<?xml version="1.0" encoding="UTF-8"?><!DOCTYPE article PUBLIC "-//NLM//DTD JATS (Z39.96) Journal Publishing DTD v1.3 20210610//EN" "https://jats.nlm.nih.gov/publishing/1.3/JATS-journalpublishing1-3.dtd"><article xml:lang="en" xmlns:xlink="http://www.w3.org/1999/xlink" xmlns:ali="http://www.niso.org/schemas/ali/1.0/" dtd-version="1.3" article-type="research-article"><front><journal-meta><journal-id journal-id-type="issn">2460-0245</journal-id><journal-title-group><journal-title>Journal of the Indonesian Mathematical Society</journal-title><abbrev-journal-title>JIMS</abbrev-journal-title></journal-title-group><issn pub-type="epub">2460-0245</issn><issn pub-type="ppub">2086-8952</issn><publisher><publisher-name>IndoMS</publisher-name><publisher-loc>Indonesia</publisher-loc></publisher></journal-meta><article-meta><article-id pub-id-type="doi">10.22342/jims.v32i3.2260</article-id><article-categories><subj-group><subject>Mathematics Subject Classification:</subject></subj-group></article-categories><title-group><article-title>Explicit Formulas for the Anti-Trace of 3-by-3 Matrix Powers via Generating Functions</article-title><subtitle>Rumus Eksplisit untuk Anti-Jejak Pangkat Matriks 3x3 melalui Fungsi Pembangkit</subtitle></title-group><contrib-group><contrib contrib-type="author"><name><surname>Gormantara</surname><given-names>Jeriko</given-names></name><address><country country="ID">Indonesia</country><email>jerikogormantara@unhas.ac.id</email></address><xref ref-type="aff" rid="AFF-1"></xref><xref ref-type="corresp" rid="cor-0"></xref></contrib><contrib contrib-type="author"><contrib-id contrib-id-type="orcid">https://orcid.org/0000-0001-8462-5080</contrib-id><name><surname>Ferdania</surname><given-names>Devi Fitri</given-names></name><address><country country="GB">United Kingdom</country><email>dxf344@student.bham.ac.uk</email></address><xref ref-type="aff" rid="AFF-2"></xref></contrib><contrib contrib-type="author"><name><surname>Yuliawan</surname><given-names>Fajar</given-names></name><address><country country="ID">Indonesia</country><email>fajar.yuliawan@itb.ac.id</email></address><xref ref-type="aff" rid="AFF-3"></xref></contrib><contrib contrib-type="author"><contrib-id contrib-id-type="orcid">https://orcid.org/0009-0001-8949-4961</contrib-id><name><surname>Garminia</surname><given-names>Hanni</given-names></name><address><country country="ID">Indonesia</country><email>garminia@itb.ac.id</email></address><xref ref-type="aff" rid="AFF-3"></xref></contrib></contrib-group><contrib-group><contrib contrib-type="editor"><name><surname>Prasetyo</surname><given-names>Puguh Wahyu</given-names></name><address><country country="ID">Indonesia</country><email>puguh.prasetyo@pmat.uad.ac.id</email></address><xref ref-type="aff" rid="EDITOR-AFF-1"></xref></contrib><contrib contrib-type="editor"><name><surname>abdurahim</surname></name><address><country country="ID">Indonesia</country><email>abdurahim@staff.unram.ac.id</email></address></contrib></contrib-group><aff id="AFF-1"><institution content-type="dept">Department of Mathematics</institution><institution-wrap><institution>Hasanuddin University</institution><institution-id institution-id-type="ror">https://ror.org/00da1gf19</institution-id></institution-wrap><country country="ID">Indonesia</country></aff><aff id="AFF-2"><institution content-type="dept">School of Mathematics</institution><institution-wrap><institution>University of Birmingham</institution><institution-id institution-id-type="ror">https://ror.org/03angcq70</institution-id></institution-wrap><country country="GB">United Kingdom</country></aff><aff id="AFF-3"><institution content-type="dept">Department of Mathematics</institution><institution-wrap><institution>Bandung Institute of Technology</institution><institution-id institution-id-type="ror">https://ror.org/00apj8t60</institution-id></institution-wrap><country country="ID">Indonesia</country></aff><aff id="EDITOR-AFF-1"><institution-wrap><institution>Universitas Ahmad Dahlan</institution><institution-id institution-id-type="ror">https://ror.org/03hn13397</institution-id></institution-wrap><country country="ID">Indonesia</country></aff><author-notes><fn fn-type="coi-statement"><label>Declarations.</label><p>The authors declare no conflict of interest.</p></fn><corresp id="cor-0">Corresponding author: Jeriko Gormantara. Email: <email>jerikogormantara@unhas.ac.id</email></corresp></author-notes><pub-date date-type="pub" iso-8601-date="2026-09-01" publication-format="electronic"><day>01</day><month>09</month><year>2026</year></pub-date><pub-date date-type="collection" iso-8601-date="2026-09-01" publication-format="electronic"><day>01</day><month>09</month><year>2026</year></pub-date><volume>32</volume><issue>3</issue><issue-title>SEPTEMBER</issue-title><fpage>1</fpage><lpage>13</lpage><elocation-id>5A15, 05A15.</elocation-id><history><date date-type="received" iso-8601-date="2025-10-03"><day>03</day><month>10</month><year>2025</year></date><date date-type="accepted" iso-8601-date="2026-03-09"><day>09</day><month>03</month><year>2026</year></date></history><permissions><copyright-statement>Copyright (c) 2026 Journal of the Indonesian Mathematical Society</copyright-statement><copyright-year>2026</copyright-year><copyright-holder>Journal of the Indonesian Mathematical Society</copyright-holder><license xlink:href="https://creativecommons.org/licenses/by-nc-nd/4.0/"><ali:license_ref xmlns:ali="http://www.niso.org/schemas/ali/1.0/">https://creativecommons.org/licenses/by-nc-nd/4.0/</ali:license_ref><license-p>This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License.</license-p></license></permissions><self-uri xlink:href="https://jims-a.org/index.php/jimsa/article/view/2260" xlink:title="2260"></self-uri><abstract><p>We study the <italic>anti-trace</italic>-the sum of anti-diagonal entries-of powers of 3-by-3 matrices. Unlike the ordinary trace, the <italic>anti-trace</italic> has no spectral characterization, so closed forms for the <italic>anti-trace</italic> of matrix powers are valuable. Starting from the Cayley-Hamilton recurrence for matrix powers and solving a trivariate generating function, we derive explicit closed-form expressions whose coefficients are signed multinomial combinations of the sums of principal minors and their antiprincipal counterparts. The formulas are purely algebraic and remain valid over any commutative ring. As a structural application, we analyze block anti-diagonal matrices of size 3m-by-3m.</p></abstract><kwd-group><kwd>anti-trace</kwd><kwd>trace</kwd><kwd>Cayley--Hamilton</kwd><kwd>generating function</kwd><kwd>block anti-diagonal matrix</kwd></kwd-group><funding-group><funding-statement>. .</funding-statement></funding-group><custom-meta-group><custom-meta><meta-name>File created by JATS Editor</meta-name><meta-value>https://jatseditor.com</meta-value></custom-meta><custom-meta><meta-name>issue-created-year</meta-name><meta-value>2026</meta-value></custom-meta></custom-meta-group></article-meta></front><body><sec id="sec-1"><title>1. INTRODUCTION</title><p>The trace of a matrix is the sum of its diagonal entries and appears throughout mathematics and its applications. In particular, traces of matrix powers arise in network analysis, number theory, dynamical systems, and control <xref ref-type="bibr" rid="BIBR-1">[1]</xref>. For instance, in a connected simple graph the total number of triangles equals <inline-formula><tex-math id="math-1"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r ( A ^ { 3 } ) / 6 \end{document} ]]></tex-math></inline-formula> , where A is the adjacency matrix <xref ref-type="bibr" rid="BIBR-2">[2]</xref>; for integer matrices one has the Euler congruence <inline-formula><tex-math id="math-2"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r (A ^ {p ^ {r}}) \equiv T r (A ^ {p ^ {r - 1}}) \pmod {p ^ {r}} \end{document} ]]></tex-math></inline-formula> for all primes <inline-formula><tex-math id="math-3"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle p \end{document} ]]></tex-math></inline-formula> and <inline-formula><tex-math id="math-4"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle r \in { \mathbb { N } } ; \end{document} ]]></tex-math></inline-formula> and in <inline-formula><tex-math id="math-5"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \mathrm { d y } \end{document} ]]></tex-math></inline-formula> namical systems and control, traces of matrix polynomials enter Lefschetz numbers and Lyapunov equations [<xref ref-type="bibr" rid="BIBR-3">3</xref>, <xref ref-type="bibr" rid="BIBR-4">4</xref>]. Because <inline-formula><tex-math id="math-6"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r ( A ) \end{document} ]]></tex-math></inline-formula>equals the sum of eigenvalues, <inline-formula><tex-math id="math-7"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r ( A ^ { n } ) \end{document} ]]></tex-math></inline-formula> is spectrally transparent: it is the sum of eigenvalues raised to the nth power.</p><p>By contrast, the <italic>anti-trace</italic><inline-formula><tex-math id="math-8"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r _ { a } ( A ) \end{document} ]]></tex-math></inline-formula> the sum of the anti-diagonal entries—has no analogous spectral identity. In the absence of such a formula, computing <inline-formula><tex-math id="math-9"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r _ { a } ( A ^ { n } ) \end{document} ]]></tex-math></inline-formula> by first forming <inline-formula><tex-math id="math-10"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A ^ { n } \end{document} ]]></tex-math></inline-formula> is expensive and obscures structure. Recent work addresses special cases for traces (e.g., tridiagonal or persymmetric matrices [ <xref ref-type="bibr" rid="BIBR-5">5</xref> , <xref ref-type="bibr" rid="BIBR-6">6</xref> , <xref ref-type="bibr" rid="BIBR-7">7</xref> ] )  and gives closed forms for 2× 2 powers in terms of trace and determinant [<xref ref-type="bibr" rid="BIBR-8">8</xref>, <xref ref-type="bibr" rid="BIBR-9">9</xref>]. For <italic>anti-trace</italic>, <xref ref-type="bibr" rid="BIBR-10">[10]</xref> obtained <inline-formula><tex-math id="math-11"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \mathrm 2 \times 2 \end{document} ]]></tex-math></inline-formula> formula involving trace, <italic>anti-trace</italic>, and determinant. To the best of our knowledge, explicit closed forms for <inline-formula><tex-math id="math-12"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r _ { a } ( A ^ { n } ) \end{document} ]]></tex-math></inline-formula> beyond <inline-formula><tex-math id="math-13"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle 2 \times 2 \end{document} ]]></tex-math></inline-formula> have not been written down in a way that is both exact and ring-valid.</p><p>To clarify the contribution of this work relative to the existing literature, we emphasize that earlier studies on traces or <italic>anti-trace</italic>s of matrix powers often rely on eigenvalue-based methods or recursive computational techniques over the real or complex numbers. In contrast, the present paper provides genuine explicit closed-form formulas with explicitly described multinomial coeficients, derived via a generating-function framework combined with the Cayley–Hamilton theorem. The approach avoids spectral arguments entirely and remains valid over arbitrary commutative rings.</p><p>Organization. Section 2 recalls basic notions (trace/<italic>anti-trace</italic> and generating functions). Section 3 develops the generating-function method and presents the closed forms. The same section records the block–anti–diagonal consequences.</p></sec><sec id="sec-2"><title>2. PRELIMINARIES</title><p>We assume familiarity with standard linear algebra. This section records the notation and basic tools used later. Throughout, <italic>R</italic> denotes a commutative ring with identity, and <inline-formula><tex-math id="math-14"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle M _ { n } ( R ) \end{document} ]]></tex-math></inline-formula> the ring of <inline-formula><tex-math id="math-15"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle n \times n \end{document} ]]></tex-math></inline-formula> matrices over <italic>R</italic>. For readability we sometimes specialize to <inline-formula><tex-math id="math-16"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle R = \mathbb { R } \end{document} ]]></tex-math></inline-formula> in examples, but every algebraic identity below holds over<italic> R.</italic></p><sec id="sec-3"><title>2.1. Trace and anti-trace.</title><p>For <inline-formula><tex-math id="math-17"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A = [ a _ { i j } ] \in M _ { n } ( R ) \end{document} ]]></tex-math></inline-formula> , the trace is</p><disp-formula id="equation-1"><tex-math id="math-18"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r (A) = \sum_ {i = 1} ^ {n} a _ {i i}. \end{document} ]]></tex-math></disp-formula><p>It is linear and cyclic: <inline-formula><tex-math id="math-19"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r ( \alpha A + \beta B ) = \alpha T r ( A ) + \beta T r ( B ) \end{document} ]]></tex-math></inline-formula> , and for any conformable matrices</p><disp-formula id="equation-2"><tex-math id="math-20"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r (X _ {1} X _ {2} \dots X _ {r}) = T r (X _ {2} \dots X _ {r} X _ {1}).\tag{1} \end{document} ]]></tex-math></disp-formula><p>When <italic>R</italic> is a field— or <inline-formula><tex-math id="math-21"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle R \subseteq F \end{document} ]]></tex-math></inline-formula> embeds in a field <italic>F</italic>—then over an algebraic closure of <inline-formula><tex-math id="math-22"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle F , T r ( A ) \end{document} ]]></tex-math></inline-formula> ) equals the sum of the eigenvalues of A counted with algebraic multiplicity; see <xref ref-type="bibr" rid="BIBR-11">[11]</xref>.</p><p>The <italic>anti-trace</italic> is the sum of the anti-diagonal entries,</p><disp-formula id="equation-3"><tex-math id="math-23"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r _ {a} (A) = a _ {1 n} + a _ {2, n - 1} + \dots + a _ {n 1}. \end{document} ]]></tex-math></disp-formula><p>Let <inline-formula><tex-math id="math-24"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle J \in M _ { n } ( R ) \end{document} ]]></tex-math></inline-formula> be the exchange (reversal) permutation matrix with ones on the anti-diagonal and zeros elsewhere. Then <inline-formula><tex-math id="math-25"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle J ^ { 2 } = I \end{document} ]]></tex-math></inline-formula> and <inline-formula><tex-math id="math-26"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle J ^ { \top } = J . \end{document} ]]></tex-math></inline-formula> and</p><disp-formula id="equation-4"><tex-math id="math-27"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r _ {a} (A) = T r (J A) = T r (A J).\tag{2} \end{document} ]]></tex-math></disp-formula><p>For <inline-formula><tex-math id="math-28"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array} { r } { n = 3 , J = \left( \begin{array} { l l l } { 0 } & { 0 } & { 1 } \\ { 0 } & { 1 } & { 0 } \\ { 1 } & { 0 } & { 0 } \end{array} \right) } \end{array} \end{document} ]]></tex-math></inline-formula> , so <inline-formula><tex-math id="math-29"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r _ { a } ( A ) = a _ { 1 3 } + a _ { 2 2 } + a _ { 3 1 } \end{document} ]]></tex-math></inline-formula> , consistent with <xref ref-type="bibr" rid="BIBR-10">[10]</xref>.</p></sec><sec id="sec-4"><title>2.2. Generating functions and multinomials.</title><p>Given a univariate sequence <inline-formula><tex-math id="math-30"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle ( f _ { n } ) _ { n \geq 0 } \end{document} ]]></tex-math></inline-formula> , its ordinary generating function (OGF) is <inline-formula><tex-math id="math-31"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \textstyle F ( x ) = \sum _ { n > 0 } f _ { n } x ^ { n } \end{document} ]]></tex-math></inline-formula> . For a three-term linear recurrence <inline-formula><tex-math id="math-32"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle f _ { n + 3 } = a f _ { n + 2 } + b f _ { n + 1 } + \end{document} ]]></tex-math></inline-formula><inline-formula><tex-math id="math-33"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle c f _ { n } \end{document} ]]></tex-math></inline-formula> with fixed <inline-formula><tex-math id="math-34"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle a , b , c , \end{document} ]]></tex-math></inline-formula> the OGF satisfies</p><disp-formula id="equation-5"><tex-math id="math-35"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \left(1 - a x - b x ^ {2} - c x ^ {3}\right) F (x) = \text { a polynomial determined by } f _ {0}, f _ {1}, f _ {2}. \end{document} ]]></tex-math></disp-formula><p>We will also use trivariate OGFs of the form <inline-formula><tex-math id="math-36"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle F (x, y, z) = \sum_ {i, j, k \geq 0} \mu_ {i, j, k} x ^ {i} y ^ {j} z ^ {k}, \end{document} ]]></tex-math></inline-formula> together with the multinomial coeficients</p><disp-formula id="equation-6"><tex-math id="math-37"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \binom {r} {i, j, k} = \frac {r !}{i ! j ! k !}, \qquad i, j, k \geq 0, i + j + k = r. \end{document} ]]></tex-math></disp-formula><p>All generating functions are formal power series (no convergence issues). These tools let us encode and solve the coeficient recurrences appearing later.</p></sec></sec><sec id="sec-5"><title>3. MAIN RESULTS</title><p>The goal of this section is to obtain an explicit closed formula for the anti–trace of powers of <inline-formula><tex-math id="math-38"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle 3 \times 3 \end{document} ]]></tex-math></inline-formula> matrices over R via generating functions, and to record structural consequences for block–anti–diagonal matrices.</p><p>Let</p><disp-formula id="equation-7"><tex-math id="math-39"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A = \left( \begin{array}{c c c} a _ {1 1} & a _ {1 2} & a _ {1 3} \\ a _ {2 1} & a _ {2 2} & a _ {2 3} \\ a _ {3 1} & a _ {3 2} & a _ {3 3} \end{array} \right) \in M _ {3} (R), \end{document} ]]></tex-math></disp-formula><p>and let <inline-formula><tex-math id="math-40"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A _ { j k } \end{document} ]]></tex-math></inline-formula> denote the submatrix obtained by deleting the jth row and kth column.</p><p>Write <inline-formula><tex-math id="math-41"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle S _ { k } \end{document} ]]></tex-math></inline-formula> for the sum of the <inline-formula><tex-math id="math-42"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle k \times k \end{document} ]]></tex-math></inline-formula> principal minors:</p><disp-formula id="equation-8"><tex-math id="math-43"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle S _ {1} = a _ {1 1} + a _ {2 2} + a _ {3 3} = T r (A), \qquad S _ {2} = \det (A _ {1 1}) + \det (A _ {2 2}) + \det (A _ {3 3}), \qquad S _ {3} = \det (A). \end{document} ]]></tex-math></disp-formula><p>Likewise, define the anti–principal minors</p><disp-formula id="equation-9"><tex-math id="math-44"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle Q _ {1} = a _ {3 1} + a _ {2 2} + a _ {1 3} = T r _ {a} (A), \quad Q _ {2} = \det (A _ {3 1}) + \det (A _ {2 2}) + \det (A _ {1 3}), \quad Q _ {3} = \det (A). \end{document} ]]></tex-math></disp-formula><p>Note that <inline-formula><tex-math id="math-45"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle S _ { 3 } = Q _ { 3 } \end{document} ]]></tex-math></inline-formula> . Now, by the characteristic equation of A, <inline-formula><tex-math id="math-46"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \lambda^ {3} - S _ {1} \lambda^ {2} + S _ {2} \lambda - S _ {3} = 0, \end{document} ]]></tex-math></inline-formula> and by Cayley–Hamilton,</p><disp-formula id="equation-10"><tex-math id="math-47"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A ^ {3} - S _ {1} A ^ {2} + S _ {2} A - S _ {3} I = 0. \end{document} ]]></tex-math></disp-formula><p>Hence, for all <inline-formula><tex-math id="math-48"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle n \in \mathbb { N } \end{document} ]]></tex-math></inline-formula></p><disp-formula id="equation-11"><tex-math id="math-49"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A ^ {n + 3} = S _ {1} A ^ {n + 2} - S _ {2} A ^ {n + 1} + S _ {3} A ^ {n}.\tag{3} \end{document} ]]></tex-math></disp-formula><p>Applying the linear functional <inline-formula><tex-math id="math-50"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r _ { a } \end{document} ]]></tex-math></inline-formula> to <xref ref-type="disp-formula" rid="equation-11">(3)</xref> yields the third–order recurrence</p><disp-formula id="equation-12"><tex-math id="math-51"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r _ {a} (A ^ {n + 3}) = S _ {1} T r _ {a} (A ^ {n + 2}) - S _ {2} T r _ {a} (A ^ {n + 1}) + S _ {3} T r _ {a} (A ^ {n}), \qquad n \geq 0.\tag{4} \end{document} ]]></tex-math></disp-formula><sec id="sec-6"><title>3.1. Anti–trace formula via generating functions.</title><p>Using <xref ref-type="disp-formula" rid="equation-12">(4)</xref> we compute the first ten values of <inline-formula><tex-math id="math-52"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r _ { a } ( A ^ { n } ) \end{document} ]]></tex-math></inline-formula></p><table-wrap id="table-1"><label>Table 1</label><caption><p><inline-formula><tex-math id="math-53"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r _ { a } ( A ^ { n } ) \end{document} ]]></tex-math></inline-formula> from the recurrence <xref ref-type="disp-formula" rid="equation-12">(4)</xref></p></caption><table><colgroup><col></col><col></col></colgroup><thead><tr><th scope="col">n</th><th scope="col"><inline-formula><tex-math id="math-54"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle Tr_a(A^n) \end{document} ]]></tex-math></inline-formula></th></tr></thead><tbody><tr><td>1</td><td><inline-formula><tex-math id="math-55"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle Q_1(1) \end{document} ]]></tex-math></inline-formula></td></tr><tr><td>2</td><td><inline-formula><tex-math id="math-56"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle Q_1(S_1)+Q_2(1)-S_2 \end{document} ]]></tex-math></inline-formula></td></tr><tr><td>3</td><td><inline-formula><tex-math id="math-57"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle Q_1(S_1^2-S_2)+Q_2(S_1)-S_2(S_1)+S_3 \end{document} ]]></tex-math></inline-formula></td></tr><tr><td>4</td><td><inline-formula><tex-math id="math-58"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle Q_1(S_1^3-2S_2(S_1)+S_3)+Q_2(S_1^2-S_2)-S_2(S_1^2)+S_2^2+S_3(S_1) \end{document} ]]></tex-math></inline-formula></td></tr><tr><td>5</td><td><inline-formula><tex-math id="math-59"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle Q_1(S_1^4-3S_2(S_1^2)+S_2^2+2S_3(S_1))+Q_2(S_1^3-2S_2(S_1)+S_3)-S_2(S_1^3)+2S_2^2(S_1)-2S_2S_3+S_3(S_1^2) \end{document} ]]></tex-math></inline-formula></td></tr><tr><td>6</td><td><inline-formula><tex-math id="math-60"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{aligned} &Q_1(S_1^5-4S_2(S_1^3)+3S_2^2(S_1)+3S_3(S_1^2)-2S_2S_3)+Q_2(S_1^4-3S_2(S_1^2)+S_2^2+2S_3(S_1))\\ &-S_2(S_1^4)+3S_2^2(S_1^2)+S_3(S_1^3)-S_3^2-4S_2S_3(S_1)+S_3^2 \end{aligned} \end{document} ]]></tex-math></inline-formula></td></tr><tr><td>7</td><td><inline-formula><tex-math id="math-61"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{aligned} &Q_1(S_1^6-5S_2(S_1^4)+6S_2^2(S_1^2)-S_2^3+4S_3(S_1^3)-6S_2S_3(S_1)+S_3^2)+Q_2(S_1^5-4S_2(S_1^3)+3S_2^2(S_1)+3S_3(S_1^2)-2S_2S_3)\\ &-S_2(S_1^5)+S_3(S_1^4)+4S_2^2S_1^3-6S_2S_3(S_1^2)-3S_2^3(S_1)+2S_3^2(S_1)+3S_2^2S_3 \end{aligned} \end{document} ]]></tex-math></inline-formula></td></tr><tr><td>8</td><td><inline-formula><tex-math id="math-62"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{aligned} &Q_1(S_1^7-6S_2(S_1^5)+10S_2^2(S_1^3)+5S_3(S_1^4)-12S_2S_3(S_1^2)\\ &-4S_2^3(S_1)+3S_3^2(S_1)+3S_2^2S_3)+Q_2(S_1^6-5S_2(S_1^4)+4S_3(S_1^3)+6S_2^2(S_1^2)-6S_2S_3(S_1)-S_2^3+S_3^2)\\ &-S_2(S_1^6)+S_3(S_1^5)+5S_2^2(S_1^4)-8S_2S_3(S_1^3)-6S_2^3(S_1^2)+3S_3^2(S_1^2)+9S_2^2S_3(S_1)+S_2^4-3S_2S_3^2 \end{aligned} \end{document} ]]></tex-math></inline-formula></td></tr><tr><td>9</td><td><inline-formula><tex-math id="math-63"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{aligned} &Q_1(S_1^8-7S_2(S_1^6)+6S_3(S_1^5)+15S_2^2(S_1^4)-20S_2S_3(S_1^3)-10S_2^3(S_1^2)+6S_3^2(S_1^2)+12S_2^2S_3(S_1)+S_2^4-3S_2S_3^2)\\ &+Q_2(S_1^7-6S_2(S_1^5)+5S_3(S_1^4)+10S_2^2(S_1^3)-12S_2S_3(S_1^2)-4S_2^3(S_1)+3S_3^2(S_1)+3S_2^2S_3)\\ &-S_2(S_1^7)+S_3(S_1^6)+6S_2^2(S_1^5)-10S_2S_3(S_1^4)-10S_2^3(S_1^3)+4S_3^2(S_1^3)+18S_2^2S_3(S_1^2)\\ &+4S_2^4(S_1)-9S_2S_3^2(S_1)-4S_2^3S_3+S_3^3 \end{aligned} \end{document} ]]></tex-math></inline-formula></td></tr><tr><td>10</td><td><inline-formula><tex-math id="math-64"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{aligned} &Q_1(S_1^9-8S_2(S_1^7)+7S_3(S_1^6)+21S_2^2(S_1^5)-30S_2S_3(S_1^4)\\ &-20S_2^3(S_1^3)+10S_3^2(S_1^3)+30S_2^2S_3(S_1^2)+5S_2^4(S_1)-12S_2S_3^2(S_1)-4S_2^3S_3+S_3^3)\\ &+Q_2(S_1^8-7S_1^6S_2+6S_1^5S_3+15S_1^4S_2^2-20S_2S_3(S_1^3)-10S_2^3(S_1^2)+6S_1^2S_3^2+12S_1S_2^2S_3+S_2^4-3S_2S_3^2)\\ &-S_2(S_1^8)+S_3(S_1^7)+7S_2^2(S_1^6)-12S_2S_3(S_1^5)-15S_2^3(S_1^4)+5S_3^2(S_1^4)+30S_2^2S_3(S_1^3)\\ &+10S_2^4(S_1^2)-18S_2S_3^2(S_1^2)-16S_2^3S_3(S_1)+3S_3^3(S_1)-S_2^5+6S_2^2S_3^2 \end{aligned} \end{document} ]]></tex-math></inline-formula></td></tr></tbody></table></table-wrap><p>The pattern suggests</p><disp-formula id="equation-13"><tex-math id="math-65"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} T r _ {a} (A ^ {n}) = Q _ {1} \left(\sum_ {i + 2 j + 3 k = n - 1} \alpha_ {i, j, k} S _ {1} ^ {i} S _ {2} ^ {j} S _ {3} ^ {k}\right) + Q _ {2} \left(\sum_ {i + 2 j + 3 k = n - 2} \beta_ {i, j, k} S _ {1} ^ {i} S _ {2} ^ {j} S _ {3} ^ {k}\right) \\ + \sum_ {i + 2 j + 3 k = n} \gamma_ {i, j, k} S _ {1} ^ {i} S _ {2} ^ {j} S _ {3} ^ {k}, \end{array}\tag{5} \end{document} ]]></tex-math></disp-formula><p>with <inline-formula><tex-math id="math-66"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \alpha _ { i , j , k } , \beta _ { i , j , k } , \gamma _ { i , j , k } \end{document} ]]></tex-math></inline-formula> depending only on <inline-formula><tex-math id="math-67"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle ( i , j , k ) \end{document} ]]></tex-math></inline-formula></p><p>• Term <inline-formula><tex-math id="math-68"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \alpha _ { i , j , k } \end{document} ]]></tex-math></inline-formula></p><p>Using equation <xref ref-type="disp-formula" rid="equation-12">(4)</xref>, the coeficients of <inline-formula><tex-math id="math-69"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle S _ { 1 } ^ { i } S _ { 2 } ^ { j } S _ { 3 } ^ { k } \end{document} ]]></tex-math></inline-formula> on both sides implies</p><disp-formula id="equation-14"><tex-math id="math-70"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \alpha_ {i, j, k} = \alpha_ {i - 1, j, k} - \alpha_ {i, j - 1, k} + \alpha_ {i, j, k - 1} \end{document} ]]></tex-math></disp-formula><p>where <inline-formula><tex-math id="math-71"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle i , j , k \geq 1 \end{document} ]]></tex-math></inline-formula> . Define a generating function <inline-formula><tex-math id="math-72"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle F _ { 1 } ( x , y , z ) \end{document} ]]></tex-math></inline-formula> as follows</p><disp-formula id="equation-15"><tex-math id="math-73"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle F _ {1} (x, y, z) = \sum_ {i, j, k \geq 0} \alpha_ {i, j, k} x ^ {i} y ^ {j} z ^ {k} \end{document} ]]></tex-math></disp-formula><p>Accordingly, we obtain</p><disp-formula id="equation-16"><tex-math id="math-74"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle F _ {1} (x, y, z) - x F _ {1} (x, y, z) + y F _ {1} (x, y, z) - z F _ {1} (x, y, z) = x - y + z\tag{6} \end{document} ]]></tex-math></disp-formula><p>The above equation is true since for <inline-formula><tex-math id="math-75"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle i + j + k \ge 2 \end{document} ]]></tex-math></inline-formula> the function <inline-formula><tex-math id="math-76"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \alpha _ { i , j , k } \end{document} ]]></tex-math></inline-formula> satisfies</p><disp-formula id="equation-17"><tex-math id="math-77"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \alpha_ {i, j, k} x ^ {i} y ^ {j} z ^ {k} - \alpha_ {i - 1, j, k} x ^ {i - 1} y ^ {j} z ^ {k} x + \alpha_ {i, j - 1, k} x ^ {i} y ^ {j - 1} z ^ {k} y - \alpha_ {i, j, k - 1} x ^ {i} y ^ {j} z ^ {k - 1} z = 0 \end{document} ]]></tex-math></disp-formula><p>Therefore, by equation <xref ref-type="disp-formula" rid="equation-16">(6)</xref> it implies</p><disp-formula id="equation-18"><tex-math id="math-78"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} F _ {1} (x, y, z) = \frac {x - y + z}{1 - x + y - z} \\ \qquad = (x - y + z) \left(1 + \sum_ {t = 1} ^ {\infty} (x - y + z) ^ {t}\right) \\ \qquad = (x - y + z) \left(1 + \sum_ {t = 1} ^ {\infty} \sum_ {k _ {1} + k _ {2} + k _ {3} = t} \binom {t} {k _ {1}, k _ {2}, k _ {3}} x ^ {k _ {1}} (- y) ^ {k _ {2}} z ^ {k _ {3}}\right) \\ \qquad = (x - y + z) \left(1 + \sum_ {t = 1} ^ {\infty} \sum_ {k _ {1} + k _ {2} + k _ {3} = t} (- 1) ^ {k _ {2}} \binom {t} {k _ {1}, k _ {2}, k _ {3}} x ^ {k _ {1}} y ^ {k _ {2}} z ^ {k _ {3}}\right) \end{array} \end{document} ]]></tex-math></disp-formula><p>Then, it follows that</p><disp-formula id="equation-19"><tex-math id="math-79"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} \alpha_ {i, j, k} = (- 1) ^ {j} \binom {i + j + k - 1} {i - 1, j, k} - (- 1) ^ {j - 1} \binom {i + j + k - 1} {i, j - 1, k} + (- 1) ^ {j} \binom {i + j + k - 1} {i, j, k - 1} \\ \qquad = (- 1) ^ {j} \left(\binom {i + j + k - 1} {i - 1, j, k} + \binom {i + j + k - 1} {i, j - 1, k} + \binom {i + j + k - 1} {i, j, k - 1}\right) \\ \qquad = (- 1) ^ {j} \left(\frac {(i + j + k - 1) !}{(i - 1) ! (j - 1) ! (k - 1) !} \left(\frac {1}{j k} + \frac {1}{i k} + \frac {1}{i j}\right)\right) \\ \qquad = (- 1) ^ {j} \left(\frac {(i + j + k - 1) !}{(i - 1) ! (j - 1) ! (k - 1) !} \left(\frac {i + j + k}{i j k}\right)\right) \\ \qquad = (- 1) ^ {j} \frac {(i + j + k) !}{i ! j ! k !} \\ \qquad = (- 1) ^ {j} \binom {i + j + k} {i, j, k} \end{array} \end{document} ]]></tex-math></disp-formula><p>As a conclusion, the function <inline-formula><tex-math id="math-80"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \alpha _ { i , j , k } \end{document} ]]></tex-math></inline-formula> can be written as</p><disp-formula id="equation-20"><tex-math id="math-81"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \alpha_ {i, j, k} = (- 1) ^ {j} \binom {i + j + k} {i, j, k}\tag{7} \end{document} ]]></tex-math></disp-formula><p>• Term <inline-formula><tex-math id="math-82"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \beta _ { i , j , k } \end{document} ]]></tex-math></inline-formula></p><p>Without loss of generality, the same technique is applied in order to find</p><p><inline-formula><tex-math id="math-83"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \beta _ { i , j , k } \end{document} ]]></tex-math></inline-formula> . It yields a similar expression as <inline-formula><tex-math id="math-84"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \alpha _ { i , j } \end{document} ]]></tex-math></inline-formula> ,k.</p><disp-formula id="equation-21"><tex-math id="math-85"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \beta_ {i, j, k} = (- 1) ^ {j} \binom {i + j + k} {i, j, k}\tag{8} \end{document} ]]></tex-math></disp-formula><p>• Term <inline-formula><tex-math id="math-86"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \gamma _ { i , j , k } \end{document} ]]></tex-math></inline-formula></p><p>Interestingly, using the same method for finding <inline-formula><tex-math id="math-87"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \gamma _ { i , j , k } \end{document} ]]></tex-math></inline-formula> difers slightly from the other cases. By comparing the coeficient of <inline-formula><tex-math id="math-88"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle S _ { 1 } ^ { i } S _ { 2 } ^ { j } S _ { 3 } ^ { k } \end{document} ]]></tex-math></inline-formula> on both sides and using equation <xref ref-type="disp-formula" rid="equation-12">(4)</xref>, this implies for <inline-formula><tex-math id="math-89"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle i , j , k \geq 1 \end{document} ]]></tex-math></inline-formula></p><disp-formula id="equation-22"><tex-math id="math-90"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \gamma_ {i, j, k} = \gamma_ {i - 1, j, k} - \gamma_ {i, j - 1, k} + \gamma_ {i, j, k - 1} \end{document} ]]></tex-math></disp-formula><p>Define the following generating function</p><disp-formula id="equation-23"><tex-math id="math-91"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle F _ {2} (x, y, z) = \sum_ {i, j, k \geq 0} \gamma_ {i, j, k} x ^ {i} y ^ {j} z ^ {k} \end{document} ]]></tex-math></disp-formula><p>Consequently, it results</p><disp-formula id="equation-24"><tex-math id="math-92"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle F _ {2} (x, y, z) - x F _ {2} (x, y, z) + y F _ {2} (x, y, z) - z F _ {2} (x, y, z) = - y + z\tag{9} \end{document} ]]></tex-math></disp-formula><p>The above equation is true since for <inline-formula><tex-math id="math-93"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle i + j + k \ge 2 \end{document} ]]></tex-math></inline-formula> the function <inline-formula><tex-math id="math-94"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \gamma _ { i , j , k } \end{document} ]]></tex-math></inline-formula> satisfies</p><disp-formula id="equation-25"><tex-math id="math-95"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \gamma_ {i, j, k} x ^ {i} y ^ {j} z ^ {k} - \gamma_ {i - 1, j, k} x ^ {i - 1} y ^ {j} z ^ {k} x + \gamma_ {i, j - 1, k} x ^ {i} y ^ {j - 1} z ^ {k} y - \gamma_ {i, j, k - 1} x ^ {i} y ^ {j} z ^ {k - 1} z = 0 \end{document} ]]></tex-math></disp-formula><p>Therefore, by equation <xref ref-type="disp-formula" rid="equation-24">(9)</xref> it implies</p><disp-formula id="equation-26"><tex-math id="math-96"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} F _ {2} (x, y, z) = \frac {- y + z}{1 - x + y - z} \\ \qquad = (- y + z) \left(1 + \sum_ {t = 1} ^ {\infty} (x - y + z) ^ {t}\right) \\ \qquad = (- y + z) \left(1 + \sum_ {t = 1} ^ {\infty} \sum_ {k _ {1} + k _ {2} + k _ {3} = t} \binom {t} {k _ {1}, k _ {2}, k _ {3}} x ^ {k _ {1}} (- y) ^ {k _ {2}} z ^ {k _ {3}}\right) \\ \qquad = (- y + z) \left(1 + \sum_ {t = 1} ^ {\infty} \sum_ {k _ {1} + k _ {2} + k _ {3} = t} (- 1) ^ {k _ {2}} \binom {t} {k _ {1}, k _ {2}, k _ {3}} x ^ {k _ {1}} y ^ {k _ {2}} z ^ {k _ {3}}\right) \end{array} \end{document} ]]></tex-math></disp-formula><p>Then, it follows that</p><disp-formula id="equation-27"><tex-math id="math-97"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} \gamma_ {i, j, k} = - (- 1) ^ {j - 1} \binom {i + j + k - 1} {i, j - 1, k} + (- 1) ^ {j} \binom {i + j + k - 1} {i, j, k - 1} \\ \qquad = (- 1) ^ {j} \left(\binom {i + j + k - 1} {i, j - 1, k} + \binom {i + j + k - 1} {i, j, k - 1}\right) \\ \qquad = (- 1) ^ {j} \left(\frac {(i + j + k - 1) !}{(i - 1) ! (j - 1) ! (k - 1) !} \left(\frac {1}{i k} + \frac {1}{i j}\right)\right) \\ \qquad = (- 1) ^ {j} \left(\frac {(i + j + k - 1) !}{(i - 1) ! (j - 1) ! (k - 1) !} \left(\frac {j + k}{i j k}\right)\right) \\ \qquad = (- 1) ^ {j} \frac {(i + j + k - 1) !}{i ! j ! k !} (j + k) \end{array} \end{document} ]]></tex-math></disp-formula><disp-formula id="equation-28"><tex-math id="math-98"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle = (- 1) ^ {j} \frac {(j + k)}{(i + j + k)} \binom {i + j + k} {i, j, k} \end{document} ]]></tex-math></disp-formula><p>As a conclusion, the function <inline-formula><tex-math id="math-99"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \gamma _ { i , j , k } \end{document} ]]></tex-math></inline-formula> can be written as</p><disp-formula id="equation-29"><tex-math id="math-100"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \gamma_ {i, j, k} = (- 1) ^ {j} \frac {(j + k)}{(i + j + k)} \binom {i + j + k} {i, j, k}\tag{10} \end{document} ]]></tex-math></disp-formula><p>For example, if <inline-formula><tex-math id="math-101"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \gamma _ { 1 , 2 , 1 } = 9 \end{document} ]]></tex-math></inline-formula> , then the coeficient of <inline-formula><tex-math id="math-102"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle S _ { 1 } S _ { 2 } ^ { 2 } S _ { 3 } \end{document} ]]></tex-math></inline-formula> is 9 in <inline-formula><tex-math id="math-103"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r _ { a } ( A ^ { 8 } ) \end{document} ]]></tex-math></inline-formula> and this is true based on <xref ref-type="table" rid="table-1">Table 1</xref>.</p><p>To sum up, by substituting equations <xref ref-type="disp-formula" rid="equation-20">(7)</xref>, <xref ref-type="disp-formula" rid="equation-21">(8)</xref>, and <xref ref-type="disp-formula" rid="equation-29">(10)</xref> into equation <xref ref-type="disp-formula" rid="equation-13">(5)</xref>, hence the explicit formula for <italic>anti-trace</italic> is obtained and stated in the following theorem.</p><p><bold>Theorem 3.1</bold>. For <inline-formula><tex-math id="math-104"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \phantom { - } \ i \ 3 \times 3 \end{document} ]]></tex-math></inline-formula> matrix A and positive integer n,</p><disp-formula id="equation-30"><tex-math id="math-105"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} T r _ {a} (A ^ {n}) = Q _ {1} \left(\sum_ {i + 2 j + 3 k = n - 1} (- 1) ^ {j} \binom {i + j + k} {i, j, k} S _ {1} ^ {i} S _ {2} ^ {j} S _ {3} ^ {k}\right) \\ \quad + Q _ {2} \left(\sum_ {i + 2 j + 3 k = n - 2} (- 1) ^ {j} \binom {i + j + k} {i, j, k} S _ {1} ^ {i} S _ {2} ^ {j} S _ {3} ^ {k}\right) \\ \quad + \sum_ {i + 2 j + 3 k = n} (- 1) ^ {j} \frac {(j + k)}{(i + j + k)} \binom {i + j + k} {i, j, k} S _ {1} ^ {i} S _ {2} ^ {j} S _ {3} ^ {k}. \end{array} \end{document} ]]></tex-math></disp-formula><p>□</p><p><bold>Example 3.2.</bold><italic>Let</italic></p><disp-formula id="equation-31"><tex-math id="math-106"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A = \left( \begin{array}{c c c} 1 & 0 & 2 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array} \right). \end{document} ]]></tex-math></disp-formula><p><italic>Then</italic></p><disp-formula id="equation-32"><tex-math id="math-107"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle S _ {1} = 6, \quad S _ {2} = \det (A _ {1 1}) + \det (A _ {2 2}) + \det (A _ {3 3}) = 1 1, \quad S _ {3} = \det (A) = 6, \end{document} ]]></tex-math></disp-formula><p><italic>and the anti–principal sums are</italic></p><disp-formula id="equation-33"><tex-math id="math-108"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle Q _ {1} = a _ {3 1} + a _ {2 2} + a _ {1 3} = 4, \quad Q _ {2} = \det (A _ {3 1}) + \det (A _ {2 2}) + \det (A _ {1 3}) = - 1. \end{document} ]]></tex-math></disp-formula><p><italic>Since A is upper triangular with (1, 3) entry 2, one checks directly that</italic></p><disp-formula id="equation-34"><tex-math id="math-109"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle (A ^ {n}) _ {1 3} = 3 ^ {n} - 1, \qquad (A ^ {n}) _ {2 2} = 2 ^ {n}, \qquad (A ^ {n}) _ {3 1} = 0, \end{document} ]]></tex-math></disp-formula><p><italic>hence</italic></p><disp-formula id="equation-35"><tex-math id="math-110"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r _ {a} (A ^ {n}) = 2 ^ {n} + 3 ^ {n} - 1 \quad f o r a l l n \geq 1. \end{document} ]]></tex-math></disp-formula><p><italic>We now compare with the theorem.</italic></p><p><italic>• For </italic><inline-formula><tex-math id="math-111"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle n = 4 \end{document} ]]></tex-math></inline-formula><italic> , the table/recurrence gives</italic></p><disp-formula id="equation-36"><tex-math id="math-112"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{r l} & T r _ {a} (A ^ {4}) = Q _ {1} (S _ {1} ^ {3} - 2 S _ {2} S _ {1} + S _ {3}) + Q _ {2} (S _ {1} ^ {2} - S _ {2}) - S _ {2} S _ {1} ^ {2} + S _ {2} ^ {2} + S _ {3} S _ {1} \\ & \qquad = 4 (2 1 6 - 2 \cdot 1 1 \cdot 6 + 6) + (- 1) (3 6 - 1 1) - 1 1 \cdot 3 6 + 1 1 ^ {2} + 6 \cdot 6 \\ & \qquad = 9 6, \end{array} \end{document} ]]></tex-math></disp-formula><p><italic>which matches </italic><inline-formula><tex-math id="math-113"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle 2 ^ { 4 } + 3 ^ { 4 } - 1 = 1 6 + 8 1 - 1 = 9 6 \end{document} ]]></tex-math></inline-formula></p><p><italic>• For </italic><inline-formula><tex-math id="math-114"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle n = 5 \end{document} ]]></tex-math></inline-formula></p><disp-formula id="equation-37"><tex-math id="math-115"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} T r _ {a} (A ^ {5}) = Q _ {1} (S _ {1} ^ {4} - 3 S _ {2} S _ {1} ^ {2} + S _ {2} ^ {2} + 2 S _ {3} S _ {1}) + Q _ {2} (S _ {1} ^ {3} - 2 S _ {2} S _ {1} + S _ {3}) \\ \qquad - S _ {2} S _ {1} ^ {3} + 2 S _ {2} ^ {2} S _ {1} - 2 S _ {2} S _ {3} + S _ {3} S _ {1} ^ {2} \\ \qquad = 4 (1 2 9 6 - 3 \cdot 1 1 \cdot 3 6 + 1 2 1 + 2 \cdot 6 \cdot 6) + (- 1) (2 1 6 - 2 \cdot 1 1 \cdot 6 + 6) \\ \qquad - 1 1 \cdot 2 1 6 + 2 \cdot 1 2 1 \cdot 6 - 2 \cdot 1 1 \cdot 6 + 6 \cdot 3 6 \\ \qquad = 2 7 4, \end{array} \end{document} ]]></tex-math></disp-formula><p><italic>which matches </italic><inline-formula><tex-math id="math-116"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle 2 ^ { 5 } + 3 ^ { 5 } - 1 = 3 2 + 2 4 3 - 1 = 2 7 4 \end{document} ]]></tex-math></inline-formula></p></sec><sec id="sec-7"><title>3.2. More results on block–anti–diagonal matrices.</title><p>We investigate the anti–trace of powers of block–anti–diagonal 3m × 3m matrices with <inline-formula><tex-math id="math-117"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle 3 \times 3 \end{document} ]]></tex-math></inline-formula> blocks.<target id="anchor-1" target-type="reference-target"/></p><p><bold>Theorem 3.3 </bold>(Even number of blocks).<italic> Let m be even and</italic></p><disp-formula id="equation-38"><tex-math id="math-118"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A = \left( \begin{array}{c c c c c} \mathbf {0} & \mathbf {0} & \dots & \mathbf {0} & A _ {m} \\ \mathbf {0} & \mathbf {0} & \dots & A _ {m - 1} & \mathbf {0} \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ \mathbf {0} & A _ {2} & \dots & \mathbf {0} & \mathbf {0} \\ A _ {1} & \mathbf {0} & \dots & \mathbf {0} & \mathbf {0} \end{array} \right), \qquad A _ {k} \in M _ {3} (R). \end{document} ]]></tex-math></disp-formula><p><italic>Then for all</italic><inline-formula><tex-math id="math-119"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle n \geq 1 . \end{document} ]]></tex-math></inline-formula></p><disp-formula id="equation-39"><tex-math id="math-120"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} (i) T r _ {a} (A ^ {2 n}) = 0. \\ (i i) T r _ {a} (A ^ {2 n + 1}) = \sum_ {i = 1} ^ {m} T r _ {a} \big ((A _ {m + 1 - i} A _ {i}) ^ {n} A _ {m + 1 - i} \big). \end{array} \end{document} ]]></tex-math></disp-formula><p><italic>Proof</italic>. A direct block multiplication shows</p><disp-formula id="equation-40"><tex-math id="math-121"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A ^ {2} = \operatorname{diag} \left(A _ {m} A _ {1}, A _ {m - 1} A _ {2}, \dots , A _ {1} A _ {m}\right). \end{document} ]]></tex-math></disp-formula><p>Hence <inline-formula><tex-math id="math-122"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A ^ { 2 n } \end{document} ]]></tex-math></inline-formula> is block–diagonal for every n. When m is even, the global anti–diagonal of the m × m block grid passes between diagonal blocks,</p><disp-formula id="equation-41"><tex-math id="math-123"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A ^ {2 n} = \left( \begin{array}{c c c c c} (A _ {m} A _ {1}) ^ {n} & \mathbf {0} & \mathbf {0} & \mathbf {0} & \mathbf {0} \\ \mathbf {0} & (A _ {m - 1} A _ {2}) ^ {n} & \dots & \mathbf {0} & \mathbf {0} \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ \mathbf {0} & \mathbf {0} & \dots & (A _ {2} A _ {m - 1}) ^ {n} & \mathbf {0} \\ \mathbf {0} & \mathbf {0} & \dots & \mathbf {0} & (A _ {1} A _ {m}) ^ {n} \end{array} \right) \end{document} ]]></tex-math></disp-formula><p>Then every entry on the (full) anti–diagonal of <inline-formula><tex-math id="math-124"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A ^ { 2 n } \end{document} ]]></tex-math></inline-formula> is zero; thus <inline-formula><tex-math id="math-125"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r _ { a } ( A ^ { 2 n } ) = 0 \end{document} ]]></tex-math></inline-formula></p><p>For odd powers, we have</p><disp-formula id="equation-42"><tex-math id="math-126"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A ^ {2 n + 1} = \left( \begin{array}{c c c c c} \mathbf {0} & \mathbf {0} & \dots & \mathbf {0} & (A _ {m} A _ {1}) ^ {n} A _ {m} \\ \mathbf {0} & \mathbf {0} & \dots & (A _ {m - 1} A _ {2}) ^ {n} A _ {m - 1} & \mathbf {0} \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ \mathbf {0} & (A _ {2} A _ {m - 1}) ^ {n} A _ {2} & \dots & \mathbf {0} & \mathbf {0} \\ (A _ {1} A _ {m}) ^ {n} A _ {1} & \mathbf {0} & \dots & \mathbf {0} & \mathbf {0} \end{array} \right) \end{document} ]]></tex-math></disp-formula><p>As we can see, <inline-formula><tex-math id="math-127"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A ^ { 2 n + 1 } \end{document} ]]></tex-math></inline-formula> has nonzero blocks exactly on the block anti–diagonal:</p><disp-formula id="equation-43"><tex-math id="math-128"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle (A ^ {2 n + 1}) _ {i, m + 1 - i} = (A _ {m + 1 - i} A _ {i}) ^ {n} A _ {m + 1 - i}. \end{document} ]]></tex-math></disp-formula><p>Along the full matrix anti–diagonal, one picks out the anti–diagonals of these blocks yield <inline-formula><tex-math id="math-129"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r _ { a } ( A ^ { 2 n + 1 } ) = \sum _ { i = 1 } ^ { m } T r _ { a } \bigl ( ( A _ { m + 1 - i } A _ { i } ) ^ { n } A _ { m + 1 - i } \bigr ) \end{document} ]]></tex-math></inline-formula> □</p><p><bold>Example 3.4</bold> (Even number of blocks). Take <inline-formula><tex-math id="math-130"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle m = 2 \end{document} ]]></tex-math></inline-formula><italic>and</italic></p><disp-formula id="equation-44"><tex-math id="math-131"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A = \left( \begin{array}{c c} \mathbf {0} & A _ {2} \\ A _ {1} & \mathbf {0} \end{array} \right), \qquad A _ {1} = \mathrm{diag} (1, 2, 3), \quad A _ {2} = \mathrm{diag} (4, 5, 6). \end{document} ]]></tex-math></disp-formula><p><italic>Then</italic></p><disp-formula id="equation-45"><tex-math id="math-132"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A ^ {2} = \operatorname{diag} (A _ {2} A _ {1}, A _ {1} A _ {2}) = \operatorname{diag} (\operatorname{diag} (4, 1 0, 1 8), \operatorname{diag} (4, 1 0, 1 8)). \end{document} ]]></tex-math></disp-formula><p><italic>Because m is even, the global anti–diagonal of </italic><inline-formula><tex-math id="math-133"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A ^ { 2 n } \end{document} ]]></tex-math></inline-formula><italic> passes through of–diagonal blocks only; hence </italic><inline-formula><tex-math id="math-134"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r _ { a } ( A ^ { 2 n } ) = 0 \end{document} ]]></tex-math></inline-formula><italic> . In particular, </italic><inline-formula><tex-math id="math-135"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r _ { a } ( A ^ { 2 } ) = 0 \end{document} ]]></tex-math></inline-formula></p><p><italic>For odd powers,</italic></p><disp-formula id="equation-46"><tex-math id="math-136"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A ^ {3} = \left( \begin{array}{c c} A _ {2} A _ {1} A _ {2} & \mathbf {0} \\ \mathbf {0} & A _ {1} A _ {2} A _ {1} \end{array} \right) \quad \Longrightarrow \quad T r _ {a} (A ^ {3}) = T r _ {a} (A _ {2} A _ {1} A _ {2}) + T r _ {a} (A _ {1} A _ {2} A _ {1}). \end{document} ]]></tex-math></disp-formula><p><italic>Since the anti–trace of a diagonal </italic><inline-formula><tex-math id="math-137"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle 3 \times 3 \end{document} ]]></tex-math></inline-formula><italic> matrix diag </italic><inline-formula><tex-math id="math-138"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle ( d _ { 1 } , d _ { 2 } , d _ { 3 } ) \end{document} ]]></tex-math></inline-formula><italic> equals </italic><inline-formula><tex-math id="math-139"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle d _ { 2 } \end{document} ]]></tex-math></inline-formula><italic> , we get</italic></p><disp-formula id="equation-47"><tex-math id="math-140"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r _ {a} (A _ {2} A _ {1} A _ {2}) = 5 0, \qquad T r _ {a} (A _ {1} A _ {2} A _ {1}) = 2 0, \qquad T r _ {a} (A ^ {3}) = 7 0, \end{document} ]]></tex-math></disp-formula><p><italic>exactly as predicted by Theorem </italic><xref ref-type="custom" custom-type="reference-target" rid="anchor-1">3.3</xref><italic> (ii) with </italic><inline-formula><tex-math id="math-141"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle n = 1 \end{document} ]]></tex-math></inline-formula><target id="anchor-2" target-type="reference-target"/></p><p><bold>Theorem 3.5</bold> (Odd number of blocks). Let m be odd and <inline-formula><tex-math id="math-142"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array} { r } { c = \frac { m + 1 } { 2 } } \end{array} \end{document} ]]></tex-math></inline-formula> . With A as above, for all <inline-formula><tex-math id="math-143"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle n \geq 1 \end{document} ]]></tex-math></inline-formula></p><disp-formula id="equation-48"><tex-math id="math-144"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle (i) T r _ {a} (A ^ {2 n}) = T r _ {a} \big ([ A _ {c} ] ^ {2 n} \big). \end{document} ]]></tex-math></disp-formula><disp-formula id="equation-49"><tex-math id="math-145"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle (ii) Tr_{a}(A^{2n + 1}) = Tr_{a}\big([A_{c}]^{2n + 1}\big) + \sum_{\substack{i = 1\\ i\neq c}}^{m}Tr_{a}\big((A_{m + 1 - i}A_{i})^{n} A_{m + 1 - i}\big). \end{document} ]]></tex-math></disp-formula><p><italic>Proof</italic>. As before, <inline-formula><tex-math id="math-146"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A ^ { 2 n } \end{document} ]]></tex-math></inline-formula> is obtained via direct block multiplication,</p><disp-formula id="equation-50"><tex-math id="math-147"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A ^ {2 n} = \left( \begin{array}{c c c c c c c} (A _ {m} A _ {1}) ^ {n} & \mathbf {0} & \dots & \dots & \dots & \mathbf {0} & \mathbf {0} \\ \mathbf {0} & (A _ {m - 1} A _ {2}) ^ {n} & \dots & \dots & \dots & \mathbf {0} & \mathbf {0} \\ \vdots & \vdots & \dots & \dots & \ddots & \vdots & \vdots \\ \vdots & \vdots & \vdots & [ A _ {c} ] ^ {2 n} & \vdots & \vdots & \vdots \\ \vdots & \vdots & \ddots & \dots & \dots & \vdots & \vdots \\ \mathbf {0} & \mathbf {0} & \dots & \dots & \dots & (A _ {2} A _ {m - 1}) ^ {n} & \mathbf {0} \\ \mathbf {0} & \mathbf {0} & \dots & \dots & \dots & \mathbf {0} & (A _ {1} A _ {m}) ^ {n} \end{array} \right) \end{document} ]]></tex-math></disp-formula><p>When m is odd, the global anti–diagonal meets only the central diagonal block, so <inline-formula><tex-math id="math-148"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r _ { a } ( A ^ { 2 n } ) \end{document} ]]></tex-math></inline-formula> reduces to the anti–trace of <inline-formula><tex-math id="math-149"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle [ A _ { c } ] ^ { 2 n } \end{document} ]]></tex-math></inline-formula></p><p>For odd powers, we have</p><disp-formula id="equation-51"><tex-math id="math-150"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A ^ {2 n + 1} = \left( \begin{array}{c c c c c c c} \mathbf {0} & \mathbf {0} & \dots & \dots & \dots & \mathbf {0} & (A _ {m} A _ {1}) ^ {n} A _ {m} \\ \mathbf {0} & \mathbf {0} & \dots & \dots & \dots & (A _ {m - 1} A _ {2}) ^ {n} A _ {m - 1} & \mathbf {0} \\ \vdots & \vdots & \dots & \dots & \ddots & \vdots & \vdots \\ \vdots & \vdots & \vdots & [ A _ {c} ] ^ {2 n + 1} & \vdots & \vdots & \vdots \\ \vdots & \vdots & \ddots & \dots & \dots & \vdots & \vdots \\ \mathbf {0} & (A _ {2} A _ {m - 1}) ^ {n} A _ {2} & \dots & \dots & \dots & \mathbf {0} & \mathbf {0} \\ (A _ {1} A _ {m}) ^ {n} A _ {1} & \mathbf {0} & \dots & \dots & \dots & \mathbf {0} & \mathbf {0} \end{array} \right) \end{document} ]]></tex-math></disp-formula><p>So, <inline-formula><tex-math id="math-151"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A ^ { 2 n + 1 } = A ^ { 2 n } A \end{document} ]]></tex-math></inline-formula> has nonzero blocks on the block anti–diagonal with</p><disp-formula id="equation-52"><tex-math id="math-152"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle (A ^ {2 n + 1}) _ {i, m + 1 - i} = \left\{ \begin{array}{l l} (A _ {m + 1 - i} A _ {i}) ^ {n} A _ {m + 1 - i}, & i \neq c, \\ [ A _ {c} ] ^ {2 n + 1}, & i = c. \end{array} \right. \end{document} ]]></tex-math></disp-formula><p>Taking the sum of the anti–traces of these blocks along the full anti–diagonal gives</p><disp-formula id="equation-53"><tex-math id="math-153"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle Tr_{a}(A^{2n + 1}) = Tr_{a}\big([A_{c}]^{2n + 1}\big) + \sum_{\substack{i = 1\\ i\neq c}}^{m}Tr_{a}\big((A_{m + 1 - i}A_{i})^{n} A_{m + 1 - i}\big) \end{document} ]]></tex-math></disp-formula><p><bold>Example 3.6</bold> (Odd number of blocks). Let <inline-formula><tex-math id="math-154"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle m = 3 \end{document} ]]></tex-math></inline-formula> with central index <inline-formula><tex-math id="math-155"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle c = 2 \end{document} ]]></tex-math></inline-formula> and</p><disp-formula id="equation-54"><tex-math id="math-156"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A = \left( \begin{array}{c c c} \mathbf {0} & \mathbf {0} & A _ {3} \\ \mathbf {0} & A _ {2} & \mathbf {0} \\ A _ {1} & \mathbf {0} & \mathbf {0} \end{array} \right), \qquad A _ {1} = \operatorname{diag} (1, 2, 3), A _ {2} = \operatorname{diag} (4, 5, 6), A _ {3} = \operatorname{diag} (7, 8, 9). \end{document} ]]></tex-math></disp-formula><p><italic>Then </italic><inline-formula><tex-math id="math-157"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A ^ {2} = \mathrm{diag} (A _ {3} A _ {1}, A _ {2} ^ {2}, A _ {1} A _ {3}) \end{document} ]]></tex-math></inline-formula><italic>so the global anti–diagonal meets only the middle block. Therefore</italic></p><disp-formula id="equation-55"><tex-math id="math-158"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r _ {a} (A ^ {2}) = T r _ {a} (A _ {2} ^ {2}) = T r _ {a} (\mathrm{diag} (1 6, 2 5, 3 6)) = 2 5, \end{document} ]]></tex-math></disp-formula><p><italic>confirming Theorem </italic><xref ref-type="custom" custom-type="reference-target" rid="anchor-2">3.5</xref><italic> (i) for </italic><inline-formula><tex-math id="math-159"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle n = 1 \end{document} ]]></tex-math></inline-formula></p><p><italic>For odd powers,</italic></p><disp-formula id="equation-56"><tex-math id="math-160"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A ^ {3} = \left( \begin{array}{c c c} A _ {3} A _ {1} A _ {3} & \mathbf {0} & \mathbf {0} \\ \mathbf {0} & A _ {2} ^ {3} & \mathbf {0} \\ \mathbf {0} & \mathbf {0} & A _ {1} A _ {3} A _ {1} \end{array} \right), \end{document} ]]></tex-math></disp-formula><p><italic>and hence, using </italic><inline-formula><tex-math id="math-161"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r _ { a } ( \mathrm { d i a g } ( d _ { 1 } , d _ { 2 } , d _ { 3 } ) ) = d _ { 2 } \end{document} ]]></tex-math></inline-formula><italic> 2</italic></p><disp-formula id="equation-57"><tex-math id="math-162"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{r l} T r _ {a} (A ^ {3}) & = T r _ {a} (A _ {2} ^ {3}) + T r _ {a} (A _ {3} A _ {1} A _ {3}) + T r _ {a} (A _ {1} A _ {3} A _ {1}) \\ & = 1 2 5 + 1 2 8 + 3 2 = 2 8 5, \end{array} \end{document} ]]></tex-math></disp-formula><p><italic>which matches Theorem</italic><xref ref-type="custom" custom-type="reference-target" rid="anchor-2"> 3.5 </xref><italic>(ii)  with </italic><inline-formula><tex-math id="math-163"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle n = 1 \end{document} ]]></tex-math></inline-formula></p></sec><sec id="sec-8"><title>3.3. Computational Remarks.</title><p>For a single numerical instance <inline-formula><tex-math id="math-164"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle ( A , n ) \end{document} ]]></tex-math></inline-formula> , computing <inline-formula><tex-math id="math-165"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A ^ { n } \end{document} ]]></tex-math></inline-formula> by exponentiation by squaring and then taking <inline-formula><tex-math id="math-166"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r _ { a } \end{document} ]]></tex-math></inline-formula> is typically faster than evaluating the closed form. <inline-formula><tex-math id="math-167"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \mathrm { ~ A ~ 3 ~ } \times \mathrm { ~ 3 ~ } \end{document} ]]></tex-math></inline-formula> matrix power by squaring uses Θ(log n) matrix multiplications; each <inline-formula><tex-math id="math-168"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle 3 \times 3 \end{document} ]]></tex-math></inline-formula> multiplication costs a fixed number of scalar operations (e.g., 27 multiplies and 18 additions in the naive scheme), so the overall cost is Θ(log n).</p><p>By contrast, the closed form</p><disp-formula id="equation-58"><tex-math id="math-169"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \begin{array}{l} T r _ {a} (A ^ {n}) = Q _ {1} \left(\sum_ {i + 2 j + 3 k = n - 1} \alpha_ {i, j, k} S _ {1} ^ {i} S _ {2} ^ {j} S _ {3} ^ {k}\right) + Q _ {2} \left(\sum_ {i + 2 j + 3 k = n - 2} \beta_ {i, j, k} S _ {1} ^ {i} S _ {2} ^ {j} S _ {3} ^ {k}\right) \\ \quad + \sum_ {i + 2 j + 3 k = n} \gamma_ {i, j, k} S _ {1} ^ {i} S _ {2} ^ {j} S _ {3} ^ {k}, \end{array} \end{document} ]]></tex-math></disp-formula><p>has <inline-formula><tex-math id="math-170"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \Theta ( n ^ { 2 } ) \end{document} ]]></tex-math></inline-formula> terms. Indeed, the number of triples <inline-formula><tex-math id="math-171"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle ( i , j , k ) \in \mathbb { N } ^ { 3 } \end{document} ]]></tex-math></inline-formula> with <inline-formula><tex-math id="math-172"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle i + 2 j + 3 k = n \end{document} ]]></tex-math></inline-formula> is</p><disp-formula id="equation-59"><tex-math id="math-173"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle N (n) = \sum_ {k = 0} ^ {\lfloor n / 3 \rfloor} \left(\left\lfloor \frac {n - 3 k}{2} \right\rfloor + 1\right) = \frac {n ^ {2}}{1 2} + O (n), \end{document} ]]></tex-math></disp-formula><p>so a direct evaluation of the multinomial sums scales quadratically in n (after a onetime extraction of <inline-formula><tex-math id="math-174"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle S _ { 1 } , S _ { 2 } , S _ { 3 } , Q _ { 1 } , Q _ { 2 } \end{document} ]]></tex-math></inline-formula> from <inline-formula><tex-math id="math-175"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A ) \end{document} ]]></tex-math></inline-formula> . This explains the observed timings: for a single large n with floating-point entries, the direct method is usually faster.</p><p>However, the closed form remains computationally useful in exact and modular arithmetic. Over general commutative rings <inline-formula><tex-math id="math-176"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle ( \mathrm { e . g . , ~ } \mathbb { Z } , \ \mathbb { Z } / p ^ { r } \mathbb { Z } \end{document} ]]></tex-math></inline-formula> , or polynomial rings), the recurrence/closed form permits exact computations and congruence arguments; forming <inline-formula><tex-math id="math-177"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A ^ { n } \end{document} ]]></tex-math></inline-formula> numerically would either be impossible or unreliable. The structural block anti-diagonal identities in Subsection 3.2 also reduce large problems to <inline-formula><tex-math id="math-178"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle 3 \times 3 \end{document} ]]></tex-math></inline-formula> blocks, which a direct power need not reveal.</p><p>In summary, for a single floating-point instance <inline-formula><tex-math id="math-179"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle ( A , n ) \end{document} ]]></tex-math></inline-formula> the direct method is often the fastest <inline-formula><tex-math id="math-180"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle \left( \Theta ( \log n ) \right) \end{document} ]]></tex-math></inline-formula> , while the closed form (and its induced recurrence) is preferable for symbolic/modular settings, when many values <inline-formula><tex-math id="math-181"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r _ { a } ( A ^ { n } ) \end{document} ]]></tex-math></inline-formula> are required, or when exploiting block structure.</p></sec></sec><sec id="sec-9"><title>4. CONCLUSION</title><p>We developed an exact, generating–function approach for computing the anti–trace <inline-formula><tex-math id="math-182"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle T r _ { a } ( A ^ { n } ) \end{document} ]]></tex-math></inline-formula> of powers of <inline-formula><tex-math id="math-183"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle 3 \times 3 \end{document} ]]></tex-math></inline-formula> matrices over <inline-formula><tex-math id="math-184"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle R , \end{document} ]]></tex-math></inline-formula> yielding closed formulas whose coeficients are multinomial combinations of principal and anti–principal minors. The key point is that anti–trace admits an algebraic treatment via the Cayley–Hamilton recurrence; no spectral information is required. In the block setting, the resulting identities are driven purely by placement along the block anti–diagonal, and they therefore remain valid over any commutative coeficient ring. Practically, the formulas avoid forming <inline-formula><tex-math id="math-185"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle A ^ { n } \end{document} ]]></tex-math></inline-formula> and the block rules let one read of anti–traces from constituent <inline-formula><tex-math id="math-186"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle 3 \times 3 \end{document} ]]></tex-math></inline-formula> blocks. Natural next steps include extending the coeficient description beyond <inline-formula><tex-math id="math-187"><![CDATA[ \documentclass{article} \usepackage{amsmath} \begin{document} \displaystyle 3 \times 3 \end{document} ]]></tex-math></inline-formula> , studying congruence refinements, and exploring additional structured classes.</p><p>No new data were created or analyzed in this study.</p><p>Jeriko Gormantara: Conceptualization, methodology, writing-original draft, writing-review &amp; editing. Devi Fitri Ferdania: Investigation, validation. Fajar Yuliawan: Formal analysis. Hanni Garminia: Supervision. All authors discussed the results and contributed to the final manuscript.</p><p>This research was supported by the ITB Research Grant under the Program PPMI FMIPA 2021, No</p><p><ext-link ext-link-type="uri" xlink:href="https://FMIPA.PPMI-1-9-2021" xlink:title="FMIPA.PPMI-1-9-2021">FMIPA.PPMI-1-9-2021</ext-link></p><p>We also thank the anonymous referees for their insightful feedback and helpful recommendations.</p></sec></body><back><sec sec-type="data-availability"><title>Data Availability Statement</title></sec><sec sec-type="author-contributions"><title>Author Contributions.</title></sec><ack><title>Acknowledgment.</title></ack><ref-list><title>REFERENCES</title><ref id="BIBR-1"><element-citation publication-type="journal"><article-title>Estimations of the trace of powers of positive selfadjoint operators by extrapolation of the moments</article-title><source>Electronic Transactions on Numerical Analysis</source><volume>39</volume><person-group person-group-type="author"><name><surname>Brezinski</surname><given-names>C.</given-names></name><name><surname>Fika</surname><given-names>P.</given-names></name><name><surname>Mitrouli</surname><given-names>M.</given-names></name></person-group><year>2012</year><page-range>144-155,</page-range><ext-link xlink:href="https://etna.ricam.oeaw.ac.at/vol.39.2012/" ext-link-type="uri" xlink:title="Website link">Website link</ext-link></element-citation></ref><ref id="BIBR-2"><element-citation publication-type="book"><article-title>Counting triangles in large graphs using randomized matrix trace estimation</article-title><source>Workshop on Large-Scale Data Mining: Theory and Applications (LDMTA</source><person-group person-group-type="author"><name><surname>Avron</surname><given-names>H.</given-names></name></person-group><year>2010</year><page-range>1-8,</page-range><ext-link xlink:href="https://www.semanticscholar" ext-link-type="uri" xlink:title="Website link">Website link</ext-link></element-citation></ref><ref id="BIBR-3"><element-citation publication-type="journal"><article-title>On congruences for the traces of powers of some matrices</article-title><source>Proceedings of the Steklov Institute of Mathematics</source><volume>263</volume><issue>1</issue><person-group person-group-type="author"><name><surname>Zarelua</surname><given-names>A.V.</given-names></name></person-group><year>2008</year><page-range>78-98,</page-range><pub-id pub-id-type="doi">10.1134/S008154380804007X</pub-id></element-citation></ref><ref id="BIBR-4"><element-citation publication-type="journal"><article-title>An algorithm for computing powers of a Hessenberg matrix and its applications</article-title><source>Linear Algebra and its Applications</source><volume>14</volume><issue>3</issue><person-group person-group-type="author"><name><surname>Datta</surname><given-names>B.N.</given-names></name><name><surname>Datta</surname><given-names>K.</given-names></name></person-group><year>1976</year><page-range>273-284,</page-range><pub-id pub-id-type="doi">10.1016/0024-3795(76)90072-0</pub-id></element-citation></ref><ref id="BIBR-5"><element-citation publication-type="journal"><article-title>Symbolic calculation of the trace of the power of a tridiagonal matrix</article-title><source>Computing</source><volume>35</volume><issue>3</issue><person-group person-group-type="author"><name><surname>Chu</surname><given-names>M.T.</given-names></name></person-group><year>1985</year><page-range>257-268,</page-range><pub-id pub-id-type="doi">10.1007/BF02240193</pub-id></element-citation></ref><ref id="BIBR-6"><element-citation publication-type="journal"><article-title>Positive integer powers of certain tridiagonal matrices</article-title><source>Applied Mathematics and Computation</source><volume>202</volume><issue>1</issue><person-group person-group-type="author"><string-name>J. 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