. 28 So, for n ≥ 2 the symmetric Barvinok rank of Cn is infinite, which certainly demonstrates the symmetric Barvinok rank can be greater than the symmetric Kapranov rank. CHAPTER 3 WHEN THE MINORS OF A SYMMETRIC MATRIX FORM A TROPICAL BASIS With the exception of r = 4, which is a special boundary case requiring a more in depth analysis, in this chapter we examine all the cases where the r ×r minors of an n×n symmetric matrix of variables do form a tropical basis. These are the cases r = 2, r = 3, and r = n. The case r = 4 is examined in Chapter 4. Before we prove this, we will want a couple of useful facts: • If A is a symmetric matrix, and we permute the rows of A by a permutation σ, and the columns of A by the same permutation, then the resulting matrix A′ will be symmetric, and A′ will have the same symmetric tropical and symmetric Kapranov rank as A. We call a permutation of the rows and columns of A by the same permutation a diagonal permutation. • If A is a symmetric matrix, and we tropically multiply row i by a constant c, and tropically multiply column i by the same constant, then the resulting matrix A′ will be symmetric, and A′ will have the same symmetric tropical and symmetric Kapranov rank as A. We call such an operation a symmetric scaling of A. In particular, we will assume without loss of generality that any symmetric matrix A has been symmetrically scaled so that every row/column has 0 as its minimal entry. 3.1 Singular Symmetric Matrices By definition, a symmetric matrix is singular if it has less than full rank, and it is a fundamental result in linear algebra that this is the case if and only if the matrix has zero determinant. Theorem 3.1. The determinant of a symmetric matrix of variables is a tropical basis for the ideal it generates. Equivalently, the n × n minor of an n × n symmetric matrix of variables forms a tropical basis. 30 Proof. The determinant of a symmetric matrix of variables is a single polynomial, and is therefore a tropical basis by Kapranov's theorem. So, an n × n symmetric matrix has symmetric tropical rank n if and only if it has symmetric Kapranov rank n, which is equivalently stated as ˜ Sn,n = Sn,n. If a symmetric matrix is symmetrically tropically singular, it has less than full tropical and Kapranov ranks, and for this symmetric matrix there exists a lift to a symmetric singular matrix over ˜K . 3.2 Rank One Symmetric Matrices The rank one case is straightforward. Theorem 3.2. A symmetric matrix has symmetric tropical rank one if and only if it has symmetric Kapranov rank one. Equivalently, the 2 × 2 minors of a symmetric matrix of variables are a tropical basis. Proof. As the symmetric tropical rank cannot be greater than the symmetric Kapranov rank, any symmetric matrix with symmetric Kapranov rank one must also have symmetric tropical rank one. If a symmetric matrix has symmetric tropical rank one, then by Proposition 2.6 it also has standard tropical rank one. This means every column of the matrix is a constant tropical multiple of the first column. If our matrix is of the form: A = a1,1 a1,2 · · · a1,n a2,1 a2,2 · · · a2,n ... .... . . ... an,1 an,2 · · · an,n , and ai represents column i of the matrix A, then ai = ci ⊙ a1 for some constant ci. By assumption A is symmetric, so ai,j = aj,i. The matrix A is the tropicalization of the matrix A˜ = ˜a1,1 ˜a1,2 · · · ˜a1,n ˜a2,1 ˜a2,2 · · · ˜a2,n ... ... . . . ... ˜an,1 ˜an,2 · · · ˜an,n , where a˜i,1 = tmi,1 , and a˜i,j = tcjm˜ i,1. The matrix A˜ has rank one by construction, and as ai,j = aj,i we have ˜ai,j = tcj ˜ai,1 = tcj+ai,1 = tai,j = taj,i = tci+aj,1 = tci taj,1 = tcia˜j,1 = a˜j,i. 31 So, A˜ is symmetric, and therefore A has Kapranov rank one. Corollary 3.3. A 3×3 symmetric matrix A has symmetric Kapranov rank two if and only if it has symmetric tropical rank two. Proof. If A has symmetric Kapranov rank two, then its symmetric tropical rank cannot be more than two, and by Theorem 3.2 its symmetric tropical rank cannot be one. If A has symmetric tropical rank two its symmetric Kapranov rank must be at least two, and by Theorem 3.1 its symmetric Kapranov rank cannot be three. 3.3 Rank Two Symmetric Matrices In this section we prove that the 3×3 minors of a symmetric n×n matrix form a tropical basis. We will a few times make the inductive assumption that, for a given natural number n, it is the case that the 3 × 3 minors of an m×m symmetric matrix form a tropical basis for m < n. The n = 3 case from Corollary 3.3 serves as the base. The proof will be built on the foundation of several lemmas. In several places the proof given here uses ideas and modifications of arguments from the corresponding proof in Section 6 of [8]. 3.3.1 Matrix Structure Lemma 3.4. Let A be a symmetric matrix of symmetric tropical rank two. After possibly a diagonal permutation A has the block structure: 0 0 0 0 0 0 B1 0 0 0 0 0 B2 0 0 0 0 0 0 C 0 0 0 CT 0 , where the matrices B1 and B2 are symmetric and positive, and the matrix C is non-negative and has no zero columns. Each 0 represents a zero matrix of the appropriate size. It is possible that A has no rows/columns with all 0 entries, and so the first row/column blocks of A may be empty. It is also possible that the matrices B1,B2 and C may have size zero. The only exceptions being A cannot be a matrix consisting of just one of the positive blocks (B1 or B2), nor can A be the zero matrix. Proof. Our proof follows the proof of Lemma 6.2 in [8], modified appropriately for symmetric matrices. In [8] they prove that if M is a real matrix normalized so that every column has 0 as its minimal entry, then if M has standard tropical rank two it has, after possibly permuting the rows and columns, the block structure: 32 0 0 0 · · · 0 0 M1 0 · · · 0 0 0 M2 · · · 0 ... ... ... . . . ... 0 0 0 · · · Mk . The matices Mi have all positive entries, each 0 represents a matrix of zeros of the appropriate size, and the first row and column blocks of M may have size zero. The block structure in the symmetric case is different because we have a modified definition of what it means for a submatrix to be singular, and we are only allowed to make diagonal permutations, not arbitrary permutaitons, of rows and columns. As defined in [9] the tropical convex hull of a set of real vectors {v1, . . . , vm} is the set of all tropical linear combinations c1 ⊙ v1 ⊕ c2 ⊙ v2 ⊕ · · · ⊕ cm ⊙ vm where c1, . . . , cm ∈ R. Theorem 4.2 from [8] states that the standard tropical rank of a real matrix is equal to one plus the dimension of the tropical convex hull of its columns. As the standard tropical rank of a matrix is equal to the standard tropical rank of its transpose, the standard tropical rank of a real matrix is also equal to one plus the dimension of the tropical convex hull of its rows. We construct a matrix A′ from A by adjoining the zero vector as the first column: A′ := 0 A . From A′ we construct the matrix A+ by adjoining the zero row as the first row: A+ := 0 A′ = 0 0 0 A . As the matrix A has symmetric tropical rank two, by Corollary 2.7 it must also have standard tropical rank two. Every row of A contains 0 as its minimal entry, and so the tropical convex hull of the columns of A′ is equal to the tropical convex hull of the columns of A. Therefore, the standard tropical rank of A′ is two. As every column of A′ contains zero as its minimal entry the tropical convex hull of the rows of A+ is equal to the tropical convex hull of the rows of A′. Therefore, the standard tropical rank of A+ is two. We derive the asserted block decomposition of A from the claim that any two columns of A+ have either equal or disjoint cosupports, where the cosupport of a column is the set of positions where it does not have a zero. To prove this, observe that if this were not so A+ would have the following submatrix, where + denotes a positive entry, ? denotes a 33 nonnegative entry, and the first column of the submatrix is taken from the first column of A+. (Recall that each column of A contains a zero entry.) 0 + + 0 0 + 0 ? 0 This 3 × 3 matrix is standard tropically nonsingular, which cannot be given A+ has standard tropical rank two. If the diagonal entry ai,i of A+ is positive, then, as A+ is symmetric, for any entry aj,i with j 6= i if aj,i is positive ai,j is as well, and this means columns i and j have equal cosupports. In particular, aj,j is positive. From this we see that the positive entries of column i, and the positive entries from columns with cosupports equal to column i, form a positive principal submatrix of A+. After possibly a diagonal permutation, this submatrix is the submatrix B1 of A+. If A+ contains additional positive diagonal entries outside of B1 then, using identical reasoning, possibly after a diagonal permutation we have the submatrix B2. There cannot be three positive diagonal blocks, for then we would be able to construct the 3 × 3 principal submatrix of A: a 0 0 0 b 0 0 0 c , where a, b, c > 0. This matrix is not symmetrically tropically singular, and this would contradict that A has symmetric tropical rank two. Note the difference here between the standard and the symmetric case. This 3×3 principal minor is not symmetrically tropically singular, but it is standard tropically singular. This is why in the standard rank two case the number of positive blocks can be arbitrarily large, while in the symmetric rank two case the number is limited to two. After possibly another diagonal permutation we can arrange the columns and rows of A+ so that, from left to right, the first columns are the zero columns, followed by the columns that contain B1, followed by the columns that contain B2. The remaining columns must all have a 0 entry on the diagonal, and a positive entry ai,j for some i 6= j. Row i obviously cannot be a zero row, nor can it intersect B1 or B2, and so must be below the submatrix B2. Denote as A′′ the submatrix formed by all columns to the right of B2, and all rows below B2. The submatrix A′′ is symmetric, does not contain a zero row/column, and has 0 along its diagonal. In particular its upper-left 1×1 principal submatrix is a zero matrix. Suppose the upper-left k × k principal submatrix of A′′ is a zero matrix. If for some column a′ i 34 all the terms in a′ i to the right of this k × k principal submatrix are 0, then the diagonal permutation that switches indices i and k + 1 will construct an upper-left (k + 1) × (k + 1) principal submatrix that is a zero matrix. We continue this process until no such column a′ i exists, in which case, given our result about either equal or disjoint cosupports, A′′, and therefore A, has our desired block decomposition. We note finally that A cannot be just a positive block, because that would violate the assumption that the minimum value in every row/column is 0. A also cannot be the zero matrix, for then it would have symmetric tropical rank one. Lemma 3.5. If A is a symmetric matrix normalized so the rows/columns have 0 as their minimal entry, and A+ is the augmented matrix A+ = 0 0 0 A , then: 1. If A has symmetric tropical rank two, so does A+. 2. If A has symmetric Kapranov rank two, so does A+. Proof Of Part (1). Suppose A has symmetric tropical rank two. We may assume that, possibly after a diagonal permutation, the matrix A has the block decomposition given in Lemma 3.4. In the proof of Lemma 3.4 we demonstrated that if A has symmetric tropical rank two, then A+ has standard tropical rank two. By Proposition 2.8 the only way a symmetric matrix can have standard tropical rank two but not symmmetric tropical rank two is if a principal 3 × 3 submatrix is standard tropically singular but not symmetrically tropically singular. By assumption, A has symmetric tropical rank two, so the only way A+ could not is if a principal 3×3 submatrix of A+ involving the initial zero row/column were tropically singular but not symmetrically tropically singular. The possible 3 × 3 principal submatrices of this type have the forms (where an element not specified as being 0 is positive): 0 0 0 0 0 0 0 0 0 , 0 0 0 0 0 0 0 0 ai,i , 0 0 0 0 ai,i 0 0 0 aj,j , 0 0 0 0 0 ai,j 0 aj,i 0 , 0 0 0 0 ai,i ai,j 0 aj,i aj,j . Of these possibilities the only one that is not obviously symmetrically tropically singular is the last one. For this possibility, if ai,i < ai,j or aj,j < ai,j then the submatrix is not 35 standard tropically singular, which cannot be. So, assume ai,j = aj,i < ai,i, aj,j . If A contains a zero row/column or the submatrix C (from Lemma 3.4) has positive size then, possibly after a diagonal permutation, A must contain the 3×3 matrix under examination as a principal submatrix, and therefore the submatrix must be symmetrically tropically singular. If A consists of two positive blocks and nothing else then A has the following 3×3 submatrix: ai,i ai,j 0 aj,i aj,j 0 0 0 ak,k , where ak,k > 0. If ai,j = aj,i < ai,i, aj,j then this matrix is not symmetrically tropically singular, which violates our assumption about A. So, A+ has symmetric tropical rank two. Part (2). If A has symmetric Kapranov rank two then there exists a rank two symmetric lift which we will call A˜. From Lemma 3.4 we know A must have two nonzero columns with disjoint cosupports. Denote as ˜ai and ˜aj the corresponding columns in ˜ A. If λ, μ ∈ ˜K have degree zero but are otherwise generic, then the vector ˜v = λ˜ai + μ˜aj has all degree zero terms. This is because as ai and aj have disjoint cosupports, the sum vk = λak,i + μak,j involves at least one term, ak,i or ak,j , of degree zero. If both have degree zero, then λ and μ being generic guarantees we do not have cancellation of leading terms. So, vk has degree zero. The matrix formed by adjoining v˜ to A˜, A˜′ := v˜ A˜ , must have rank two. If we augment A˜′ by adding a row formed by the linear combination of rows i and j of A˜′ multiplied by λ and μ, respectively, then as A˜ is symmetric we get the symmetric matrix A˜+ := ˜a0,0 ˜vT ˜v A . This matrix has rank two. The entry ˜a0,0 is: ˜a0,0 = λvi + λvk = λ(λai,i + μai,j) + μ(λaj,i + μaj,j) = λ2ai,i + 2λμai,j + μ2aj,j . 36 For the final equality we use ai,j = aj,i. At least one of ai,i, ai,j , aj,j has degree zero. As λ, μ are generic we cannot have cancellation of leading terms, and therefore ˜a0,0 has degree zero. So, the above matrix is a rank two symmetric lift of A+, and therefore A+ has symmetric Kapranov rank two. 3.3.2 Kapranov and Tropical Rank We now state the major theorem of this chapter, which has two implications. The proof of the simpler implication is given first. The proof of the more difficult implication is the subject of the rest of this chapter. An outline of the proof of this more difficult implication is provided below, followed by the complete proof. Theorem 3.6. A symmetric matrix A has symmetric tropical rank two if and only if it has symmetric Kapranov rank two. Symmetric Kapranov rank two implies symmetric tropical rank two. If A has symmetric Kapra-nov rank two then by Theorem 3.2 it cannot have symmetric tropical rank one. The symmetric tropical rank cannot be greater than the symmetric Kapranov rank, and so A must have tropical rank two. Outline that symmetric tropical rank two implies symmetric Kapranov rank two. Our method of proof for this implication is to first prove some special cases, and then use these special cases to construct our general proof. We may assume that if A has symmetric tropical rank two then it has the block decomposition given by Lemma 3.4. We prove the following special cases: Case 1 - Suppose A has the form 0 0 C 0 0 0 CT 0 0 , where C is nonnegative and has no zero column. If A has symmetric tropical rank two it has symmetric Kapranov rank two. Case 2 - Suppose A has the form B1 0 0 0 0 0 0 0 B2 , 37 where B1,B2 are positive symmetric matrices of positive size. If A has symmetric tropical rank two it has symmetric Kapranov rank two. Combining these two results, we prove: Case 3 - Suppose A has the form B1 0 0 0 0 0 B2 0 0 0 0 0 0 0 0 0 0 0 0 C 0 0 0 CT 0 , where B1,B2 are symmetric and positive, C is nonnegative and does not contain a zero column, and either C or both B1 and B2 have positive size. If A has symmetric tropical rank two it has symmetric Kapranov rank two. With this third case proven we will complete the proof of the theorem with a simple induction argument. The rest of this chapter is devoted to completing the proof sketched by this outline. 3.3.3 Supporting Lemmas We now prove the cases above as a series of lemmas. Lemma 3.7. Suppose A is a matrix of the form 0 0 C 0 0 0 CT 0 0 , where C is nonnegative and has no zero column. If A has symmetric tropical rank two, it has symmetric Kapranov rank two. Proof. We number the rows and columns of A from −k to l, where k × k and l × l are the dimensions of the upper-left and bottom-right zero matrices, respectively. So, the upper-left zero matrix is the submatrix of nonpositive indices, and the bottom-right zero matrix is the submatrix of nonnegative indices. The row and column indexed zero consists of all zeroes. Further, in A the rows and columns in the upper-left zero matrix will be referred to \"in reverse.\" So, the first and second columns of the upper-left zero matrix are indexed 0 and −1 in A. Elements from C or CT will be represented with an indexed lower-case \"c,\" while other elements will be represented with an indexed lower-case \"a.\" 38 As C does not contain a zero column we may, possibly after a diagonal permutation, assume the entries C−1,1 = C1,−1 are positive. We now construct a symmetric rank two lifting A˜ of A. The upper-right submatrix AUR = 0 C 0 0 has (standard) tropical rank two, and so by Theorem 6.5 from [8] there exists a rank two lift A˜UR of this submatrix.1 As C does not contain the zero column, the first two columns of A˜UR must be linearly independent, and every other column of A˜UR can be written as a linear combination of these first two columns: λja0 + μja1 = aj . The relation λja0,0 + μja0,1 = a0,j implies the degrees of λj and μj cannot both be positive, if one has positive degree the other must have degree zero, and if their degrees are both nonpositive they must be equal. If both λj and μj had negative degrees, then given C does not contain the zero column C would have a negative entry, but this is not allowed as C is nonnegative. If μj had positive degree then λj would have degree zero, but this cannot be as then C would contain the zero column. So, we must have deg(λj) ≥ deg(μj) = 0. We use this lift A˜UR, and its transpose, for the upper-right and bottom left submatrices of A˜. We must complete the lift with entries ai,j for every i, j with ij > 0, such that deg(ai,j) = 0, ai,j = aj,i, and the entire matrix A˜ has rank two. We begin this task with the 3 × 3 central minor: a−1,−1 a−1,0 c−1,1 a0,−1 a0,0 a0,1 c1,−1 a1,0 a1,1 . We pick a1,1 such that deg(a1,1) = 0, but otherwise generically. We want this matrix to be singular, and so once a1,1 has been picked a−1,−1 is determined. As a1,1 is generic, a−1,−1 is as well. If deg(a−1,−1) < 0, then in order for the above 3×3 matrix to be singular the leading terms in a0,0a1,1 − a0,1a1,0 would need to cancel, which is impossible if a1,1 is generic. If deg(a−1,−1) > 0, then as deg(c−1,1) = deg(c1,−1) > 0 there would only be a single degree zero term, a−1,0a0,−1a1,1, in the determinant of the 3 × 3 matrix, which would make it impossible for it to be singular. So, deg(a−1,−1) = 0. 1Theorem 6.5 from [8] relies upon Corollary 6.4 from the same paper, and Corollary 6.4 contains an error in its proof. A correction for this error is given in the first appendix of this dissertation. 39 From here every term ai,1 and ai,−1, with i > 1 or i < 1, respectively, is determined by the need for the matrices a−1,−1 a−1,0 c−1,1 a0,−1 a0,0 a0,1 ci,−1 ai,0 ai,1 and ai,−1 ai,0 ci,1 a0,−1 a0,0 a0,1 c1,−1 a1,0 a1,1 to be, respectively, singular, and that a1,1 and a−1,−1 are generic ensures all these terms are generic and have degree zero. The remaining entries i, j > 1 in the bottom-right zero matrix are determined by the relations: λjai,0 + μjai,1 = ai,j . As ai,1 is generic, deg(ai,j) = 0 even if deg(λj) = deg(μj). The degree zero upper-left entries are determined similarly. It remains to be proven that our lift is symmetric. We first prove a1,i = ai,1, with i > 1. We examine the matrices a−1,−1 a−1,0 c−1,i a0,−1 a0,0 a0,i c1,−1 a1,0 a1,i and a−1,−1 a−1,0 c−1,1 a0,−1 a0,0 a0,1 ci,−1 ai,0 ai,1 . By construction a−1,0 = a0,−1, a1,0 = a0,1, c−1,1 = c1,−1, and c−1,i = ci,−1. So, the formula for the determinant of the first matrix is the same as the formula for the determinant of the second with a1,i replaced by ai,1. As both matrices are singular we must have a1,i = ai,1. For the remaining terms verifying symmetry is a straightforward calculation (here i, j > 1): aj,i = λiaj,0 + μiaj,1 = λia0,j + μia1,j = λi(λja0,0 + μja0,1) + μi(λja1,0 + μja1,1) = λj(λia0,0 + μia1,0) + μj(λia0,1 + μia1,1) = λj(λia0,0 + μia0,1) + μj(λia1,0 + μia1,1) = λja0,i + μja1,i = λjai,0 + μjai,1 = ai,j . The verification of symmetry for i, j < −1 is essentially identical. So, we have constructed a symmetric rank two lift A˜ of A, and therefore A has Kapranov rank two. Lemma 3.8. Suppose A is a matrix of the form: 40 B1 0 0 0 0 0 0 0 B2 where B1 and B2 are positive symmetric matrices of positive size. If A has symmetric tropical rank two, then it has symmetric Kapranov rank two. Proof. As in the previous lemma we number the rows and columns from −k to l, where k × k and l × l are the dimensions of B1 and B2, respectively. Also, as in the previous lemma, we refer to the rows and columns of A \"in reverse.\" Terms from B1 or B2 will be represented with an indexed lower-case \"b,\" while all other terms will be represented with an indexed lower-case \"a.\" By induction we may assume the matrices B1 0 0 0 and 0 0 0 B2 have symmetric rank two lifts ˜B 1 and ˜B 2, respectively, and after possibly scaling we may assume the bottom-right entry of ˜B 1 is equal to the top-left entry of ˜B 2. We now construct a symmetric rank two lift ˜ A of A. We begin with the lifts ˜B 1 and ˜B 2, and construct the entries in the upper-right zero matrix. Like in Lemma 3.7 we start with the 3 × 3 central minor: b−1,−1 a−1,0 a−1,1 a0,−1 a0,0 a0,1 a1,−1 a1,0 b1,1 . We need this matrix to be singular and symmetric. That we can find degree zero en-tries a−1,1 = a1,−1 that make this true follows from applying Kapranov's theorem to the determinant of the matrix b−1,−1 a−1,0 x a0,−1 a0,0 a0,1 x a1,0 b1,1 . Note, we cannot assume a−1,1 = a1,−1 is generic, but that will not be necessary. Also, note that as the 3×3 central minor is singular, there cannot be cancellation of leading terms for either of its minors: a−1,0 a−1,1 a0,0 a0,1 , or a0,−1 a0,0 a1,−1 a1,0 . If in either of these minors we had cancellation of leading terms there would be no way the leading terms could all cancel for the determinant of the entire 3 × 3 matrix. Every term ai,1 with i < −1 and ai,−1 with i > 1 is determined by the need for the matrices 41 b−1,−1 a−1,0 a−1,1 a0,−1 a0,0 a0,1 ai,−1 ai,0 bi,1 and bi,−1 ai,0 ai,1 a0,−1 a0,0 a0,1 a1,−1 a1,0 b1,1 to be, respectively, singular. That every such term has degree zero follows from the leading terms of the 2 × 2 minors discussed above not canceling. Every column in ˜B 2 can be written as a linear combination of the first two: λjb0 + μjb1 = bj . We use these relations to define the entries ai,j with i < 0 and j > 0: λjai,0 + μjai,1 = ai,j . We similarly use the first two columns of ˜B 1 to define the terms ai,j with i > 0, j < 0. This determines a rank two matrix A˜. We must verify the matrix is symmetric, and is a lift of A. Suppose i < 0. We must verify that all terms ai,j with j > 1 have degree zero. We can write column j as a linear combination of columns −1 and 1: σja−1 + ρja1 = aj . As all the terms in row 0 have degree zero, it cannot be that σj and ρj both have positive degree, and if their degrees were negative they must be equal. If the degrees were negative this would imply elements in ˜B 2 with negative degree, which cannot be. If deg(ρj) > 0 while deg(σj) = 0, then ˜B 2 would have a column outside the first where all elements have degree zero, which cannot be. So, we must have 0 = deg(ρj) ≤ deg(σj). As ai,−1 has positive degree and ai,1 has degree zero it must be that ai,j has degree zero as well. Identical reasoning gives us that all terms ai,j with j < −1 and i > 0 also have degree zero. It remains to be proven that ˜ A is symmetric. As ˜B 1 and ˜B 2 are symmetric, we must only prove ai,j = aj,i when ij < 0. Suppose j > 1, and examine the two matrices b−1,−1 a−1,0 a−1,j a0,−1 a0,0 a0,j a1,−1 a1,0 b1,j , and b−1,−1 a−1,0 a−1,1 a0,−1 a0,0 a0,1 aj,−1 aj,0 bj,1 . By construction a−1,0 = a0,−1, a0,1 = a1,0, a0,j = aj,0, and b1,j = bj,1. As the above matrices are also singular we must have a−1,j = aj,−1. The proof that a1,j = aj,1 for j < −1 is essentially identical. From here verifying symmetry is a calculation: 42 aj,i = σiaj,−1 + ρiaj,1 = σia−1,j + ρia1,j = σi(σja−1,−1 + ρja−1,1) + μi(σja1,−1 + ρja1,1) = σj(σia−1,−1 + ρia1,−1) + ρj(σia−1,1 + ρia1,1) = σj(σia−1,−1 + ρia−1,1) + ρj(σia1,−1 + ρia1,1) = σja−1,i + ρja1,i = σjai,−1 + ρjai,1 = ai,j . So, A˜ is a rank two symmetric lift of A, and therefore A has symmetric Kapranov rank two. As outlined in Theorem 3.6 above, we combine Lemma 3.7 and Lemma 3.8 in our proof of the next lemma. Lemma 3.9. Suppose A has the form B1 0 0 0 0 0 B2 0 0 0 0 0 0 0 0 0 0 0 0 C 0 0 0 CT 0 , where B1,B2 are symmetric and positive, C is nonnegative and does not contain a zero column, and either C or both B1 and B2 have positive size. If A has symmetric tropical rank two it has symmetric Kapranov rank two. Proof. If B1 and B2 both have size zero, this is Lemma 3.7. If C has size zero, this is Lemma 3.8. So, suppose C has positive size, and at least one of B1 and B2 have positive size. The method of proof here is similar to the method used for the previous two lemmas. By induction we may find a rank two symmetric lift for the upper-left matrix B1 0 0 0 B2 0 0 0 0 , and the lower-right matrix 0 0 0 0 0 C 0 CT 0 . Call these lifts ˜B and ˜ C, respectively. After possibly some scaling we may assume the bottom-right entry of ˜B coincides with the top-left entry of ˜ C. The lifts ˜B and ˜ C will be, respectively, the upper-left and lower-right parts of the lift A˜ we wish to construct. We number the rows and columns of A˜ in a manner similar to Lemmas 3.7 and 3.8, with the a0,0 entry being the degree zero entry that must match up 43 for the two lifts. We will refer to any element of A with an indexed lower-case A, and not distinguish among elements in B1,B2,C,CT , or outside these submatrices. We must complete the lift A˜ by finding entries for the terms ai,j with ij < 0. We again start with the 3 × 3 central submatrix: a−1,−1 a−1,0 a−1,1 a0,−1 a0,0 a0,1 a1,−1 a1,0 a1,1 . We pick a−1,1 and a1,−1 such that this matrix is singular and a−1,1 = a1,−1. Every term ai,1 for i < −1, and ai,−1 for i > 1, is then determined by the need for the matrices a−1,−1 a−1,0 a−1,1 a0,−1 a0,0 a0,1 ai,−1 ai,0 ai,1 and ai,−1 ai,0 ai,1 a0,−1 a0,0 a0,1 a1,−1 a1,0 a1,1 to be, respectively, singular. Every column of ˜ C can be written as a linear combination of the first two: λjc0 + μjc1 = cj . We use these relations to define the entries ai,j with i < 0 and j > 1: λjai,0 + μjai,1 = ai,j . We similarly use the first two columns of ˜B to define the terms ai,j with i > 0, j < −1. This determines a rank two matrix A˜. We must verify the matrix is symmetric, and is a lift of A. We first prove A˜ is symmetric. By construction all terms of the form ai,j with ij ≥ 0 satisfy ai,j = aj,i. Also, by construction a1,−1 = a−1,1. Using these facts we note the matrices ai,−1 ai,0 x a0,−1 a0,0 a0,1 a1,−1 a1,0 a1,1 and a−1,i a−1,0 a−1,1 a0,i a0,0 a0,1 x a1,0 a1,1 are transposes. Therefore, ai,1, the unique value of x that matrix the matrix on the left singular, is equal to a1,i, the unique value of x that makes the matrix on the right singular. Using these equalities we note the matrices ai,i ai,0 x a0,i a0,0 a0,j a1,i a1,0 a1,j and ai,i ai,0 ai,1 a0,i a0,0 a0,1 x aj,0 aj,1 are also transposes. So, ai,j , the unique value of x that makes the matrix on the left singular, is equal to aj,i, the unique value of x that makes the matrix on the right singular. So, the matrix A˜ is symmetric. 44 It remains to be proven that each ai,j with ij < 0 has degree zero. Suppose i < 0, j > 0. That ai,j has degree zero follows because the matrix ai,i ai,0 ai,j a0,i a0,0 a0,j aj,i aj,0 aj,j is singular, ai,i has positive degree, and all other terms that are not ai,j = aj,i have degree zero. The only way this matrix could possibly be singular is if ai,j has degree zero. As our matrix is symmetric this completes the proof. 3.3.4 Completed Theorem We now have all the tools we need to complete the proof of Theorem 3.6. Symmetric tropical rank two implies symmetric Kapranov rank two. Suppose A is a sym-metric matrix with symmetric tropical rank two. We may assume A is in the form given by Lemma 3.4. If A has only one zero row/column then by Lemma 3.9 A has symmetric Kapranov rank two. If A has no zero row/column then the matrix A+ = 0 0 0 A has symmetric tropical rank two by Lemma 3.5, and therefore symmetric Kapranov rank two by Lemma 3.9. If A+ has symmetric Kapranov rank two, then by eliminating the first row/column from the lift we see A has symmetric Kapranov rank two as well. If A has more than one zero row/column we may proceed by induction on the number of such columns. In particular, A must have the form A = 0 0 0 A− , where A− is a symmetric matrix with symmetric tropical rank two, with one fewer zero row/column than A, and therefore by induction A− has symmetric Kapranov rank two. By Lemma 3.5 A has symmetric Kapranov rank two as well. Combining Theorem 3.2 and Theorem 3.6 we see that the r × r minors of a symmetric n × n matrix form a tropical basis for r = 2 and r = 3. CHAPTER 4 SYMMETRIC TROPICAL RANK THREE The r × r minors of an m × n matrix of variables form a tropical basis if r = 2, 3, or min(m, n). They do not form a tropical basis if 4 < r < min(m, n). The r = 4 case is special. The 4 × 4 minors of an m × n matrix of variables form a tropical basis if min(m, n) ≤ 6, but otherwise not. Tropical rank three is exceptional for symmetric matrices as well. In this chapter we prove that the 4 × 4 minors of a symmetric 5 × 5 matrix of variables form a tropical basis, and in proving this develop a method that might generalize to larger matrices. In Chapter 5 we will prove that the 4×4 minors of a symmetric n×n matrix of variables do not form a tropical basis if n > 12. Whether the 4 × 4 minors of an n × n matrix of variables form a tropical basis for 5 < n < 13 remains unknown. Note that throughout this chapter we will frequently be dealing with submatrices of a given matrix. Unless stated otherwise, the columns and rows of a submatrix inherit their labels from the larger matrix. So, if A is a 5 × 5 matrix the principal submatrix A33 has columns and rows labeled sequentially 1, 2, 4, 5. 4.1 Definitions and Assumptions Before we get to the meat of the proof we will need to justify a few assumptions we will want to make in order to simplify things. 4.1.1 Symmetric Scaling We will make frequent use of the facts from Chapter 3 that neither the symmetric tropical rank nor symmetric Kapranov rank of a matrix changes as a result of a symmetric scaling or a diagonal permutation. Proposition 4.1. If A is a 5 × 5 symmetric matrix and σ is a permutation that realizes the tropical determinant, then there exists a matrix A′ such that A′ can be obtained from A through a sequence of symmetric scalings, every entry in A′ is nonnegative, and ai,σ(i) = 0 for all 1 ≤ i ≤ n. 46 Proof. Note that within this proof, and only within this proof, if we are talking about the \"form\" of a matrix a blank entry can have any value, positive or negative. If σ = id then we can form A′ by scaling each row/column i by −ai,i to obtain a matrix with the form 0 0 0 0 0 . The tropical determinant must be realized by σ = id, and the matrix must be symmetric, which clearly implies all the off-diagonal elements must be nonnegative. If σ is a 5-cycle then we can assume without loss of generality that σ = (12345). Scale row / column 2 by −a1,2, row / column 3 by −a2,3, and so on until row / column 5. Next, scale all the rows / columns with odd labels by an amount equal to −a1,5/2, and scale all the rows / columns with even labels by an amount equal to a1,5/2. The matrix A′ obtained from this scaling must have the form 0 0 0 0 0 0 0 0 0 0 , and its tropical determinant must be realized by σ = (12345), which means its tropical determinant must be 0. If any blank entry, and its symmetric counterpart, above were negative the tropical determinant of the matrix would be negative, which would violate our assumption. If σ is a 3-cycle we can assume without loss of generality that σ = (123). Scale row / column 2 by −a1,2, row / column 3 by −a2,3. Then, scale rows / columns 1 and 3 by −a1,3/2, and row / column 2 by a1,3/2. Scale row / column 4 by −a4,4, and row / column 5 by −a5,5. This scaled matrix will have the form 0 0 0 0 0 0 0 0 , and its tropical determinant must be 0. As in the previous example, if any blank entry, and its symmetric counterpart, were negative the matrix would have negative determinant, which would violate our assumption. 47 If σ is a 3-cycle and a tranposition we can assume without loss of generality that σ = (123)(45). We can scale the first three indices exactly as we did in the previous paragraph. If we then scale both rows / columns 4 and 5 by −a4,5/2 we construct a matrix A′′ of the form 0 0 0 0 0 0 0 0 . where the determinant is 0. If any entry along the top three diagonal terms were negative the determinant of the matrix would be negative, which is not allowed. If a′′ 4,4 < 0 then we can scale row / column 4 by −a′′ 4,4, and row / column 5 by the opposite amount. This keeps the matrix in the form above, but ensures the lower two diagonal entries are nonnegative. Exactly the same reasoning applies if a′′ 5,5 < 0. If any other entry were negative we can assume without loss of generality that the minimum entry in the matrix is a′′ 3,4 and its symmetric counterpart a′′ 4,3. If more than these two entries are negative the matrix would have negative determinant, which cannot be. If we scale row / column 4 by −a′′ 3,4, and row / column 5 by the opposite, then the matrix maintains the form above, but with all terms nonnegative. So, we have constructed A′. Finally, if σ is the product of two transpositions we can assume without loss of generality that σ = (12)(34). Scale row / column 1 by −a1,2/2 and row / column 2 by −a1,2/2. If after this scaling either of the top two diagonal terms are negative we can scale as we did in the previous paragraph to keep the off-diagonal terms 0 and make the diagonal terms nonnegative. The same can be done for the 2×2 block corresponding with the transposition (34). Scale row /column 5 by −a5,5 to get the matrix 0 0 0 0 0 . It is quick to check that, given the determinant of this matrix is 0, any negative terms can be scaled away. We will assume without loss of generality that all 5×5 matrices have been symmetrically scaled to satisfy the properties of Proposition 4.1. 48 4.1.2 The Form Matrix We will also want to deal with all matrices that have a certain structure, and this structure will be captured by the form of the matrix, defined below. Definition 4.2. A form matrix is a matrix in which every entry is either blank, a non-negative constant, or the symbol ′′+′′. A nonnegative matrix A has the form of a form matrix A′ if everywhere A′ has a constant, A has the same constant, and everywhere A′ has a ′′+′′, A has a positive entry. For example, the matrix 0 2 1 2 0 3 1 3 0 has any of the following forms: 0 + + + 0 + + + 0 , 0 2 2 0 + + 0 , + + + 0 + + + 0 , 0 2 1 2 0 3 1 3 0 . It does not, however, have the form + + + + 0 + + + 0 , because it has a 0 as its upper-left entry. 4.2 The Method of Joints We now define the \"method of joints,\" which will be the primary method by which we prove our theorem. 4.2.1 The Definition of Joints Definition 4.3. Suppose A is a symmetric matrix, and there are distinct indices i and j (assume without loss of generality i < j) such that: • The principal submatrix Aii is symmetrically tropically singular, and there are distinct minimizing monomials Xσ1 ,Xσ2 , such that the variables in Xσ1 involving the index j are not the same as the variables in Xσ2 involving the index j. • The same is true with i and j reversed. • The submatrix Aji is symmetrically tropically singular, and there are two minimizing monomials Xτ1 ,Xτ2 such that Xτ1 contains the variable Xi,j , while Xτ2 does not. 49 The indices i and j are joints of the matrix A. If the submatrix Aii satisfies the first condition above, we say it satisfies the joint requirement for joints i and j. Similarly for the submatrix Ajj . For example, consider a matrix A of the form 0 0 0 0 0 0 0 0 0 0 0 . We will demonstrate this matrix has joints 4 and 5. The principal submatrix A44 has the form 0 0 0 0 0 0 . This submatrix is symmetrically tropically singular, with minimizing monomials X2 1,2X3,3X5,5 and X2 1,2X2 3,5. In particular, the only variable in the first monomial involving the index 5 is X5,5, while the second monomial contains the variable X3,5. So, A44 satisfies the joint requirement for joints 4 and 5. Identical reasoning can be applied to the principal submatrix A55. The submatrix A54 has the form 0 0 0 0 0 0 . The submatrix is symmetrically tropically singular, with minimizing monomials X2 1,2X3,3X4,5 and X2 1,2X3,4X3,5. One of these minimizing monomials contains the variable X4,5, while the other does not. Therefore, A has joints 4 and 5. 4.2.2 Joints and Kapranov Rank Our proof that the 4×4 minors of a symmetric 5×5 matrix form a tropical basis is based upon first proving that every symmetric matrix over R with joints has symmetric Kapranov rank of at most three. We then prove an exceptional case of a 5×5 symmetric matrix over R that does not have joints, but still has symmetric Kapranov rank three. Finally, we prove that if the 4 × 4 submatrices of a 5 × 5 symmetric matrix are all symmetrically tropically singular then either A has joints, or A has the form of the exceptional case. 50 Proposition 4.4. If a 5×5 symmetric matrix A has joints, then it has symmetric Kapranov rank of at most three. Proof. We will construct a symmetric rank three lift A˜ of A. After possibly a diagonal permutation we may assume A has joints 4 and 5. We define the matrices: X55 := A1,1 A1,2 A1,3 X1,4 A1,2 A2,2 A2,3 X2,4 A1,3 A2,3 A3,3 X3,4 X1,4 X2,4 X3,4 X4,4 , and ˜X 55 = a1,1 a1,2 a1,3 x1,4 a1,2 a2,2 a2,3 x2,4 a1,3 a2,3 a3,3 x3,4 x1,4 x2,4 x3,4 x4,4 , where the Ai,j are the same as the corresponding terms in the matrix A, and the ai,j terms are constants in the field ˜K such that deg(ai,j) = Ai,j , but are otherwise generic. As the ai,j are generic, the tropicalization of the determinant of ˜X 55 is the tropical determinant of X55. By Kapranov's theorem if (A1,4,A2,4,A3,4,A4,4) is a point on the tropical hypersurface given by the tropical determinant of X55, then there is a lift to a point (a1,4, a2,4, a3,4, a4,4) in ˜K 4 on the hypersurface given by the determinant of ˜X 55. This lift gives us a singular 4 × 4 matrix A˜55 := a1,1 a1,2 a1,3 a1,4 a1,2 a2,2 a2,3 a2,4 a1,3 a2,3 a3,3 a3,4 a1,4 a2,4 a3,4 a4,4 , that tropicalizes to the submatrix A55 of A. An identical argument can be used to construct a singular lift of A44 A˜44 = a1,1 a1,2 a1,3 a1,5 a1,2 a2,2 a2,3 a2,5 a1,3 a2,3 a3,3 a3,5 a1,5 a2,5 a3,5 a5,5 , where the top-left 3 × 3 submatrics of A˜44 and A˜55 are identical. We note that if we multiply the fourth column and the fourth row of A˜44 by the same degree zero generic constant that we will still have a singular symmetric lift of A44. So, we can assume the terms ai,4 and aj,5 for any i, j ≤ 5 are generic relative to each other (except for a4,5 and a5,4, which we have not yet determined, and which must, of course, be equal). All the entries in a lift of A have now been determined now except a4,5 = a5,4. To get a4,5 we examine the matrices: 51 X54 := A1,1 A1,2 A1,3 A1,5 A1,2 A2,2 A2,3 A2,5 A1,3 A2,3 A3,3 A3,5 A1,4 A2,4 A3,4 X4,5 , and ˜X 54 := a1,1 a1,2 a1,3 a1,5 a1,2 a2,2 a2,3 a2,5 a1,3 a2,3 a3,3 a3,5 a1,4 a2,4 a3,4 x4,5 . The determinant of ˜X 54 is a linear function in the variable x4,5, and the tropical determinant of X54 is a tropical linear function in the variable X4,5. As the terms in the upper-left 3×3 submatrix of ˜X 54 are generic, and the constant terms in the rightmost column of ˜X 54 are generic with respect to the constant terms in the bottom row, the tropicalization of the determinant of ˜X 54 is the determinant of X54. Again, by Kapranov's theorem, if A4,5 is on the tropical hypersurface given by the tropical determinant of X54, then it lifts to a point on the determinant of ˜X 54. In other words, if the tropical determinant of the matrix A1,1 A1,2 A1,3 A1,5 A1,2 A2,2 A2,3 A2,5 A1,3 A2,3 A3,3 A3,5 A1,4 A2,4 A3,4 A4,5 is realized by two minimizing monomials, one involving the variable X4,5 and the other not, then there exists a value a4,5 ∈ ˜K that makes the matrix a1,1 a1,2 a1,3 a1,5 a1,2 a2,2 a2,3 a2,5 a1,3 a2,3 a3,3 a3,5 a1,4 a2,4 a3,4 a4,5 a singular lift of A54. The requirements for our three applications of Kapranov's theorem are exactly the requirements that 4 and 5 are joints of A. So, if A has joints 4 and 5 then we have now determined all the elements in a lift of the matrix A: A˜ := a1,1 a1,2 a1,3 a1,4 a1,5 a1,2 a2,2 a2,3 a2,4 a2,5 a1,3 a2,3 a3,3 a3,4 a3,5 a1,4 a2,4 a3,4 a4,4 a4,5 a1,5 a2,5 a3,5 a4,5 a5,5 . 52 It remains to be proven that such a lift has rank three. We do this by first proving there is a linear combination of the first three columns equal to the fourth. As the entries in the upper-left 3 × 3 submatrix were chosen generically, this submatrix has rank three, and therefore there is a unique set of coefficients c1, c2, c3 ∈ ˜K such that c1 a1,1 a1,2 a1,3 + c2 a1,2 a2,2 a2,3 + c3 a1,3 a2,3 a3,3 = a1,4 a2,4 a3,4 . That this unique set of coefficients also satisfy c1a1,4 + c2a2,4 + c3a3,4 = a4,4, and c1a1,5 + c2a2,5 + c3a3,5 = a4,5 follows immediately from the singularity of A˜45 and A˜55, respectively. Identical reasoning proves that there exists a linear combination of the first three columns of A˜ equal to the fifth, using the singularity of A˜54 (which, as it is the transpose of A˜45, follows from the singularity of A˜45) and A˜44. Therefore A˜ is a rank three lift of A, and so A has symmetric Kapranov rank at most three. 4.3 The Exceptional Case In our analysis of 5 × 5 symmetric matrices with symmetric tropical rank three or less, there is one possible form that does not have joints, but which still has symmetric Kapranov rank three. Proposition 4.5. If a symmetric tropical matrix A has the form: 0 0 + + N 0 0 + + + + + 0 0 P + + 0 0 P N + P P 0 , with N, P > 0 and N ⊗P less than any element in the 2×2 submatrix determined by rows 1 and 2, and columns 3 and 4, then A has symmetric Kapranov rank three. Proof. The principal submatrix formed from the columns and rows with indices 1, 3, and 5 has the form 0 + N + 0 P N P 0 . 53 Any matrix with this form is symmetrically tropically nonsingular, and therefore A must have symmetric tropical rank three. Consequently, its symmetric Kapranov rank must be at least three. We first note that columns 1, 3, and 5 of A cannot be tropically dependendent, and therefore A must have Kapranov rank at least three. We augment the matrix A, producing a matrix A′ with the form 0 0 0 + + N 0 0 0 + + + 0 0 0 P P 0 + + P 0 0 P + + P 0 0 P N + 0 P P 0 , such that A′ 33 = A. If A′ has a lift A˜′ to a symmetric rank three matrix, then A˜′ 33 will be a symmetric rank three lift of A. So, it is sufficient to prove that A′ has symmetric Kapranov rank three. The upper-right 4 × 4 submatrix of A′ is tropically singular, and therefore has a lift to a singular 4 × 4 matrix: a1,3 a1,4 a1,5 a1,6 a2,3 a2,4 a2,5 a2,6 a3,3 a3,4 a3,5 a3,6 a4,3 a4,4 a4,5 a4,6 . As deg(a3,4) = deg(a4,3) we can multiply the first column of this matrix by a degree zero constant so that a3,4 = a4,3, and the matrix is still singular. We will use this singular 4× 4 matrix with a3,4 = a4,3 to construct a lift for columns 3 through 6 of A′: a1,3 a1,4 a1,5 a1,6 a2,3 a2,4 a2,5 a2,6 a3,3 a3,4 a3,5 a3,6 a3,4 a4,4 a4,5 a4,6 a3,5 a4,5 a5,5 a5,6 a3,6 a4,6 a5,6 a6,6 where a5,5, a5,6, and a6,6 have not yet been determined. We know there is a linear combi-nation of columns a3, a4, and a6 (the third, fourth, and sixth columns of A˜′) such that for rows 1 through 4: αa3 + βa4 + γa6 = a5. If we pick a6,6 such that deg(a6,6) = 0 but otherwise generically, then this relation uniquely determines a5,6 and a5,5 in such a way that deg(a5,5) = 0 and deg(a5,6) = P. We will pick x5,5, x5,6, and x6,6 such that this is true on all rows, and the values tropicalize appropriately. 54 Define M to be the minimum element in the 2 × 2 submatrix of A′ determined by rows 1 and 2, and columns 4 and 5. If deg(α) were minimal out of deg(α), deg(β), and deg(γ), then in order for the linear relation above to hold on the third row we would need either deg(α) = P or deg(α) = deg(γ) < P. In the first case the linear relation on the fourth row would be impossible. In the second case, given P ⊗ N < M, the linear relation on the first row would be impossible. So, deg(α) cannot be minimal. If deg(γ) < deg(α) were minimal, then for the linear relation on the third row to work out we would need deg(γ) = P. This would make the linear relation on the fourth row impossible. So, deg(β) must be uniquely minimal. In order for the linear relation on the fourth row to work out we must have deg(β) = 0, and in order for the linear relation on the third row to work out we must have deg(α), deg(γ) ≥ P. If deg(α) = P, then, again given P ⊗ N < M, the linear relation on the first row would be impossible. So, deg(α) > P. With a6,6 determined the linear relations define a5,6 and a5,5 as a5,6 = αa3,6 + βa4,6 + γa6,6, a5,5 = αa3,5 + βa4,5 + γa5,6. Given the required degrees of α, β, γ, the known degrees of the terms from the lift of the upper-right 4× submatrix of A′, and the assumption that a6,6 satisfies deg(a6,6) = 0 but is otherwise generic, we must have deg(a5,6) = P, and deg(a5,5) = 0. What remains is to find values for x1,1, x1,2, x2,2 such that the evaluation of the matrix x1,1 x1,2 a1,3 a1,4 a1,5 a1,6 x1,2 x2,2 a2,3 a2,4 a2,5 a2,6 a1,3 a2,3 a3,3 a3,4 a3,5 a3,6 a1,4 a2,4 a3,4 a4,4 a4,5 a4,6 a1,5 a2,5 a3,5 a4,5 a5,5 a5,6 a1,6 a2,6 a3,6 a4,6 a5,6 a6,6 has rank three and tropicalizes to A′. If we examine the submatrix formed by columns 1, 3, 4 and 6, x1,1 a1,3 a1,4 a1,6 x1,2 a2,3 a2,4 a2,6 a1,3 a3,3 a3,4 a3,6 a1,4 a3,4 a4,4 a4,6 a1,5 a3,5 a4,5 a5,6 a1,6 a3,6 a4,6 a6,6 , then we note that, as there is a linear combination of columns a3, a4, a6 equal to column a5 there is linear combination of rows 3 4, and 6 in the above 6×4 matrix equal to row 5. We pick x1,1 = a1,1 so that the matrix 55 a1,1 a1,3 a1,4 a1,6 a1,3 a3,3 a3,4 a3,6 a1,4 a3,4 a4,4 a4,6 a1,6 a3,6 a4,6 a6,6 is singular. Given the known degrees of the elements in the matrix, and that a6,6 is generic, we must have deg(a1,1) = 0. We can use an identical method to construct x1,2 = a1,2 of the appropriate degree. Therefore every row of the above 6 × 4 matrix can be constructed from rows 3, 4, and 6, and so the matrix has rank three. In particular, this means the third column of A˜′ can be constructed as a linear combination of the first, fourth, and sixth columns. What remains to be proven is that x2,2 can be chosen with the appropriate degree so that the second column of A˜′ can be written as a linear combination of the first, fourth, and sixth columns. To do this we examine the 6 × 4 submatrix a1,1 a1,2 a1,4 a1,6 a1,2 x2,2 a2,4 a2,6 a1,3 a2,3 a3,4 a3,6 a1,4 a2,4 a4,4 a4,6 a1,5 a2,5 a4,5 a5,6 a1,6 a2,6 a4,6 a6,6 . We already know rows 3 and 5 of this matrix can be written as a linear combination of rows 1, 4, and 6. To prove this is also true for row 2 we examine the submatrix a1,1 a1,2 a1,4 a1,6 a1,2 x2,2 a2,4 a2,6 a1,4 a2,4 a4,4 a4,6 a1,6 a2,6 a4,6 a6,6 , and note that we can pick x2,2 = a2,2 such that the matrix is singular and deg(a2,2) = 0. This means that every row of A˜′ can be written as a linear combination of rows 1, 4, and 6, and therefore A˜′ has a rank three lift. As A′ has a symmetric rank three lift, so does A, and our proof is complete. 4.4 Searching for Joints The proof that, with one exception, if every 4×4 submatrix of a 5×5 symmetric matrix is symmetrically tropically singular then the matrix must have joints involves the analysis of a number of cases, and will be broken down into many lemmas. We will, throughout, assume that A is a symmetric matrix with symmetric tropical rank at most three. 56 4.4.1 There Must Be a Transposition Before we go through these cases, we will need an additional fact concerning the per-mutations that realize the tropical determinant of a symmetrically singular 5 × 5 matrix. Lemma 4.6. If A is a 5 × 5 symmetrically tropically singular matrix, then there is a permutation with a tranposition in its cycle decomposition realizing the tropical determinant. Proof. If σ realizes the tropical determinant of A, then if σ has a 2-cycle in its cycle decomposition there is nothing to prove. If the cycle decomposition of σ has a 4-cycle then by the proof of Proposition 2.10 there must also be a permutation realizing the tropical determinant that is the product of two transpositions. As for the other possibilities, after perhaps a diagonal permutation, the matrix A must have one of the following forms: Identity : 0 0 0 0 0 , 3-cycle : 0 0 0 0 0 0 0 0 , 5-cycle : 0 0 0 0 0 0 0 0 0 0 . If A is symmetrically singular then each of these matrices must have an additional 0 term that is not specified above, and for any of these possibilities an additional 0 term will introduce a permutation realizing the tropical determinant with a cycle decomposition that includes a transposition. 4.4.2 Not Two Transpositions After possibly a diagonal permutation, we may assume the upper-left 2 × 2 submatrix of A has the form: 0 0 , and A has a permutation that realizes the tropical determinant whose disjoint cycle decom-position includes the transposition (12). 57 Lemma 4.7. If A has symmetric tropical rank three, and does not have a permutation realizing the determinant whose disjoint cycle decomposition is a product of transpositions, then A has joints. Proof. As A must have a permutation realizing the determinant that involves the transpo-sition (12), the only possibilities for this minimizing permutation are (12) and (12)(345), which would give A the form: 0 0 0 + + + 0 + + + 0 , or 0 0 + 0 0 0 + 0 0 0 + . In the first possibility the submatrix A12 has the form: 0 0 + + + 0 + + + 0 . The submatrix A12 must be singular, and so there must be another 0 term in the first row, and a corresponding 0 term in the first column. By corresponding, we mean that if the 0 in the first row of A12 is in the ith column, then the 0 in the first column of A12 must be in the ith row. Taking this into account, after possibly a diagonal permutation, A will have the form: 0 0 0 0 0 0 0 + + + 0 + + + 0 . The submatrix A11 is 0 0 0 + + + 0 + + + 0 . As this submatrix must be symmetrically tropically singular we can see from its form that there must be two permutations realizing the tropical determinant, one (noting A11 inherits its indices from A) whose disjoint cycle decomposition contains the transposition (23), and another whose disjoint cycle decomposition does not. The same will be true, mutatis mutandis, of the submatrix A22. From this we can see A has joints 1 and 2. As for the second possibility, the submatrix A12 will have the form: 58 0 + 0 0 0 + 0 0 0 + . A12 must be symmetrically tropically singular, and so there must be an additional 0 term in the first row, and an additional 0 term in the first column. Noting this, after possibly a diagonal permutation, the matrix A must have one of the forms: 0 0 0 0 0 0 + 0 0 0 + 0 0 0 + , or 0 0 0 0 0 + 0 0 0 0 + 0 0 0 + . Using essentially identical reasoning as in the first possibility, we find that A will have joints 1 and 2. 4.4.3 The Case with Five Zeros So, we may assume A has a permutation that realizes the determinant with a disjoint cycle decomposition that is the product of two transpositions. After possibly a diagonal permutation, we may assume this disjoint cycle decomposition is (12)(34). Lemma 4.8. Suppose the matrix A has the form: + 0 0 + + 0 0 + 0 . Then A has joints. Proof. The submatrix A55 must be symmetrically tropically singular, and therefore, after possibly a diagonal permutation, it must have the form + 0 0 0 + 0 0 + 0 0 0 + . After another diagonal permutation A55 can be arranged to have the form 0 0 0 0 0 0 0 0 . The matrix A will have the corresponding form 59 0 0 0 0 0 0 0 0 0 . This is a form that will come up as a possibility in other cases, and we will refer to it as off-diagonal form. We will complete our lemma by proving that any matrix in off-diagonal form must have joints. If A has off-diagonal form, the submatrix A11 will have the form: 0 0 0 0 0 . This submatrix must be symmetrically tropically singular. Denote by M the minimal element in the 2 × 2 submatrix formed by rows 3 and 4, and columns 3 and 4 (recall A11 inherits its indices from A), and denote by N the minimal element in the 2 × 1 submatrix formed by rows 3 and 4, and column 5. Suppose M < 2N. Given A11 is symmetrically tropically singular it must, up to a diagonal permutation, have one of the two forms: 0 0 0 M 0 M 0 , 0 0 0 M M 0 M 0 . In either case the submatrix A11 satisfies the joint requirement for joints 1 and 3. If M = 2N then, again given A11 is symmetrically tropically singular, it must have, up to a diagonal permutation, one of the three forms: 0 0 0 2N 0 N N 0 , 0 0 0 2N N 0 N 0 , 0 0 0 2N N 0 2N N 0 . In either case the submatrix A11 again satisfies the joint requirement for joints 1 and 3. Finally, if M > 2N then as A11 is symmetrically tropically singular it must have the form: 0 0 0 N 0 N N N 0 . In this case, again, the submatrix A11 satisfies the joint requirement for joints 1 and 3. In each of these six possibilities A11 satisfies the joint requirement for joints 1 and 3. An identical analysis can be performed on the submatrix A33, and from this we can get that A has joints 1 and 3. So, any matrix with off-diagonal form has joints. 60 4.4.4 The Case with Six Zeros Lemma 4.9. Suppose A has the form: + 0 0 + + 0 0 0 0 . Then A has joints. Proof. The submatrix A55 must be symmetrically tropically singular, and this means either there is a diagonal permutation that will put A in off-diagonal form, in which case we are done, or A55 has the form: + 0 0 + 0 + 0 + 0 0 + 0 + + 0 0 . In this case A33 must have the form: + 0 + 0 + + + + 0 0 . As A33 must be symmetrically tropically singular, it must have one of the following two forms: + 0 + 0 0 + + 0 + + 0 0 0 0 , or + 0 + 0 + + + + 0 0 0 0 . In the first possibility A has the form: + 0 0 + 0 0 + 0 + 0 0 0 + 0 + + 0 0 0 0 0 . This form has joints 1 and 2. In the second possibility A has the form: + 0 0 + 0 + 0 + 0 0 + 0 + + 0 0 0 0 0 . The submatrix A44 has the form: 61 + 0 0 0 + 0 0 0 + 0 . This submatrix must be symmetrically tropically singular and therefore, up to a diagonal permutation, must have the form: + 0 0 0 0 + 0 0 0 + 0 0 . The corresponding form for A is: + 0 0 + 0 0 + 0 + 0 0 + 0 + + 0 0 0 0 0 0 . Any matrix of this form has joints 1 and 2. 4.4.5 The Case with Seven Zeros Lemma 4.10. Suppose A has the form: + 0 0 + 0 0 0 0 0 . Then A has joints. Proof. Suppose A has the form + 0 0 + 0 0 + 0 0 + + + 0 . The submatrices A33 and A44 will have the form: + 0 0 + 0 + + 0 . For these submatrices to be symmetrically tropically singular they must have, up to a diagonal permutation, one of the two forms: 62 + 0 0 0 + 0 0 0 0 + + 0 , or + 0 0 0 + 0 0 0 + 0 + 0 . Examining the possibilities and what they imply for the form of A we get that A, possibly after a diagonal permutation, must either have off-diagonal form, in which case we are done, or have one of the following two forms: + 0 0 0 0 + 0 0 0 0 0 0 0 0 0 , or + 0 0 0 + 0 0 0 0 0 0 0 0 . The first possibility has joints 3 and 4. The second possibility requires more analysis. Suppose A has the form of our second possibility above. Denote by M the minimal off-diagonal term in A that is not necessarily 0. If M is in the 2 × 2 submatrix formed by rows 1 and 2, and columns 3 and 4 then, after possibly a diagonal permutation, A will have the f"}]},"highlighting":{"196852":{"ocr_t":[]}}}