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Early Evidence of Vibe-Proving with Consumer LLMs: A Case Study on Spectral Region Characterization with ChatGPT-5.2 (Thinking)
Brecht Verbeken, Brando Vagenende, Marie-Anne Guerry, Andres Algaba, Vincent Ginis
Intelligence
Status: succeeded | Model: Gemma-4-26B-A4B | Prompt: intel-v1 | Confidence: 92%
Last extracted: 7/20/2026, 9:04:24 PM
Summary
This paper presents a case study on 'vibe-proving,' an iterative workflow where human researchers collaborate with consumer-grade Large Language Models (specifically ChatGPT-5.2 Thinking) to solve research-level mathematics problems. The study details the resolution of Conjecture 20 by Ran and Teng (2024) regarding the exact nonreal spectral region of a 4-cycle row-stochastic nonnegative matrix family. The authors document an auditable process of generating, refereeing, and repairing proof drafts, concluding that while LLMs are effective for high-level proof search and structural suggestions, human experts remain essential for correctness-critical verification and handling logical gaps.
Entities (8)
Relation Signals (7)
Ran and Teng → proposed → Conjecture 20
confidence 95% · Conjecture 20 of Ran and Teng (2024) on the exact nonreal spectral region...
ChatGPT-5.2 (Thinking) → usedfor → Vibe-Proving
confidence 95% · We present early evidence for vibe-proving with a consumer subscription LLM... We analyze seven shareable ChatGPT-5.2 (Thinking) threads...
Human Experts → essentialfor → Correctness-Critical Closure
confidence 93% · human experts remain essential for correctness-critical closure.
ChatGPT-5.2 (Thinking) → resolved → Conjecture 20
confidence 92% · resolves Conjecture 20 of Ran and Teng (2024)... through an auditable case study
ChatGPT-5.2 (Thinking) → mostusefulfor → High-Level Proof Search
confidence 91% · The model is most useful for high-level proof search, while human experts remain essential for correctness-critical closure.
Karpelevich Theorem → characterizes → Karpelevich Region
confidence 90% · the definitive result is the Karpelevich theorem (1951) [37], which characterizes the region Kn of possible eigenvalues...
Dimitriev and Dynkin → developed → Trigonometric Method
confidence 88% · Dimitriev and Dynkin (1946) [20] developed a geometric method for characterizing the eigenvalue regions... which we call the 'trigonometric method'
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Abstract
Abstract:Large Language Models (LLMs) are increasingly used as scientific copilots, but evidence on their role in research-level mathematics remains limited, especially for workflows accessible to individual researchers. We present early evidence for vibe-proving with a consumer subscription LLM through an auditable case study that resolves Conjecture 20 of Ran and Teng (2024) on the exact nonreal spectral region of a 4-cycle row-stochastic nonnegative matrix family. We analyze seven shareable ChatGPT-5.2 (Thinking) threads and four versioned proof drafts, documenting an iterative pipeline of generate, referee, and repair. The model is most useful for high-level proof search, while human experts remain essential for correctness-critical closure. The final theorem provides necessary and sufficient region conditions and explicit boundary attainment constructions. Beyond the mathematical result, we contribute a process-level characterization of where LLM assistance materially helps and where verification bottlenecks persist, with implications for evaluation of AI-assisted research workflows and for designing human-in-the-loop theorem proving systems.
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- Source: https://arxiv.org/abs/2602.18918v1
- Canonical: https://arxiv.org/abs/2602.18918v1
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Early Evidence of Vibe-Proving with Consumer LLMs: A Case Study on Spectral Region Characterization with ChatGPT-5.2 (Thinking) Brecht Verbeken1,2, 0000-0002-7506-3298 &Brando Vagenende1 0000-0002-0573-8093 &Marie-Anne Guerry1 0000-0001-5842-8905 Andres Algaba1,2 0000-0002-0532-3066 &Vincent Ginis1,2,3 0000-0003-0063-9608 1Data Analytics Lab, Vrije Universiteit Brussel, Pleinlaan 5, 1050 Brussel, Belgium 2imec-SMIT, Vrije Universiteit Brussel, Pleinlaan 9, 1050 Brussels, Belgium 3School of Engineering and Applied Sciences, Harvard University, Cambridge, Massachusetts 02138, USA Abstract Large Language Models (LLMs) are increasingly used as scientific copilots, but evidence on their role in research-level mathematics remains limited, especially for workflows accessible to individual researchers. We present early evidence for vibe-proving with a consumer subscription LLM through an auditable case study that resolves Conjecture 20 of Ran and Teng (2024) on the exact nonreal spectral region of a 44-cycle row-stochastic nonnegative matrix family. We analyze seven shareable ChatGPT-5.2 (Thinking) threads and four versioned proof drafts, documenting an iterative pipeline of generate, referee, and repair. The model is most useful for high-level proof search, while human experts remain essential for correctness-critical closure. The final theorem provides necessary and sufficient region conditions and explicit boundary attainment constructions. Beyond the mathematical result, we contribute a process-level characterization of where LLM assistance materially helps and where verification bottlenecks persist, with implications for evaluation of AI-assisted research workflows and for designing human-in-the-loop theorem proving systems. †footnotetext: Corresponding author: brecht.verbeken@vub.be Keywords AI-assisted mathematics ⋅· Karpelevich region ⋅· large language models ⋅· stochastic matrices 1 Introduction Vibe-coding, where programmers steer Large Language Models (LLMs) to generate software through high-level natural-language intent rather than line-by-line specification, has rapidly transformed code generation [55, 51, 15]. Underpinning this shift is the sustained capability growth of frontier LLMs across successive systems and evaluations [9, 16, 43, 58]. As these models grow more capable, they are increasingly deployed as scientific collaborators: generating candidate research ideas in controlled studies [33, 57], acting as agentic research assistants that plan, search, and iterate over literature and experiments [3, 29, 31, 44, 49, 56, 69], and accelerating evidence synthesis and literature screening [19, 42, 59]. At field scale, LLM tooling is associated with measurable shifts in scientific production and communication [39], but it remains unclear how reliably these systems support auditable, research-level mathematical workflows under consumer-access conditions [10, 22]. Beyond collaboration, LLMs increasingly execute end-to-end scientific workflows across domains. In the social sciences, models serve as simulated subjects and predictive surrogates [35, 46, 72]. In the natural sciences, agentic systems couple LLM reasoning to domain tools, autonomously running chemistry campaigns, designing biomolecules, and accelerating experimental protocols [7, 45, 71, 61, 53, 60, 68], alongside work on test-time discovery and automated search [70, 67]. Mathematics provides an unusually crisp stress test for whether vibe-coding’s paradigm can extend beyond software: correctness is, in principle, checkable, and performance on benchmark suites has improved rapidly [34, 4, 14, 47, 27, 30, 26]. Recent systems reach gold-medalist-level performance on Olympiad geometry [65, 12], and a gold-medal score on International Mathematical Olympiad (IMO) problems [11]. At the research frontier, program search and large-scale exploration pipelines generate new mathematical artefacts [54, 63, 28], case studies document substantive human–LLM co-working on active problems [10, 8], and community infrastructures curate open problem sets and evaluation tasks [17, 13, 6], including initiatives that explicitly invite LLM attempts [1]. Semi-autonomous pipelines have even been applied at scale to open mathematics problems [22, 5]. Together, these results indicate that end-to-end AI-assisted mathematical discovery is becoming practically feasible. These Erdős-scale and benchmark-scale results show what “specialized” LLM systems can achieve, but they leave a practical question unresolved: can individual researchers produce mathematically substantive progress with consumer-access models and explicit human verification? This question is central for vibe-proving, i.e., sustained, iterative theorem development through conversational interaction rather than one-shot outputs. Unlike vibe-coding, where software has a runtime oracle, vibe-proving faces an intrinsic verification bottleneck because every logical step must be checked and a single hidden gap can invalidate the argument. Our case study began when we encountered Conjecture 20 of [52] on nonreal eigenvalue regions of a structured 4×44× 4 row-stochastic family of matrices. The project did not start from a pre-registered protocol, but from “vibing” with ChatGPT-5.1 (Thinking) on an active research problem. During that initial exploratory phase, we did not systematically retain our earliest interactions. Only after recognizing that the reasoning approach suggested by the LLM was not only interesting but also potentially viable, did we transition from informal experimentation to systematic documentation and structured iteration. Consequently, we reconstructed these initial prompts using ChatGPT-5.2 (Thinking). All subsequent documented interactions were conducted directly with ChatGPT-5.2 (Thinking). We therefore ground the paper in auditable artefacts: seven share links (Transcript 1–Transcript 7) and four proof drafts (Appendix Versions A–D), and we explicitly mark where early prompt content is reconstructed rather than preserved verbatim. Within this artefact set, we show that a stable interaction pattern of iterating between generate, referee, and repair resolves the conjecture into a complete characterization theorem with explicit boundary attainment. The model is strongest at proposing global structure (reparameterization and extremal strategy), whereas human effort is concentrated on correctness-critical verification (quadrant control, branch tracking, endpoint admissibility, and dense algebraic expansions). This framing complements recent mathematics results obtained with specialized or (semi-)autonomous stacks, including dedicated theorem-discovery pipelines [23, 22], as well as domain-specific outputs in arithmetic geometry, combinatorics, and robust control [25, 24, 41, 2]. Our contribution is a subscription-level, conversation-auditable account where the revision trajectory is inspectable end-to-end. The remainder of the paper is organized as follows. Section 2 introduces the mathematical setting and Conjecture 20. Section 3 analyzes the interaction workflow with ChatGPT and documents the iterative proof-development process. We then discuss the division of labor between model and human verification, limitations, and implications for AI-assisted research workflows, and finally provide transcript-linked appendices and versioned proof artefacts for full auditability. 2 Mathematical Background 2.1 Spectral Regions of Stochastic Matrices A fundamental problem in matrix theory is to characterize which complex numbers can appear as eigenvalues of matrices with specific structural constraints. For row-stochastic matrices (nonnegative matrices with row sums equal to 1), the definitive result is the Karpelevich theorem (1951) [37], which characterizes the region KnK_n of possible eigenvalues of n×n× n row-stochastic matrices. For subclasses, such characterizations remain largely open. Karpelevich’s original proof is notoriously difficult [36]. This difficulty has motivated significant recent work revisiting and demystifying the Karpelevich theorem [48], as well as the search for alternative approaches for subclasses of stochastic matrices where simpler characterizations might exist [32, 38, 66]. 2.2 Dimitriev and Dynkin’s Trigonometric Method Before Karpelevich’s general theorem, Dmitriev and Dynkin (1946) [20] developed a geometric method for characterizing the eigenvalue regions KnK_n for small n. Their approach, which we call the “trigonometric method,” is particularly elegant: 1. For an eigenvalue λ with eigenvector v, derive a multiplicative constraint from the eigenvalue equations. 2. Reparametrize using arguments (u=Arg(z+t)u=Arg(z+t)) to convert the constraint into a trigonometric optimization problem. 3. Use convexity and majorization arguments to characterize the boundary. Dmitriev and Dynkin successfully characterized K2,K3,K4K_2,K_3,K_4, and K5K_5 using this approach. Their method has the advantage of being more elementary than Karpelevich’s and more amenable to adaptation for matrix subclasses with specific zero patterns. 2.3 Conjecture 20 of Ran and Teng Recently, Ran and Teng (2024) [52] studied eigenvalue regions for matrices with prescribed zero patterns, obtaining complete characterizations for the 3×33× 3 case. They also posed two conjectures for dimension four, of which Conjecture 20 concerns what we will call the “4-cycle” pattern. Consider the family of 4×44× 4 row-stochastic matrices: A(α,β,γ,δ)=(α1−α000β1−β000γ1−γ1−δ00δ),α,β,γ,δ∈(0,1).A(α,β,γ,δ)= pmatrixα&1-α&0&0\\ 0&β&1-β&0\\ 0&0&γ&1-γ\\ 1-δ&0&0&δ pmatrix, α,β,γ,δ∈(0,1). (1) This matrix has a so called cyclic zero pattern: state 1 transitions to states 1 or 2; state 2 to 2 or 3; state 3 to 3 or 4; state 4 to 4 or 1. Such patterns arise naturally in the study of Markov chains with restricted transitions. Ran and Teng (2024) [52] defined the region R as: R=z=a+bi∈K4:a>0,G(a,b)>0R=\z=a+bi∈ K_4:a>0,G(a,b)>0\ (2) where K4K_4 is the Karpelevich region for 4×44× 4 matrices and G(a,b)=(b2+a2+a)2+2a2−b2.G(a,b)=(b^2+a^2+a)^2+2a^2-b^2. (3) Conjecture 20 (Ran and Teng, 2024 [52], Conjecture 20). For any irreducible matrix A of the form (1), the non-real eigenvalues of A lie in the region R. Ran and Teng (2024) [52] provided substantial numerical evidence for this conjecture and proved that the curve G(a,b)=0G(a,b)=0 is indeed attained by the one-parameter subfamily AL(α)A_L(α) where β=γ=δ=0β=γ=δ=0: AL(α)=(α1−α00001000011000).A_L(α)= pmatrixα&1-α&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0 pmatrix. (4) They conjectured that this curve forms the left boundary of the non-real part of the spectral region. 2.4 Contribution We observed that Ran and Teng’s (2024) [52] Conjecture 20 could be used to provide a complete characterization. Using the Karpelevich theorem and results on realizing boundary arcs (specifically, that the right boundary of K4K_4 is already attained within our restricted class), we provided a target theorem statement and asked ChatGPT for a proof strategy. The collaboration produced a proof of a theorem characterizing all non-real eigenvalues of the matrix family (1): necessary and sufficient conditions, plus explicit boundary attainment. The full proof, developed through extended dialogue with ChatGPT, will soon be available on the ArXiv. For this paper, the goal is to describe the process, not the final product. 3 The Conversations with ChatGPT 3.1 Casual testing of GPT’s capabilities The interaction with ChatGPT did not begin with a research plan. We were “vibing” by casually testing the “thinking extended” mode on problems from our research area. During these informal explorations, we observed that the reasoning approach suggested by the model was not only interesting but also potentially viable. Because these initial tests were exploratory, we did not systematically document our earliest interactions with ChatGPT-5.1 (Thinking). To compensate for the missing conversations with ChatGPT-5.1 (Thinking), we reran the initial prompts with ChatGPT-5.2 (Thinking) in Transcript 1 and Transcript 2. From this point onward, all formal testing and documented interactions were performed exclusively using ChatGPT-5.2 (Thinking), starting from Transcript 3. What follows is a reconstruction with the versioned drafts (Appendix Versions A–D). 3.2 Setting Up the Theorem We provided the Ran and Teng paper [52] and explained our observation about the right boundary being already characterized by Karpelevich [37]. This resulted in the characterization Theorem below. Afterwards we provided ChatGPT with this theorem statement and prompted the model towards the direction of thinking in line with the trigonometric method of Dimitriev and Dynkin [21], and asked for a proof strategy (see Transcript 1 for the exact statement we used in the prompt): Theorem (Spectral region for a 4-cycle row-stochastic matrix). Let A(α,β,γ,δ)=(α1−α000β1−β000γ1−γ1−δ00δ),α,β,γ,δ∈[0,1].A(α,β,γ,δ)= pmatrixα&1-α&0&0\\ 0&β&1-β&0\\ 0&0&γ&1-γ\\ 1-δ&0&0&δ pmatrix, α,β,γ,δ∈[0,1]. Let λ=a+ib∈σ(A(α,β,γ,δ))λ=a+ib∈σ(A(α,β,γ,δ)) with b≠0b≠ 0, and write b+=|b|b_+=|b|. Then: 1. 0≤a≤10≤ a≤ 1; 2. a+b+≤1a+b_+≤ 1; 3. with G(a,b+):=(b+2+a2+a)2+2a2−b+2,G(a,b_+):=(b_+^2+a^2+a)^2+2a^2-b_+^2, we have G(a,b+)≥0G(a,b_+)≥ 0. Conversely, if b>0b>0 and 0≤a≤1,a+b<1,G(a,b)≥0,0≤ a≤ 1, a+b<1, G(a,b)≥ 0, then there exist α,β,γ,δ∈[0,1]α,β,γ,δ∈[0,1] such that λ∈σ(A(α,β,γ,δ))λ∈σ(A(α,β,γ,δ)). (The lower half-plane follows by conjugation.) Moreover: • the segment CR:λ=1−x+ixC_R:\ λ=1-x+ix (x∈[0,1]x∈[0,1]) is attained by α=β=γ=δ=1−xα=β=γ=δ=1-x; • the curve portion CLC_L of G(a,b)=0G(a,b)=0 in the upper half-plane joining i to 0 is attained by AL(α)=(α1−α00001000011000),α∈[0,1];A_L(α)= pmatrixα&1-α&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0 pmatrix, α∈[0,1]; • every point strictly inside b>0, 0≤a≤1,a+b<1,G(a,b)≥0\b>0,\ 0≤ a≤ 1,\ a+b<1,\ G(a,b)≥ 0\ is attained. 3.3 The Proof Strategy ChatGPT then proposed a trigonometric roadmap (angle parametrization plus Jensen/Karamata) inspired by [21, 62] for the remaining constraints (see Transcript 1). The next prompt drew on our knowledge of the classical literature (see Transcript 1). The response outlined the trigonometric approach: the multiplicative constraint from the eigenvalue equations, reparametrization by arguments u=Arg(z+t)u=Arg(z+t), the convex function F, and the use of Jensen’s inequality and Karamata’s inequality for the optimization. We then requested a hypercritical review, which surfaced a fatal gap in the constrained Karamata/majorization step (see Transcript 2). 3.4 The Iterations Appendix Version A contained the right structure but had gaps (see Transcript 3): incorrect quadrant handling for arctan , a flawed inequality in what became Lemma 2, and incomplete algebra connecting the optimization to the curve G(a,b)=0G(a,b)=0. The first correctness-critical bug was a feasibility claim in what became Lemma 2: the model’s original construction fails in the near-endpoint regime M↓π/2M π/2. The repair was to localize the argument near the endpoint configuration u4↑Mu_4 M, and to require only the strict interior condition needed for that regime (namely u1<Mu_1<M) rather than uniform slack up to M. This “local feasibility by continuity” patch is documented in Appendix Version B and corroborated by an independent referee pass (see Transcript 6). ChatGPT helped simplify and confirm the factorization |λ|6≥N(a,b)|λ|^6≥ N(a,b) (see Transcript 5). After some manipulation, it produced: |λ|6−N(a,b)=|λ−1|2G(a,b).|λ|^6-N(a,b)=|λ-1|^2G(a,b). (5) A disproportionate share of the iteration time went into Lemma 4 (see Transcript 5), which forces the “tight regime” 3m+M>2π3m+M>2π from the algebraic condition G(a,b)≤0G(a,b)≤ 0. To close this step, we wrote a fully expanded Lamport-style draft of Lemma 4 and ran a dedicated “referee” thread focused on finding silent branch/sign errors rather than proposing new structure. This pass largely validated the backbone of the argument (quadratic-in-s=b2s=b^2 analysis, the deduction b2>3a2b^2>3a^2 locating m in the correct range, reduction of 3m+M>2π3m+M>2π to a tangent inequality, and the load-bearing algebraic identity |λ|6−N(a,b)=|λ−1|2G(a,b)|λ|^6-N(a,b)=|λ-1|^2G(a,b)), but it also surfaced several obligations that earlier drafts had left implicit: (i) the discriminant condition for G as a quadratic in s is Δ≥0⇔−12≤a≤16 ≥ 0 - 12≤ a≤ 16, so the conclusion a≤16a≤ 16 must be explicitly combined with the already-established constraint a≥0a≥ 0; (i) the Arg/arctan/tanArg/ / branch conventions and the monotonicity interval for tan must be stated once to justify the conversion of 3m+M>2π3m+M>2π into a tangent inequality; and (i) every squaring/cross-multiplication step must be preceded by a one-line positivity check (notably 1−a>01-a>0 and denominator signs). We incorporated these items as explicit guard lines and reorganized Lemma 4 into a “sign-check → transform → square/cross-multiply” micro-step schedule, which made the lemma locally checkable and prevented regressions in later rewrites. Appendix Version C is the first draft where the correctness-critical obligations are made explicit and discharged end-to-end (see Transcript 4). Appendix Version D then primarily improves auditability: it expands the load-bearing algebraic derivations (notably the factorization |λ|6−N(a,b)=|λ−1|2G(a,b)|λ|^6-N(a,b)=|λ-1|^2G(a,b); see Transcript 5) and rewrites the argument in a Lamport-style dependency structure (see Transcript 7). 3.5 From chronology to workflow-level lessons The remainder of the paper abstracts the above chronology into workflow-level observations about verification, patch search, and division of labor. To keep Section 3 chronological, we consolidate those methodological takeaways in the Discussion (Section 4), with explicit pointers to the corresponding transcripts and appendix versions. 4 Discussion 4.1 What This Case Suggests In this case study, a conversational workflow with ChatGPT-5.2 (Thinking) produced a complete and checkable proof of Ran and Teng (2024) [52] Conjecture 20. The LLM supplied a viable global architecture early (characterization theorem + trigonometric strategy), while the main human work consisted of identifying and discharging correctness obligations (quadrant/branch handling, endpoint cases, and long algebra). Success in this case used the following features of the problem and workflow: 1. Prior scaffolding: The target was an explicit region characterization with known boundary information (Karpelevich right boundary and the conjectured left boundary G(a,b)=0G(a,b)=0), and a classical proof template existed via a Dmitriev–Dynkin-style trigonometric reduction. 2. Two anchoring prompts: The interaction was organized around (i) providing the target characterization theorem statement and boundary context and (i) deriving a trigonometric/majorization-based proof strategy specialized to the 4-cycle matrix family. 3. Untrusted outputs: Early drafts contained correctness-critical errors (e.g., arctan quadrant handling and a false inequality regime in Lemma 2), so every model-generated step was treated as a candidate and checked before inclusion. 4. Independent patch search: When a gap was identified, we queried multiple independent sessions and compared proposed fixes, adopting only patches that could be verified and rejecting incompatible derivations. We unpack the concrete verification workflow and division of labor in Sections 4.2–4.3. 4.2 Workflow and Verification Strategy Across the seven threads, we converged on a stable loop of generate, referee, repair: generate candidate steps and proof skeletons; run referee-style passes to surface hidden obligations (branch conventions, positivity checks before squaring/cross-multiplying, endpoint admissibility); and then repair targeted subclaims without rewriting unrelated parts. Three concrete practices materially improved reliability: • Parallel patch search. When a gap was identified, we opened multiple independent sessions with the same obligation and treated their outputs as competing patches. We adopted only patches that were independently checkable (or where independent sessions converged), and rejected fixes that relied on silent case splits or unverifiable algebraic compression. • Fresh-session referee passes (bounded). At checkpoints, we pasted the then-current draft into a fresh thread and asked for a gap list (missing assumptions, unjustified monotonicity/branch steps, unsafe squaring). Once major gaps were fixed, repeated “hypercritical” prompting showed diminishing returns (stylistic noise and occasional false positives), so we used it as a bounded diagnostic rather than an endless loop. • Regression control via versioning and dependency visibility. Proof rewrites sometimes reintroduced errors elsewhere (notation drift, dropped sign constraints). We therefore kept versioned drafts aligned with transcript excerpts and re-checked downstream dependencies after each patch. We also frequently requested Lamport-style claim decomposition [40] to make dependencies explicit and to localize verification (see Transcript 7). 4.3 Division of labor and the verification bottleneck In this project, the model’s highest leverage was proposing global structure (reparameterizations, extremal strategies, candidate factorizations) and producing fast symbolic manipulations; human time was dominated by correctness-critical closure (branch/quadrant tracking, sign checks before squaring, endpoint admissibility, and long expansions). Two observations were consistent across iterations: 1. The bottleneck is narrow but expensive. Most iteration time concentrated on a small number of obligations (notably the tight-regime step around Lemma 4 and a few load-bearing factorizations), where a single silent sign/branch error can invalidate the argument. 2. Mechanization would directly target the slowest work. The dominant verification load was algebraic expansion/simplification and inequality-domain checking. These are well matched to CAS and certified inequality/interval checkers, suggesting a practical hybrid pipeline: LLMs propose patches for explicit obligations; mechanized tooling validates the algebra/inequality substeps. For completeness, we summarize the division of contributions (with transcript anchors): Human contributions (verification and orchestration). • Problem selection and target theorem specification (see Transcript 1). • Error detection and obligation listing (quadrant/branch handling, Lemma 2 feasibility, endpoint issues). • Orchestration of independent patch search and regression control via versioned drafts. • Final correctness closure on sign/branch conditions and expanded algebra (see Transcript 5 and Transcript 6). ChatGPT contributions (structure and candidate derivations). • Early global roadmap (trigonometric reduction + extremal strategy; Transcript 1). • Algebraic manipulation support, including confirming the key factorization (Transcript 5). • Structured Lamport-style rewrite that made dependencies explicit (Transcript 7). • Targeted referee passes that surfaced missing guard conditions when prompted appropriately (Transcript 2). 4.4 Formal Proof Assistants: The Verification Alternative A natural question is how vibe proving with LLMs compares to formal proof assistants such as Lean, Coq, or Isabelle [64, 18]. These systems offer something LLMs cannot: mechanically verified correctness relative to a small logical kernel. When a Lean proof checks, the result is certain (modulo trust in the kernel, which is small but not zero, as bugs have been found in proof assistant kernels, and complete formal verification of the kernel itself remains an open problem) [50]. We did not attempt a formalization of the present proof. A mechanically verified proof would require encoding the trigonometric optimization, convexity/majorization steps, and the inequality-heavy algebra within a proof assistant’s libraries. This would yield stronger guarantees, but it would also change the artefact (proof script rather than narrative proof) and require additional engineering beyond the scope of this case study. Our output is a conventional, human-checked proof accompanied by an auditable chat trail. 4.5 Limitations 1. Novelty: The proof strategy stayed within a classical Dmitriev–Dynkin/Jensen–Karamata template. The model’s main contribution was assembling and adapting this template to the 4-cycle matrix family rather than introducing a fundamentally new method. 2. Error modes: We observed recurrent correctness failures in early drafts, including inverse-trigonometric branch/quadrant mistakes, missing sign conditions before squaring, and “compressed” algebra that omitted required intermediate steps. 3. Verification bottleneck: The slowest component was discharging a small set of technical obligations (notably Lemma 4’s inequality/expansion chain and endpoint admissibility), rather than generating candidate derivations. 4. Scope: Evidence here is limited to one structured spectral-region characterization problem with strong prior scaffolding (known conjectured boundary and known trigonometric reduction). We do not test problems lacking such structure. 4.6 Recommendations for Practitioners The workflow-level observations in Sections 4.2–4.3 suggest the following checklist for vibe proving: 1. Start from scaffolding. Prefer problems where you can state a concrete target theorem and where a recognizable reduction/template exists. 2. Turn critique into obligations. Convert “this seems wrong” into an explicit obligation list (domains, branch conventions, positivity checks before squaring, endpoints). 3. Use parallel patch search. Treat independent sessions as competing patch generators; adopt only patches that you can verify locally. 4. Control regressions. Keep versioned drafts and re-check downstream dependencies after each patch; prefer Lamport-style decomposition to expose dependencies. 5. Mechanize the bottleneck. Offload expansions and inequality-domain checks to CAS / certified checkers; reserve human time for conceptual choices and boundary cases. Acknowledgements This research was supported by funding from the Flemish Government under the “Onderzoeksprogramma Artificiële Intelligentie (AI) Vlaanderen” program. Andres Algaba acknowledges support from the Francqui Foundation (Belgium) through a Francqui Start-Up Grant and a fellowship from the Research Foundation Flanders (FWO) under Grant No.1286924N. Vincent Ginis acknowledges support from Research Foundation Flanders under Grant No.G032822N and G0K9322N. Author contributions All authors collaboratively conceived the main idea of the study. Brecht Verbeken co-constructed the proof with the assistance of ChatGPT-5.1 (Thinking) and ChatGPT-5.2 (Thinking). Brecht Verbeken, Brando Vagenende, and Marie-Anne Guerry reviewed and validated the final proof. Andres Algaba and Brecht Verbeken drafted the manuscript. All authors collaboratively revised the manuscript and provided critical feedback. Data and code availability Conversations • Transcript 1 (rerun): setup prompt + roadmap; includes follow-up algebra sanity checks (via ChatGPT’s Python tool). • Transcript 2 (rerun): hypercritical review; identifies the majorization/Karamata gap. • Transcript 3:referee pass on the first full proof attempt (Appendix Version A) and early gap list (branch/quadrant handling; Lemma 2 feasibility near M; missing algebra). • Transcript 4 corresponds to the first essentially complete draft (Appendix Version C). • Transcript 5: referee pass on Lemma 4 (tight-regime step); checks branch/sign conditions and validates the factorization |λ|6−N(a,b)=|λ−1|2G(a,b)|λ|^6-N(a,b)=|λ-1|^2G(a,b). • Transcript 6: independent fresh-session critique corroborating the branch/quadrant and Lemma 2 issues/fix. • Transcript 7: Lamport-style rewrite used for the final exposition (Appendix Version D). Note that the following link template can be used to inspect meta-data from each conversation: https://chatgpt.com/backend-api/share/<share-id>. • Transcript 1 (rerun): https://chatgpt.com/share/699464b2-c81c-8002-9ba2-bad952e6414a • Transcript 2 (rerun): https://chatgpt.com/share/69946540-e684-8002-a8ca-feb45e4da7be • Transcript 3: https://chatgpt.com/share/697b2d1-0418-8007-8f2a-474e3e6430fa • Transcript 4: https://chatgpt.com/share/697b2bd-5450-8007-9115-92589d95e1ba • Transcript 5: https://chatgpt.com/share/697b29a-c190-8007-b37e-87b390e9d9f • Transcript 6: https://chatgpt.com/share/697b288-010c-8007-972b-ec676e9a16f7 • Transcript 7: https://chatgpt.com/share/697b2ab-f83c-8007-a8fc-9439d3d7488 References [1] M. Abouzaid, A. J. Blumberg, M. Hairer, J. Kileel, T. G. Kolda, P. D. Nelson, D. Spielman, N. Srivastava, R. Ward, S. Weinberger, and L. Williams (2026) First proof. External Links: 2602.05192, Link Cited by: §1. [2] A. Asadi, K. Chatterjee, E. Goharshady, M. Karrabi, A. 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Cited by: §1. Here, we show the subsequent versions of the proof. Appendix A Appendix First version Theorem 21 (Precise spectral region for a 4-cycle row-stochastic matrix). Let A(α,β,γ,δ)=(α1−α000β1−β000γ1−γ1−δ00δ),α,β,γ,δ∈[0,1).A(α,β,γ,δ)= pmatrixα&1-α&0&0\\ 0&β&1-β&0\\ 0&0&γ&1-γ\\ 1-δ&0&0&δ pmatrix, α,β,γ,δ∈[0,1). Let λ=a+ib∈σ(A(α,β,γ,δ))λ=a+ib∈σ(A(α,β,γ,δ)) with b≠0b≠ 0, and write b+=|b|b_+=|b|. Then: 1. 0≤a≤10≤ a≤ 1. 2. a+b+≤1a+b_+≤ 1. 3. with G(a,b+):=(b+2+a2+a)2+2a2−b+2,G(a,b_+):=(b_+^2+a^2+a)^2+2a^2-b_+^2, we have G(a,b+)≥0G(a,b_+)≥ 0. Conversely, if b>0b>0 and 0≤a≤1,a+b<1,G(a,b)>0,0≤ a≤ 1, a+b<1, G(a,b)>0, then there exist α,β,γ,δ∈[0,1)α,β,γ,δ∈[0,1) such that λ∈σ(A(α,β,γ,δ))λ∈σ(A(α,β,γ,δ)). (The lower half-plane follows by conjugation.) Moreover: • the segment CR:λ=1−x+ixC_R:\ λ=1-x+ix (x∈[0,1]x∈[0,1]) is attained by α=β=γ=δ=1−xα=β=γ=δ=1-x; • the curve portion CLC_L of G(a,b)=0G(a,b)=0 in the upper half-plane joining i to 0 is attained by AL(α)=(α1−α00001000011000),α∈[0,1);A_L(α)= pmatrixα&1-α&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0 pmatrix, α∈[0,1); • every point strictly inside b>0, 0≤a≤1,a+b<1,G(a,b)>0\b>0,\ 0≤ a≤ 1,\ a+b<1,\ G(a,b)>0\ is attained. Proof. 1) Eigenvalue equation. Let v≠0v≠ 0 satisfy Av=λvAv=λ v. Row-by-row, (λ−α)v1=(1−α)v2,(λ−β)v2=(1−β)v3,(λ−γ)v3=(1−γ)v4,(λ−δ)v4=(1−δ)v1.(λ-α)v_1=(1-α)v_2, (λ-β)v_2=(1-β)v_3, (λ-γ)v_3=(1-γ)v_4, (λ-δ)v_4=(1-δ)v_1. Since 1−α,…,1−δ>01-α,…,1-δ>0, if any vk=0v_k=0 then all vj=0v_j=0, contradiction. Hence v1v2v3v4≠0v_1v_2v_3v_4≠ 0, and multiplying gives (λ−α)(λ−β)(λ−γ)(λ−δ)=(1−α)(1−β)(1−γ)(1−δ).(λ-α)(λ-β)(λ-γ)(λ-δ)=(1-α)(1-β)(1-γ)(1-δ). (1) Put t1=1−α,…,t4=1−δ∈(0,1]t_1=1-α,…,t_4=1-δ∈(0,1] and z=λ−1z=λ-1. Then λ−(1−tk)=z+tkλ-(1-t_k)=z+t_k, so Equation 1 becomes (z+t1)(z+t2)(z+t3)(z+t4)=t1t2t3t4.(z+t_1)(z+t_2)(z+t_3)(z+t_4)=t_1t_2t_3t_4. (2) 2) Angle parametrization and the function F. Assume b>0b>0 (the case b<0b<0 follows by conjugation since A is real). Write z=x+iy,x=a−1,y=b.z=x+iy, x=a-1,\ y=b. For t>0t>0, z+tz+t lies in the upper half-plane, so define u(t)=arg(z+t)∈(0,π)u(t)=arg(z+t)∈(0,π). Elementary trigonometry gives the inverse relation t=t(u)=ycotu−x,|z+t(u)|=ycscu.t=t(u)=y u-x, |z+t(u)|=y u. (3) Define F(u):=log|z+t(u)|−logt(u)=log(ycscu)−log(ycotu−x).F(u):= |z+t(u)|- t(u)= (y u)- (y u-x). (4) Let m:=arg(z+1)=arg(λ),M:=arg(z)=arg(λ−1).m:=arg(z+1)=arg(λ), M:=arg(z)=arg(λ-1). (5) As t↓0t 0, u(t)↑Mu(t) M; as t↑1t 1, u(t)↓mu(t) m. Thus t∈(0,1]⟺u∈[m,M).t∈(0,1] u∈[m,M). (6) Lemma 1 (Log-modulus/argument reformulation). A nonreal λ satisfies Equation 2 for some tk∈(0,1]t_k∈(0,1] iff there exist uk∈[m,M)u_k∈[m,M) such that u1+u2+u3+u4=2π,u_1+u_2+u_3+u_4=2π, (7) F(u1)+F(u2)+F(u3)+F(u4)=0.F(u_1)+F(u_2)+F(u_3)+F(u_4)=0. (8) Proof. Write z+tk=|z+tk|eiukz+t_k=|z+t_k|e^iu_k with uk∈(0,π)u_k∈(0,π). Taking arguments in Equation 2 gives ∑uk≡0(mod2π)Σ u_k≡ 0 2π; since ∑uk∈(0,4π)Σ u_k∈(0,4π) this forces ∑uk=2πΣ u_k=2π, i.e. Equation 7. Taking moduli and logs gives Equation 8. Condition Equation 6 is exactly tk∈(0,1]t_k∈(0,1]. Conversely, Equation 7 and Equation 8 give equality of arguments and moduli, hence Equation 2. ∎ Define :=(u1,…,u4)∈[m,M)4:u1+⋯+u4=2π,Ψ(u1,…,u4):=∑k=14F(uk).P:= \(u_1,…,u_4)∈[m,M)^4:\ u_1+·s+u_4=2π \, (u_1,…,u_4):= _k=1^4F(u_k). Then λ is a nonreal eigenvalue of some A(α,β,γ,δ)A(α,β,γ,δ) iff ≠∅P≠ and 0∈Ψ()0∈ (P). 3) Necessarily 0≤a≤10≤ a≤ 1 (indeed a<1a<1 if b≠0b≠ 0). Every (u1,…,u4)∈(u_1,…,u_4) has average π/2π/2, so π/2∈[m,M)π/2∈[m,M), i.e. m≤π2<M.m≤ π2<M. (9) For b>0b>0, m≤π/2⇔a≥0m≤π/2 a≥ 0. Also M>π/2⇔a<1M>π/2 a<1 because ℜ(λ−1)=a−1<0 (λ-1)=a-1<0 iff a<1a<1. Thus ≠∅P≠ implies 0≤a<1,0≤ a<1, (10) hence in particular 0≤a≤10≤ a≤ 1. 4) Strict convexity of F. Differentiate Equation 4 using Equation 3. A direct computation yields F′(u)=x2+y2(x−ycotu)2sin2u>0(u∈(0,π)),F (u)= x^2+y^2(x-y u)^2 ^2u>0 (u∈(0,π)), (11) so F is strictly convex on (0,π)(0,π), hence on [m,M)[m,M). 5) Right boundary: a+b≤1a+b≤ 1, and attainment of CRC_R. P is convex and connected; Ψ is continuous, hence Ψ() (P) is an interval. By Jensen and strict convexity, Ψ(u1,…,u4)≥4F(u1+⋯+u44)=4F(π/2). (u_1,…,u_4)≥ 4F\! ( u_1+·s+u_44 )=4F(π/2). (12) Thus 0∈Ψ()0∈ (P) forces F(π/2)≤0F(π/2)≤ 0. Now F(π/2)=log(y)−log(−x)=log(b1−a),F(π/2)= (y)- (-x)= \! ( b1-a ), so F(π/2)≤0⇔b≤1−a⇔a+b≤1F(π/2)≤ 0 b≤ 1-a a+b≤ 1. This proves item (2) (for b>0b>0; for b<0b<0 replace b by |b||b|). Attainment of CRC_R: if t1=t2=t3=t4=x∈(0,1]t_1=t_2=t_3=t_4=x∈(0,1], then Equation 2 becomes (z+x)4=x4(z+x)^4=x^4, i.e. λ=1−x±ixλ=1-x± ix, attained by α=β=γ=δ=1−xα=β=γ=δ=1-x. 6) A max principle on P. There are two regimes. Lemma 2 (Unbounded supremum). If 3m+M≤2π,3m+M≤ 2π, (13) then supΨ=+∞ _P =+∞. Proof. Take u4↑Mu_4 M and u1=u2=u3=(2π−u4)/3u_1=u_2=u_3=(2π-u_4)/3. Condition Equation 13 ensures u1≥mu_1≥ m. Also u4<Mu_4<M and u1≤(2π−M)/3≤Mu_1≤(2π-M)/3≤ M since M≥π/2⇒2π≤4M≥π/2 2π≤ 4M. Hence (u1,…,u4)∈(u_1,…,u_4) . As u4↑Mu_4 M, t(u4)=ycotu4−x↓0t(u_4)=y u_4-x 0, so −logt(u4)→+∞- t(u_4)→+∞ while log(ycscu4) (y u_4) stays bounded; thus F(u4)→+∞F(u_4)→+∞ and Ψ→+∞ →+∞. ∎ Lemma 3 (Finite maximum in the tight regime). If 3m+M>2π,U:=2π−3m<M,3m+M>2π, U:=2π-3m<M, (14) then maxΨ=3F(m)+F(U), _P =3F(m)+F(U), (15) and equality holds iff (u1,…,u4)(u_1,…,u_4) is a permutation of (U,m,m,m)(U,m,m,m). Proof. Order uku_k decreasing: v1≥v2≥v3≥v4v_1≥ v_2≥ v_3≥ v_4. Since each vj≥mv_j≥ m and ∑vj=2πΣ v_j=2π, v1=2π−(v2+v3+v4)≤2π−3m=U.v_1=2π-(v_2+v_3+v_4)≤ 2π-3m=U. Thus (v1,…,v4)∈[m,U]4(v_1,…,v_4)∈[m,U]^4 with sum 2π2π. The vector (U,m,m,m)(U,m,m,m) majorizes (v1,…,v4)(v_1,…,v_4). By Karamata (convex F), ∑j=14F(vj)≤F(U)+3F(m), _j=1^4F(v_j)≤ F(U)+3F(m), with equality only at permutations of (U,m,m,m)(U,m,m,m) since F is strictly convex. ∎ 7) The inequality G(a,b)≥0G(a,b)≥ 0. We prove necessity for every nonreal eigenvalue. 7.1 Geometry lemma: G(a,b)≤0⇒3m+M>2πG(a,b)≤ 0 3m+M>2π Lemma 4. Assume b>0b>0 and 0≤a≤10≤ a≤ 1. If G(a,b)≤0G(a,b)≤ 0, then 3m+M>2π3m+M>2π. Proof. Set s=b2s=b^2. Then G(a,b)=s2+s(2a2+2a−1)+(a2+a)2+2a2.G(a,b)=s^2+s(2a^2+2a-1)+(a^2+a)^2+2a^2. (16) This is quadratic in s with discriminant Δ=(2a2+2a−1)2−4((a2+a)2+2a2)=−(2a+1)(6a−1). =(2a^2+2a-1)^2-4 ((a^2+a)^2+2a^2 )=-(2a+1)(6a-1). Thus G(a,b)≤0G(a,b)≤ 0 forces Δ>0 >0, hence a<1/6a<1/6, and s lies between the two real roots; in particular s≥s−(a):=12−a−a2−12(1−6a)(1+2a).s≥ s_-(a):= 12-a-a^2- 12 (1-6a)(1+2a). (17) Case 1: a=0a=0. Then m=arg(ib)=π/2m=arg(ib)=π/2, and M=arg(−1+ib)=π−arctan(b)M=arg(-1+ib)=π- (b). Hence 3m+M=3π2+π−arctan(b)=5π2−arctan(b)>2π.3m+M= 3π2+π- (b)= 5π2- (b)>2π. Case 2: a>0a>0. From Equation 17 we get s>s−(a)≥3a2s>s_-(a)≥ 3a^2, with strictness because s−(a)−3a2=12−a−4a2−12(1−6a)(1+2a)s_-(a)-3a^2= 12-a-4a^2- 12 (1-6a)(1+2a) satisfies (LHS)2−(RHS)2=32a3(2a+1)>0(LHS)^2-(RHS)^2=32a^3(2a+1)>0 for a>0a>0. Hence b2>3a2b^2>3a^2, i.e. tanm=b/a>3 m=b/a> 3, so m>π/3m>π/3 and 3m−π∈(0,π/2)3m-π∈(0,π/2). Write ϕ:=arctan(b1−a)∈(0,π/2)φ:= \! ( b1-a )∈(0,π/2), so M=π−ϕM=π-φ. Then 3m+M>2π⟺ 3m−π>ϕ⟺tan(3m)>b1−a,3m+M>2π\ \ 3m-π>φ\ \ (3m)> b1-a, (18) since tan is increasing on (0,π/2)(0,π/2) and tan(3m−π)=tan(3m) (3m-π)= (3m). Let t=tanm=b/at= m=b/a (allowed since a>0a>0). The triple-angle identity gives tan(3m)=3t−t31−3t2 (3m)= 3t-t^31-3t^2, and a direct simplification yields tan(3m)−b1−a=b(4a3−3a2−4ab2+b2)a(a−1)(a2−3b2). (3m)- b1-a= b\,(4a^3-3a^2-4ab^2+b^2)a(a-1)(a^2-3b^2). (19) Here a>0a>0, a−1<0a-1<0, and a2−3b2<0a^2-3b^2<0 (since b2>3a2b^2>3a^2), so the denominator in Equation 19 is positive. Hence the sign equals the sign of N(a,b):=4a3−3a2−4ab2+b2=(1−4a)b2+a2(4a−3).N(a,b):=4a^3-3a^2-4ab^2+b^2=(1-4a)b^2+a^2(4a-3). (20) For a<1/6a<1/6, 1−4a>01-4a>0 and 4a−3<04a-3<0, so N(a,b)N(a,b) is strictly increasing in b2b^2 and has a unique zero at s0(a)=a2(3−4a)1−4a.s_0(a)= a^2(3-4a)1-4a. (21) One checks that s−(a)>s0(a)s_-(a)>s_0(a) for every a∈(0,1/6)a∈(0,1/6): after multiplying by 1−4a>01-4a>0 and squaring, the difference becomes 256a6>0256a^6>0. Therefore b2≥s−(a)>s0(a)b^2≥ s_-(a)>s_0(a) implies N(a,b)>0N(a,b)>0. By Equation 19–18, this gives 3m+M>2π3m+M>2π. ∎ 7.2 Tight regime: maxΨ≥0⇔G≥0 ≥ 0 G≥ 0 Assume 3m+M>2π3m+M>2π, so Lemma 6.2 applies with U=2π−3m∈(0,π)U=2π-3m∈(0,π) and maximizer (U,m,m,m)(U,m,m,m). Since u(1)=arg(z+1)=mu(1)=arg(z+1)=m, Equation 6 gives t(m)=1t(m)=1, hence F(m)=log|z+1|−log1=log|λ|.F(m)= |z+1|- 1= |λ|. (22) Also t(U)=bcotU−(a−1)=bcotU+1−at(U)=b U-(a-1)=b U+1-a and |z+t(U)|=bcscU|z+t(U)|=b U, so by Equation 15 maxΨ=3log|λ|+log(bcscU)−log(bcotU+1−a)=log(|λ|3bcscUbcotU+1−a). _P =3 |λ|+ (b U)- (b U+1-a)= \! ( |λ|^3\,b Ub U+1-a ). (23) Thus maxΨ≥0 _P ≥ 0 is equivalent to |λ|3bcscU≥bcotU+1−a.|λ|^3\,b U\ ≥\ b U+1-a. (24) Now U=2π−3mU=2π-3m, so sinU=−sin(3m) U=- (3m) and cosU=cos(3m) U= (3m). Since λ3=(a+ib)3=(a3−3ab2)+i(3a2b−b3)λ^3=(a+ib)^3=(a^3-3ab^2)+i(3a^2b-b^3), we have cos(3m)=ℜ(λ3)|λ|3=a(a2−3b2)|λ|3,sin(3m)=ℑ(λ3)|λ|3=b(3a2−b2)|λ|3. (3m)= (λ^3)|λ|^3= a(a^2-3b^2)|λ|^3, (3m)= (λ^3)|λ|^3= b(3a^2-b^2)|λ|^3. (25) Hence sinU=b(b2−3a2)|λ|3,cosU=a(a2−3b2)|λ|3. U= b(b^2-3a^2)|λ|^3, U= a(a^2-3b^2)|λ|^3. (26) In the tight regime, necessarily m>π/3m>π/3 (since M≤πM≤π), hence b2−3a2>0b^2-3a^2>0, so sinU>0 U>0 and bcscU=|λ|3b2−3a2,bcotU=a(a2−3b2)b2−3a2.b U= |λ|^3b^2-3a^2, b U= a(a^2-3b^2)b^2-3a^2. (27) Substitute Equation 27 into Equation 24 and clear the positive denominator b2−3a2b^2-3a^2: |λ|6≥a(a2−3b2)+(1−a)(b2−3a2).|λ|^6\ ≥\ a(a^2-3b^2)+(1-a)(b^2-3a^2). (28) Using |λ|2=a2+b2|λ|^2=a^2+b^2, a direct expansion and factorization yields |λ|6−(a(a2−3b2)+(1−a)(b2−3a2))=((a−1)2+b2)G(a,b).|λ|^6- (a(a^2-3b^2)+(1-a)(b^2-3a^2) )= ((a-1)^2+b^2 )\,G(a,b). (29) Since (a−1)2+b2=|λ−1|2>0(a-1)^2+b^2=|λ-1|^2>0, Equation 28 is equivalent to G(a,b)≥0G(a,b)≥ 0. Therefore, in the tight regime, maxΨ≥0⟺G(a,b)≥0. _P ≥ 0 G(a,b)≥ 0. (30) 7.3 Necessity of G≥0G≥ 0 for all nonreal eigenvalues Let λ be a nonreal eigenvalue of some A(α,β,γ,δ)A(α,β,γ,δ). Then ≠∅P≠ and 0∈Ψ()0∈ (P), hence supΨ≥0 _P ≥ 0. If 3m+M≤2π3m+M≤ 2π, the previous lemma (contrapositive) gives G(a,b)≥0G(a,b)≥ 0. If 3m+M>2π3m+M>2π, then supΨ=maxΨ = and Equation 30 gives G(a,b)≥0G(a,b)≥ 0. This proves item (3) in the theorem. 8) Boundary identification and sharpness. 8.1 Right boundary CRC_R Already established in §5. 8.2 Left boundary CLC_L: attainment by AL(α)A_L(α), and G=0G=0 For AL(α)=(α1−α00001000011000),α∈[0,1),A_L(α)= pmatrixα&1-α&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0 pmatrix, α∈[0,1), we have β=γ=δ=0β=γ=δ=0, so Equation 1 reduces to λ3(λ−α)=1−α.λ^3(λ-α)=1-α. (31) Eliminate α: α=λ4−1λ3−1.α= λ^4-1λ^3-1. (32) Thus α∈ℝα iff ℑ(λ4−1λ3−1)=0 \! ( λ^4-1λ^3-1 )=0. Compute ℑ(λ4−1λ3−1)=ℑ((λ4−1)(λ3¯−1))|λ3−1|2. \! ( λ^4-1λ^3-1 )= ((λ^4-1)( λ^3-1) )|λ^3-1|^2. A direct expansion and factorization in λ=a+ibλ=a+ib gives, for b≠0b≠ 0, ℑ(λ4−1λ3−1)=b|λ−1|2G(a,b)|λ3−1|2. \! ( λ^4-1λ^3-1 )= b\,|λ-1|^2\,G(a,b)|λ^3-1|^2. (33) Hence, for b≠0b≠ 0, α∈ℝ⟺G(a,b)=0.α G(a,b)=0. (34) So every nonreal eigenvalue of AL(α)A_L(α) lies on G=0G=0. Moreover, AL(α)A_L(α) always has a nonreal conjugate pair: rewrite Equation 31 as λ4−αλ3−(1−α)=0⟺(λ−1)(λ3+(1−α)(λ2+λ+1))=0.λ^4-αλ^3-(1-α)=0 (λ-1) (λ^3+(1-α)(λ^2+λ+1) )=0. Let c=1−α∈(0,1]c=1-α∈(0,1] and f(λ)=λ3+c(λ2+λ+1)f(λ)=λ^3+c(λ^2+λ+1). Then f′(λ)=3λ2+2cλ+cf (λ)=3λ^2+2cλ+c has discriminant 4c(c−3)<04c(c-3)<0, so f is strictly increasing on ℝR. Also f(0)=c>0f(0)=c>0 and f(λ)→−∞f(λ)→-∞ as λ→−∞λ→-∞, hence f has exactly one real root. Therefore the remaining two roots of f form a nonreal conjugate pair, giving the desired pair for AL(α)A_L(α). Endpoints: at α=0α=0, Equation 31 becomes λ4=1λ^4=1, giving λ=iλ=i. As α↑1α 1, Equation 31 tends to λ3(λ−1)=0λ^3(λ-1)=0; the conjugate pair among the three non-11 roots tends to 0. Thus α∈[0,1)α∈[0,1) traces a connected arc of G=0G=0 from i to 0. Finally, every point on that arc is attained as follows. Let λ=a+ibλ=a+ib with b>0b>0, 0≤a≤10≤ a≤ 1, a+b≤1a+b≤ 1, and G(a,b)=0G(a,b)=0. By Lemma 7.1 (since G=0≤0G=0≤ 0), we are in the tight regime 3m+M>2π3m+M>2π, so Lemma 6.2 applies and the maximizer is (U,m,m,m)(U,m,m,m) with U=2π−3m∈[m,M)U=2π-3m∈[m,M). By §7.2, G=0G=0 implies maxΨ=0 _P =0, hence Ψ(U,m,m,m)=0 (U,m,m,m)=0. By Lemma 2.1 this produces t1=t(U)∈(0,1]t_1=t(U)∈(0,1] and t2=t3=t4=t(m)=1t_2=t_3=t_4=t(m)=1, satisfying Equation 2. Therefore the corresponding parameters α=1−t(U)∈[0,1),β=γ=δ=0α=1-t(U)∈[0,1), β=γ=δ=0 yield exactly AL(α)A_L(α) and give λ as an eigenvalue. This identifies CLC_L and proves it is attained. 9) Existence for every strict interior point. Fix λ=a+ibλ=a+ib with b>0b>0 such that 0≤a≤1,a+b<1,G(a,b)>0.0≤ a≤ 1, a+b<1, G(a,b)>0. (35) Define z,m,M,F,,Ψz,m,M,F,P, as above. Then ≠∅P≠ by §3. We show 0∈Ψ()0∈ (P). By Jensen, Ψ(π/2,π/2,π/2,π/2)=4F(π/2)=4log(b1−a)<0 (π/2,π/2,π/2,π/2)=4F(π/2)=4 \! ( b1-a )<0 since a+b<1⇔b<1−a+b<1 b<1-a. To produce a positive value: if 3m+M≤2π3m+M≤ 2π, Lemma 6.1 gives supΨ=+∞ _P =+∞, so Ψ takes positive values. If 3m+M>2π3m+M>2π, Lemma 6.2 gives maxΨ=3F(m)+F(U) _P =3F(m)+F(U); by §7.2, G(a,b)>0G(a,b)>0 implies maxΨ>0 _P >0, so Ψ takes positive values. Since P is connected and Ψ is continuous, Ψ() (P) is an interval containing both negative and positive values, hence contains 0. Thus there exists (u1,…,u4)∈(u_1,…,u_4) with Ψ(u1,…,u4)=0 (u_1,…,u_4)=0. By Lemma 2.1 this yields tk=t(uk)∈(0,1]t_k=t(u_k)∈(0,1] satisfying Equation 2. Set α=1−t1,β=1−t2,γ=1−t3,δ=1−t4∈[0,1),α=1-t_1, β=1-t_2, γ=1-t_3, δ=1-t_4∈[0,1), and then Equation 1 holds, so λ∈σ(A(α,β,γ,δ))λ∈σ(A(α,β,γ,δ)). Conjugation gives the lower half-plane. ∎ Appendix B Second version Theorem 22 (Spectral region for a 4-cycle row-stochastic matrix). Let A(α,β,γ,δ)=(α1−α000β1−β000γ1−γ1−δ00δ),α,β,γ,δ∈[0,1).A(α,β,γ,δ)= pmatrixα&1-α&0&0\\ 0&β&1-β&0\\ 0&0&γ&1-γ\\ 1-δ&0&0&δ pmatrix, α,β,γ,δ∈[0,1). Let λ=a+ib∈σ(A(α,β,γ,δ))λ=a+ib∈σ(A(α,β,γ,δ)) with b≠0b≠ 0, and write b+=|b|b_+=|b|. Then: 1. 0≤a≤10≤ a≤ 1; 2. a+b+≤1a+b_+≤ 1; 3. with G(a,b+):=(b+2+a2+a)2+2a2−b+2,G(a,b_+):=(b_+^2+a^2+a)^2+2a^2-b_+^2, we have G(a,b+)≥0G(a,b_+)≥ 0. Conversely, if b>0b>0 and 0≤a≤1,a+b<1,G(a,b)>0,0≤ a≤ 1, a+b<1, G(a,b)>0, then there exist α,β,γ,δ∈[0,1)α,β,γ,δ∈[0,1) such that λ∈σ(A(α,β,γ,δ))λ∈σ(A(α,β,γ,δ)). (The lower half-plane follows by conjugation.) Moreover: • the segment CR:λ=1−x+ixC_R:\ λ=1-x+ix (x∈[0,1]x∈[0,1]) is attained by α=β=γ=δ=1−xα=β=γ=δ=1-x; • the curve portion CLC_L of G(a,b)=0G(a,b)=0 in the upper half-plane joining i to 0 is attained by AL(α)=(α1−α00001000011000),α∈[0,1);A_L(α)= pmatrixα&1-α&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0 pmatrix, α∈[0,1); • every point strictly inside b>0, 0≤a≤1,a+b<1,G(a,b)>0\b>0,\ 0≤ a≤ 1,\ a+b<1,\ G(a,b)>0\ is attained. Proof. 1) Eigenvalue equation. Let v≠0v≠ 0 satisfy Av=λvAv=λ v. Row-by-row, (λ−α)v1=(1−α)v2,(λ−β)v2=(1−β)v3,(λ−γ)v3=(1−γ)v4,(λ−δ)v4=(1−δ)v1.(λ-α)v_1=(1-α)v_2, (λ-β)v_2=(1-β)v_3, (λ-γ)v_3=(1-γ)v_4, (λ-δ)v_4=(1-δ)v_1. Since 1−α,…,1−δ>01-α,…,1-δ>0, if any vk=0v_k=0 then all vj=0v_j=0, contradiction. Hence v1v2v3v4≠0v_1v_2v_3v_4≠ 0, and multiplying gives (λ−α)(λ−β)(λ−γ)(λ−δ)=(1−α)(1−β)(1−γ)(1−δ).(λ-α)(λ-β)(λ-γ)(λ-δ)=(1-α)(1-β)(1-γ)(1-δ). (1) Put t1=1−α,…,t4=1−δ∈(0,1]t_1=1-α,…,t_4=1-δ∈(0,1] and z=λ−1z=λ-1. Then λ−(1−tk)=z+tkλ-(1-t_k)=z+t_k, so (1) becomes (z+t1)(z+t2)(z+t3)(z+t4)=t1t2t3t4.(z+t_1)(z+t_2)(z+t_3)(z+t_4)=t_1t_2t_3t_4. (2) 2) Angle parametrization and the function F. Assume b>0b>0 (the case b<0b<0 follows by conjugation since A is real). Write z=x+iy,x=a−1,y=b.z=x+iy, x=a-1,\ y=b. For t>0t>0, z+tz+t lies in the upper half-plane, so define u(t)=Arg(z+t)∈(0,π)u(t)=Arg(z+t)∈(0,π). Elementary trigonometry gives the inverse relation t=t(u)=ycotu−x,|z+t(u)|=ycscu.t=t(u)=y u-x, |z+t(u)|=y u. (3) Define F(u):=log|z+t(u)|−logt(u)=log(ycscu)−log(ycotu−x).F(u):= |z+t(u)|- t(u)= (y u)- (y u-x). (4) Let m:=Arg(z+1)=Arg(λ),M:=Arg(z)=Arg(λ−1).m:=Arg(z+1)=Arg(λ), M:=Arg(z)=Arg(λ-1). (5) As t↓0t 0, u(t)↑Mu(t) M; as t↑1t 1, u(t)↓mu(t) m. Thus t∈(0,1]⟺u∈[m,M).t∈(0,1] u∈[m,M). (6) Lemma 5 (Log-modulus/argument reformulation). A nonreal λ satisfies (2) for some tk∈(0,1]t_k∈(0,1] iff there exist uk∈[m,M)u_k∈[m,M) such that u1+u2+u3+u4=2π,u_1+u_2+u_3+u_4=2π, (7) F(u1)+F(u2)+F(u3)+F(u4)=0.F(u_1)+F(u_2)+F(u_3)+F(u_4)=0. (8) Proof. Write z+tk=|z+tk|eiukz+t_k=|z+t_k|e^iu_k with uk∈(0,π)u_k∈(0,π). Taking arguments in (2) gives ∑uk≡0(mod2π)Σ u_k≡ 0 2π; since ∑uk∈(0,4π)Σ u_k∈(0,4π) this forces ∑uk=2πΣ u_k=2π, i.e. (7). Taking moduli and logs gives (8). Condition (6) is exactly tk∈(0,1]t_k∈(0,1]. Conversely, (7) and (8) give equality of arguments and moduli, hence (2). ∎ Define :=(u1,…,u4)∈[m,M)4:u1+⋯+u4=2π,Ψ(u1,…,u4):=∑k=14F(uk).P:= \(u_1,…,u_4)∈[m,M)^4:\ u_1+·s+u_4=2π \, (u_1,…,u_4):= _k=1^4F(u_k). (9) Then λ is a nonreal eigenvalue of some A(α,β,γ,δ)A(α,β,γ,δ) iff ≠∅P≠ and 0∈Ψ()0∈ (P). 3) Necessarily 0≤a≤10≤ a≤ 1 (indeed a<1a<1 if b≠0b≠ 0). Every (u1,…,u4)∈(u_1,…,u_4) has average π/2π/2, so π/2∈[m,M)π/2∈[m,M), i.e. m≤π2<M.m≤ π2<M. (10) For b>0b>0, m≤π/2⇔a≥0m≤π/2 a≥ 0. Also M>π/2⇔a<1M>π/2 a<1 because ℜ(λ−1)=a−1<0 (λ-1)=a-1<0 iff a<1a<1. Thus ≠∅P≠ implies 0≤a<1,0≤ a<1, (11) hence in particular 0≤a≤10≤ a≤ 1. 4) Strict convexity of F. Differentiate (4) using (3). A direct computation yields F′(u)=x2+y2(x−ycotu)2sin2u>0(u∈(0,π)),F (u)= x^2+y^2(x-y u)^2 ^2u>0 (u∈(0,π)), (12) so F is strictly convex on (0,π)(0,π), hence on [m,M)[m,M). 5) Right boundary: a+b≤1a+b≤ 1, and attainment of CRC_R. P is convex and connected; Ψ is continuous, hence Ψ() (P) is an interval. By Jensen and strict convexity, Ψ(u1,…,u4)≥4F(u1+⋯+u44)=4F(π/2). (u_1,…,u_4)≥ 4F\! ( u_1+·s+u_44 )=4F(π/2). (13) Thus 0∈Ψ()0∈ (P) forces F(π/2)≤0F(π/2)≤ 0. Now F(π/2)=log(y)−log(−x)=log(b1−a),F(π/2)= (y)- (-x)= \! ( b1-a ), so F(π/2)≤0⇔b≤1−a⇔a+b≤1F(π/2)≤ 0 b≤ 1-a a+b≤ 1. This proves item (2) in the theorem (for b>0b>0; for b<0b<0 replace b by |b||b|). Attainment of CRC_R: if t1=t2=t3=t4=x∈(0,1]t_1=t_2=t_3=t_4=x∈(0,1], then (2) becomes (z+x)4=x4(z+x)^4=x^4, i.e. λ=1−x±ixλ=1-x± ix, attained by α=β=γ=δ=1−xα=β=γ=δ=1-x. 6) A max principle on P. There are two regimes. Lemma 6 (Unbounded supremum). If 3m+M≤2π,3m+M≤ 2π, (14) then supΨ=+∞ _P =+∞. Proof. Take u4↑Mu_4 M and u1=u2=u3=(2π−u4)/3u_1=u_2=u_3=(2π-u_4)/3. Condition (14) ensures u1≥mu_1≥ m. Also u4<Mu_4<M and u1≤(2π−M)/3≤Mu_1≤(2π-M)/3≤ M since M≥π/2⇒2π≤4M≥π/2 2π≤ 4M. Hence (u1,…,u4)∈(u_1,…,u_4) . As u4↑Mu_4 M, t(u4)=ycotu4−x↓0t(u_4)=y u_4-x 0, so −logt(u4)→+∞- t(u_4)→+∞ while log(ycscu4) (y u_4) stays bounded; thus F(u4)→+∞F(u_4)→+∞ and Ψ→+∞ →+∞. ∎ Lemma 7 (Finite maximum in the tight regime). If 3m+M>2π,U:=2π−3m<M,3m+M>2π, U:=2π-3m<M, (15) then maxΨ=3F(m)+F(U), _P =3F(m)+F(U), (16) and equality holds iff (u1,…,u4)(u_1,…,u_4) is a permutation of (U,m,m,m)(U,m,m,m). Proof. Order uku_k decreasing: v1≥v2≥v3≥v4v_1≥ v_2≥ v_3≥ v_4. Since each vj≥mv_j≥ m and ∑vj=2πΣ v_j=2π, v1=2π−(v2+v3+v4)≤2π−3m=U.v_1=2π-(v_2+v_3+v_4)≤ 2π-3m=U. Thus (v1,…,v4)∈[m,U]4(v_1,…,v_4)∈[m,U]^4 with sum 2π2π. The vector (U,m,m,m)(U,m,m,m) majorizes (v1,…,v4)(v_1,…,v_4). By Karamata (convex F), ∑j=14F(vj)≤F(U)+3F(m), _j=1^4F(v_j)≤ F(U)+3F(m), with equality only at permutations of (U,m,m,m)(U,m,m,m) since F is strictly convex. ∎ 7) The inequality G(a,b)≥0G(a,b)≥ 0. We prove necessity for every nonreal eigenvalue. 7.1 Geometry lemma: G(a,b)≤0⇒3m+M>2πG(a,b)≤ 0 3m+M>2π Lemma 8. Assume b>0b>0 and 0≤a≤10≤ a≤ 1. If G(a,b)≤0G(a,b)≤ 0, then 3m+M>2π3m+M>2π. Proof. Set s=b2s=b^2. Then G(a,b)=s2+s(2a2+2a−1)+(a2+a)2+2a2.G(a,b)=s^2+s(2a^2+2a-1)+(a^2+a)^2+2a^2. (17) This is quadratic in s with discriminant Δ=(2a2+2a−1)2−4((a2+a)2+2a2)=−(2a+1)(6a−1). =(2a^2+2a-1)^2-4 ((a^2+a)^2+2a^2 )=-(2a+1)(6a-1). Thus G(a,b)≤0G(a,b)≤ 0 forces Δ>0 >0, hence a<1/6a<1/6, and s lies between the two real roots; in particular s≥s−(a):=12−a−a2−12(1−6a)(1+2a).s≥ s_-(a):= 12-a-a^2- 12 (1-6a)(1+2a). (18) Case 1: a=0a=0. Then m=Arg(ib)=π/2m=Arg(ib)=π/2 and M=Arg(−1+ib)=π−arctan(b)M=Arg(-1+ib)=π- (b). Hence 3m+M=3π2+π−arctan(b)=5π2−arctan(b)>2π.3m+M= 3π2+π- (b)= 5π2- (b)>2π. Case 2: a>0a>0. From (18) one gets s>s−(a)≥3a2s>s_-(a)≥ 3a^2, with strictness because s−(a)−3a2=12−a−4a2−12(1−6a)(1+2a)s_-(a)-3a^2= 12-a-4a^2- 12 (1-6a)(1+2a) satisfies (LHS)2−(RHS)2=32a3(2a+1)>0(LHS)^2-(RHS)^2=32a^3(2a+1)>0 for a>0a>0. Hence b2>3a2b^2>3a^2, i.e. tanm=b/a>3 m=b/a> 3, so m>π/3m>π/3 and 3m−π∈(0,π/2)3m-π∈(0,π/2). Write ϕ:=arctan(b1−a)∈(0,π/2)φ:= \! ( b1-a )∈(0,π/2), so M=π−ϕM=π-φ. Then 3m+M>2π⟺ 3m−π>ϕ⟺tan(3m)>b1−a,3m+M>2π\ \ 3m-π>φ\ \ (3m)> b1-a, (19) since tan is increasing on (0,π/2)(0,π/2) and tan(3m−π)=tan(3m) (3m-π)= (3m). Let t=tanm=b/at= m=b/a (allowed since a>0a>0). The triple-angle identity gives tan(3m)=3t−t31−3t2 (3m)= 3t-t^31-3t^2, and a direct simplification yields tan(3m)−b1−a=b(4a3−3a2−4ab2+b2)a(a−1)(a2−3b2). (3m)- b1-a= b\,(4a^3-3a^2-4ab^2+b^2)a(a-1)(a^2-3b^2). (20) Here a>0a>0, a−1<0a-1<0, and a2−3b2<0a^2-3b^2<0 (since b2>3a2b^2>3a^2), so the denominator in (20) is positive. Hence the sign equals the sign of N(a,b):=4a3−3a2−4ab2+b2=(1−4a)b2+a2(4a−3).N(a,b):=4a^3-3a^2-4ab^2+b^2=(1-4a)b^2+a^2(4a-3). (21) For a<1/6a<1/6, 1−4a>01-4a>0 and 4a−3<04a-3<0, so N(a,b)N(a,b) is strictly increasing in b2b^2 and has a unique zero at s0(a)=a2(3−4a)1−4a.s_0(a)= a^2(3-4a)1-4a. (22) One checks that s−(a)>s0(a)s_-(a)>s_0(a) for every a∈(0,1/6)a∈(0,1/6): after multiplying by 1−4a>01-4a>0 and squaring, the difference becomes 256a6>0256a^6>0. Therefore b2≥s−(a)>s0(a)b^2≥ s_-(a)>s_0(a) implies N(a,b)>0N(a,b)>0. By (20) and (19), this gives 3m+M>2π3m+M>2π. ∎ 7.2 Tight regime: maxΨ≥0⇔G≥0 ≥ 0 G≥ 0 Assume 3m+M>2π3m+M>2π, so Lemma 7 applies with U=2π−3m∈(0,π)U=2π-3m∈(0,π) and maximizer (U,m,m,m)(U,m,m,m). Since u(1)=Arg(z+1)=mu(1)=Arg(z+1)=m, (6) gives t(m)=1t(m)=1, hence F(m)=log|z+1|−log1=log|λ|.F(m)= |z+1|- 1= |λ|. (23) Also t(U)=bcotU−(a−1)=bcotU+1−at(U)=b U-(a-1)=b U+1-a and |z+t(U)|=bcscU|z+t(U)|=b U, so by (16) maxΨ=3log|λ|+log(bcscU)−log(bcotU+1−a)=log(|λ|3bcscUbcotU+1−a). _P =3 |λ|+ (b U)- (b U+1-a)= \! ( |λ|^3\,b Ub U+1-a ). (24) Thus maxΨ≥0 _P ≥ 0 is equivalent to |λ|3bcscU≥bcotU+1−a.|λ|^3\,b U\ ≥\ b U+1-a. (25) Now U=2π−3mU=2π-3m, so sinU=−sin(3m) U=- (3m) and cosU=cos(3m) U= (3m). Since λ3=(a+ib)3=(a3−3ab2)+i(3a2b−b3)λ^3=(a+ib)^3=(a^3-3ab^2)+i(3a^2b-b^3), we have cos(3m)=ℜ(λ3)|λ|3=a(a2−3b2)|λ|3,sin(3m)=ℑ(λ3)|λ|3=b(3a2−b2)|λ|3. (3m)= (λ^3)|λ|^3= a(a^2-3b^2)|λ|^3, (3m)= (λ^3)|λ|^3= b(3a^2-b^2)|λ|^3. (26) Hence sinU=b(b2−3a2)|λ|3,cosU=a(a2−3b2)|λ|3. U= b(b^2-3a^2)|λ|^3, U= a(a^2-3b^2)|λ|^3. (27) In the tight regime one necessarily has m>π/3m>π/3 (since M≤πM≤π), hence b2−3a2>0b^2-3a^2>0, so sinU>0 U>0 and bcscU=|λ|3b2−3a2,bcotU=a(a2−3b2)b2−3a2.b U= |λ|^3b^2-3a^2, b U= a(a^2-3b^2)b^2-3a^2. (28) Substitute (28) into (25) and clear the positive denominator b2−3a2b^2-3a^2: |λ|6≥a(a2−3b2)+(1−a)(b2−3a2).|λ|^6\ ≥\ a(a^2-3b^2)+(1-a)(b^2-3a^2). (29) Using |λ|2=a2+b2|λ|^2=a^2+b^2, a direct expansion and factorization yields |λ|6−(a(a2−3b2)+(1−a)(b2−3a2))=((a−1)2+b2)G(a,b).|λ|^6- (a(a^2-3b^2)+(1-a)(b^2-3a^2) )= ((a-1)^2+b^2 )\,G(a,b). (30) Since (a−1)2+b2=|λ−1|2>0(a-1)^2+b^2=|λ-1|^2>0, (29) is equivalent to G(a,b)≥0G(a,b)≥ 0. Therefore, in the tight regime, maxΨ≥0⟺G(a,b)≥0. _P ≥ 0 G(a,b)≥ 0. (31) 7.3 Necessity of G≥0G≥ 0 for all nonreal eigenvalues Let λ be a nonreal eigenvalue of some A(α,β,γ,δ)A(α,β,γ,δ). Then ≠∅P≠ and 0∈Ψ()0∈ (P), hence supΨ≥0 _P ≥ 0. If 3m+M≤2π3m+M≤ 2π, Lemma 8 (contrapositive) gives G(a,b)≥0G(a,b)≥ 0. If 3m+M>2π3m+M>2π, then supΨ=maxΨ = and (31) gives G(a,b)≥0G(a,b)≥ 0. This proves item (3) in the theorem. 8) Boundary identification and sharpness. 8.1 Right boundary CRC_R Already established in §5. 8.2 Left boundary CLC_L: attainment by AL(α)A_L(α), and G=0G=0 For AL(α)=(α1−α00001000011000),α∈[0,1),A_L(α)= pmatrixα&1-α&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0 pmatrix, α∈[0,1), we have β=γ=δ=0β=γ=δ=0, so (1) reduces to λ3(λ−α)=1−α.λ^3(λ-α)=1-α. (32) Eliminate α: α=λ4−1λ3−1.α= λ^4-1λ^3-1. (33) Thus α∈ℝα iff ℑ(λ4−1λ3−1)=0 \! ( λ^4-1λ^3-1 )=0. Compute ℑ(λ4−1λ3−1)=ℑ((λ4−1)(λ3¯−1))|λ3−1|2. \! ( λ^4-1λ^3-1 )= ((λ^4-1)( λ^3-1) )|λ^3-1|^2. A direct expansion and factorization in λ=a+ibλ=a+ib gives, for b≠0b≠ 0, ℑ(λ4−1λ3−1)=b|λ−1|2G(a,b)|λ3−1|2. \! ( λ^4-1λ^3-1 )= b\,|λ-1|^2\,G(a,b)|λ^3-1|^2. (34) Hence, for b≠0b≠ 0, α∈ℝ⟺G(a,b)=0.α G(a,b)=0. (35) So every nonreal eigenvalue of AL(α)A_L(α) lies on G=0G=0. Moreover, AL(α)A_L(α) always has a nonreal conjugate pair: rewrite (32) as λ4−αλ3−(1−α)=0⟺(λ−1)(λ3+(1−α)(λ2+λ+1))=0.λ^4-αλ^3-(1-α)=0 (λ-1) (λ^3+(1-α)(λ^2+λ+1) )=0. Let c=1−α∈(0,1]c=1-α∈(0,1] and f(λ)=λ3+c(λ2+λ+1)f(λ)=λ^3+c(λ^2+λ+1). Then f′(λ)=3λ2+2cλ+cf (λ)=3λ^2+2cλ+c has discriminant 4c(c−3)<04c(c-3)<0, so f is strictly increasing on ℝR. Also f(0)=c>0f(0)=c>0 and f(λ)→−∞f(λ)→-∞ as λ→−∞λ→-∞, hence f has exactly one real root. Therefore the remaining two roots of f form a nonreal conjugate pair. Endpoints: at α=0α=0, (32) becomes λ4=1λ^4=1, giving λ=iλ=i. As α↑1α 1, (32) tends to λ3(λ−1)=0λ^3(λ-1)=0; the conjugate pair among the three non-11 roots tends to 0. Thus α∈[0,1)α∈[0,1) traces a connected arc of G=0G=0 from i to 0. Finally, every point on that arc is attained as follows. Let λ=a+ibλ=a+ib with b>0b>0, 0≤a≤10≤ a≤ 1, a+b≤1a+b≤ 1, and G(a,b)=0G(a,b)=0. By Lemma 8 (since G=0≤0G=0≤ 0), we are in the tight regime 3m+M>2π3m+M>2π, so Lemma 7 applies and the maximizer is (U,m,m,m)(U,m,m,m) with U=2π−3m∈[m,M)U=2π-3m∈[m,M). By (31), G=0G=0 implies maxΨ=0 _P =0, hence Ψ(U,m,m,m)=0 (U,m,m,m)=0. By Lemma 5 this produces t1=t(U)∈(0,1]t_1=t(U)∈(0,1] and t2=t3=t4=t(m)=1t_2=t_3=t_4=t(m)=1, satisfying (2). Therefore the corresponding parameters α=1−t(U)∈[0,1),β=γ=δ=0α=1-t(U)∈[0,1), β=γ=δ=0 yield exactly AL(α)A_L(α) and give λ as an eigenvalue. This identifies CLC_L and proves it is attained. 9) Existence for every strict interior point. Fix λ=a+ibλ=a+ib with b>0b>0 such that 0≤a≤1,a+b<1,G(a,b)>0.0≤ a≤ 1, a+b<1, G(a,b)>0. (36) Define z,m,M,F,,Ψz,m,M,F,P, as above. Then ≠∅P≠ by (10)–(11). We show 0∈Ψ()0∈ (P). By Jensen (13), Ψ(π/2,π/2,π/2,π/2)=4F(π/2)=4log(b1−a)<0 (π/2,π/2,π/2,π/2)=4F(π/2)=4 \! ( b1-a )<0 since a+b<1⇔b<1−a+b<1 b<1-a. To produce a positive value: if 3m+M≤2π3m+M≤ 2π, Lemma 6 gives supΨ=+∞ _P =+∞, so Ψ takes positive values. If 3m+M>2π3m+M>2π, Lemma 7 gives maxΨ=3F(m)+F(U) _P =3F(m)+F(U); by (31), G(a,b)>0G(a,b)>0 implies maxΨ>0 _P >0, so Ψ takes positive values. Since P is connected and Ψ is continuous, Ψ() (P) is an interval containing both negative and positive values, hence contains 0. Thus there exists (u1,…,u4)∈(u_1,…,u_4) with Ψ(u1,…,u4)=0 (u_1,…,u_4)=0. By Lemma 5 this yields tk=t(uk)∈(0,1]t_k=t(u_k)∈(0,1] satisfying (2). Set α=1−t1,β=1−t2,γ=1−t3,δ=1−t4∈[0,1),α=1-t_1, β=1-t_2, γ=1-t_3, δ=1-t_4∈[0,1), and then (1) holds, so λ∈σ(A(α,β,γ,δ))λ∈σ(A(α,β,γ,δ)). Conjugation gives the lower half-plane. ∎ Appendix C Third version Throughout, fix A(α,β,γ,δ)=(α1−α000β1−β000γ1−γ1−δ00δ),α,β,γ,δ∈[0,1).A(α,β,γ,δ)= pmatrixα&1-α&0&0\\ 0&β&1-β&0\\ 0&0&γ&1-γ\\ 1-δ&0&0&δ pmatrix, α,β,γ,δ∈[0,1). This is row-stochastic (row sums are 11) and real. We write λ=a+ibλ=a+ib, and b+=|b|b_+=|b|. Theorem (Spectral region for a 4-cycle row-stochastic matrix) Let λ=a+ib∈σ(A(α,β,γ,δ))λ=a+ib∈σ(A(α,β,γ,δ)) with b≠0b≠ 0. Then: 1. 0≤a≤10≤ a≤ 1. 2. a+b+≤1a+b_+≤ 1. 3. Define for b≥0b≥ 0: G(a,b):=(b2+a2+a)2+2a2−b2.G(a,b):=(b^2+a^2+a)^2+2a^2-b^2. Then G(a,b+)≥0G(a,b_+)≥ 0. Conversely, if b>0b>0 and 0≤a≤1,a+b<1,G(a,b)>0,0≤ a≤ 1, a+b<1, G(a,b)>0, then there exist α,β,γ,δ∈[0,1)α,β,γ,δ∈[0,1) such that λ∈σ(A(α,β,γ,δ))λ∈σ(A(α,β,γ,δ)). (The lower half-plane follows by conjugation.) Moreover: • The segment CR:λ=1−x+ix(x∈[0,1])CR:\ λ=1-x+ix\ (x∈[0,1]) is attained by α=β=γ=δ=1−xα=β=γ=δ=1-x. • The curve portion CLCL in the upper half-plane joining i to 0 defined by G(a,b)=0G(a,b)=0 is attained by AL(α)=(α1−α00001000011000),α∈[0,1).A_L(α)= pmatrixα&1-α&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0 pmatrix, α∈[0,1). • Every point strictly inside b>0, 0≤a≤1,a+b<1,G(a,b)>0\b>0,\ 0≤ a≤ 1,\ a+b<1,\ G(a,b)>0\ is attained. Proof Because A is real, λ∈σ(A)⇒λ¯∈σ(A)λ∈σ(A) λ∈σ(A). So it suffices to prove all necessary conditions for b>0b>0, and then replace b by |b||b| for the final statements. The constructive (“converse”) part will also be done for b>0b>0. 1) Eigenvalue equation ⇒ multiplicative constraint Let v=(v1,v2,v3,v4)⊤≠0v=(v_1,v_2,v_3,v_4) ≠ 0 satisfy Av=λvAv=λ v. Writing the eigenvalue equation row-by-row gives: αv1+(1−α)v2 α v_1+(1-α)v_2 =λv1, =λ v_1, βv2+(1−β)v3 β v_2+(1-β)v_3 =λv2, =λ v_2, γv3+(1−γ)v4 γ v_3+(1-γ)v_4 =λv3, =λ v_3, (1−δ)v1+δv4 (1-δ)v_1+δ v_4 =λv4. =λ v_4. Rearrange each: (λ−α)v1 (λ-α)v_1 =(1−α)v2, =(1-α)v_2, (λ−β)v2 (λ-β)v_2 =(1−β)v3, =(1-β)v_3, (λ−γ)v3 (λ-γ)v_3 =(1−γ)v4, =(1-γ)v_4, (λ−δ)v4 (λ-δ)v_4 =(1−δ)v1. =(1-δ)v_1. Claim: none of v1,v2,v3,v4v_1,v_2,v_3,v_4 is zero. Because α,β,γ,δ∈[0,1)α,β,γ,δ∈[0,1), we have 1−α,1−β,1−γ,1−δ>01-α,1-β,1-γ,1-δ>0. If v1=0v_1=0, then from the first equation (1−α)v2=0⇒v2=0(1-α)v_2=0 v_2=0. Then second gives v3=0v_3=0, third gives v4=0v_4=0, contradicting v≠0v≠ 0. The same cyclically holds for any vk=0v_k=0. Hence v1v2v3v4≠0v_1v_2v_3v_4≠ 0. Now multiply the four equations in (1.1). The left-hand side is (λ−α)(λ−β)(λ−γ)(λ−δ)v1v2v3v4,(λ-α)(λ-β)(λ-γ)(λ-δ)\,v_1v_2v_3v_4, and the right-hand side is (1−α)(1−β)(1−γ)(1−δ)v1v2v3v4.(1-α)(1-β)(1-γ)(1-δ)\,v_1v_2v_3v_4. Cancel v1v2v3v4≠0v_1v_2v_3v_4≠ 0 to obtain the necessary condition: (λ−α)(λ−β)(λ−γ)(λ−δ)=(1−α)(1−β)(1−γ)(1−δ).(λ-α)(λ-β)(λ-γ)(λ-δ)=(1-α)(1-β)(1-γ)(1-δ). Conversely, if (1.2) holds and additionally λ≠α,β,γ,δλ≠α,β,γ,δ, one can solve the relations in (1.1) recursively for v and check the final equation closes exactly when (1.2) holds. In our nonreal case λ∉ℝλ , while α,β,γ,δ∈ℝα,β,γ,δ , so automatically λ≠α,β,γ,δλ≠α,β,γ,δ. Thus (1.2) is equivalent to existence of a nonzero eigenvector for nonreal λ. 2) Substitute tk=1−parametert_k=1-parameter and shift z=λ−1z=λ-1 Define t1=1−α,t2=1−β,t3=1−γ,t4=1−δ.t_1=1-α,\ t_2=1-β,\ t_3=1-γ,\ t_4=1-δ. Since each parameter lies in [0,1)[0,1), each tk∈(0,1]t_k∈(0,1]. Let z=λ−1z=λ-1. Then λ−(1−tk)=λ−1+tk=z+tk.λ-(1-t_k)=λ-1+t_k=z+t_k. Also (1−α)(1−β)(1−γ)(1−δ)=t1t2t3t4.(1-α)(1-β)(1-γ)(1-δ)=t_1t_2t_3t_4. So (1.2) becomes: (z+t1)(z+t2)(z+t3)(z+t4)=t1t2t3t4.(z+t_1)(z+t_2)(z+t_3)(z+t_4)=t_1t_2t_3t_4. Write z=x+iyz=x+iy with x=a−1x=a-1 and y=by=b. We assume b>0b>0 in the rest of the necessity direction; later we replace by |b||b|. 3) Argument parametrization: t↔u=arg(z+t)t u= (z+t) Fix z=x+iyz=x+iy with y>0y>0. For any t>0t>0, the complex number z+t=(x+t)+iyz+t=(x+t)+iy lies in the upper half-plane, so its argument u(t):=arg(z+t)u(t):= (z+t) is well-defined in (0,π)(0,π). 3.1 Inverse formulas t=t(u)t=t(u), |z+t(u)||z+t(u)| Let u∈(0,π)u∈(0,π). Consider z+t=(x+t)+iyz+t=(x+t)+iy with argument u. Then tanu=ℑ(z+t)ℜ(z+t)=yx+t. u= (z+t) (z+t)= yx+t. Since y>0y>0, this determines x+tx+t uniquely: x+t=ycotu⇒t=ycotu−x.x+t=y u t=y u-x. Also |z+t|2=(x+t)2+y2=(ycotu)2+y2=y2(cot2u+1)=y2csc2u,|z+t|^2=(x+t)^2+y^2=(y u)^2+y^2=y^2( ^2u+1)=y^2 ^2u, So |z+t(u)|=ycscu.|z+t(u)|=y u. (3.2) 3.2 Endpoints m=arg(z+1)m= (z+1), M=arg(z)M= (z), and monotone correspondence Define m:=arg(z+1)=arg(λ),M:=arg(z)=arg(λ−1).m:= (z+1)= (λ), M:= (z)= (λ-1). Claim: u(t)u(t) is strictly decreasing in t>0t>0. Indeed, u(t)=arctan(yx+t)u(t)= ( yx+t ), and for y>0y>0 the function yx+t yx+t is strictly decreasing in t, hence arctan of it is strictly decreasing. Therefore as t↓0t 0, u(t)↑u(0)=arg(z)=Mu(t) u(0)= (z)=M; and as t↑1t 1, u(t)↓u(1)=arg(z+1)=mu(t) u(1)= (z+1)=m. Hence the mapping t↦u(t)t u(t) is a continuous strictly decreasing bijection from (0,1](0,1] onto [m,M)[m,M). Equivalently: t∈(0,1]⟺u∈[m,M).t∈(0,1] u∈[m,M). 4) The function F and reformulation in (uk)(u_k) Define for u∈[m,M)u∈[m,M): F(u):=log|z+t(u)|−logt(u).F(u):= |z+t(u)|- t(u). Using (3.1)–(3.2) this becomes explicitly F(u)=log(ycscu)−log(ycotu−x).F(u)= (y u)- (y u-x). Note t(u)=ycotu−x>0t(u)=y u-x>0 on [m,M)[m,M) by construction. 5) Lemma 1 (exact equivalence of eigenvalue condition) Lemma 9. A nonreal λ satisfies (2.1) for some t1,t2,t3,t4∈(0,1]t_1,t_2,t_3,t_4∈(0,1] if and only if there exist u1,u2,u3,u4∈[m,M)u_1,u_2,u_3,u_4∈[m,M) such that u1+u2+u3+u4=2π,u_1+u_2+u_3+u_4=2π, F(u1)+F(u2)+F(u3)+F(u4)=0.F(u_1)+F(u_2)+F(u_3)+F(u_4)=0. Proof. (⇒ ) Suppose (2.1) holds with tk∈(0,1]t_k∈(0,1]. For each k, let uk=arg(z+tk)∈(0,π)u_k= (z+t_k)∈(0,π). Because tk∈(0,1]t_k∈(0,1], by (3.4) we have uk∈[m,M)u_k∈[m,M). Write z+tk=|z+tk|eiukz+t_k=|z+t_k|e^iu_k. Then the left side of (2.1) has argument u1+⋯+u4u_1+·s+u_4 (mod 2π2π), while the right side t1t2t3t4>0t_1t_2t_3t_4>0 has argument 0 (mod 2π2π). Thus u1+⋯+u4≡0(mod2π)u_1+·s+u_4≡ 0 2π. But each uk∈(0,π)u_k∈(0,π), so the sum lies strictly in (0,4π)(0,4π). The only multiple of 2π2π in (0,4π)(0,4π) is 2π2π. Hence u1+⋯+u4=2πu_1+·s+u_4=2π, proving (5.1). Taking moduli in (2.1): |z+t1|⋯|z+t4|=t1t2t3t4|z+t_1|·s|z+t_4|=t_1t_2t_3t_4. Take log of both sides: ∑k=14log|z+tk|=∑k=14logtk⇔∑k=14(log|z+tk|−logtk)=0. _k=1^4 |z+t_k|= _k=1^4 t_k _k=1^4 ( |z+t_k|- t_k )=0. But log|z+tk|−logtk=F(uk) |z+t_k|- t_k=F(u_k) by definition, so (5.2) holds. (⇐ ) Conversely, suppose uk∈[m,M)u_k∈[m,M) satisfy (5.1)–(5.2). Define tk=t(uk)t_k=t(u_k). Then by (3.4), tk∈(0,1]t_k∈(0,1]. Also z+tk=|z+tk|eiukz+t_k=|z+t_k|e^iu_k. Multiply: ∏k=14(z+tk)=(∏k=14|z+tk|)exp(i∑k=14uk)=(∏k=14|z+tk|)ei2π=∏k=14|z+tk|. _k=1^4(z+t_k)= ( _k=1^4|z+t_k| ) (i _k=1^4u_k )= ( _k=1^4|z+t_k| )e^i2π= _k=1^4|z+t_k|. So the product ∏(z+tk)Π(z+t_k) is positive real, with modulus ∏|z+tk|Π|z+t_k|. Equation (5.2) implies ∑log|z+tk|=∑logtkΣ |z+t_k|=Σ t_k, hence ∏|z+tk|=∏tkΠ|z+t_k|=Π t_k. Therefore ∏k=14(z+tk)=∏k=14tk _k=1^4(z+t_k)= _k=1^4t_k, which is exactly (2.1). ∎ 6) The feasible set P, its convexity/connectedness, and Ψ(P) (P) is an interval Let P:=(u1,u2,u3,u4)∈[m,M)4:u1+u2+u3+u4=2π,P:= \(u_1,u_2,u_3,u_4)∈[m,M)^4:\ u_1+u_2+u_3+u_4=2π \, and Ψ(u1,u2,u3,u4):=∑k=14F(uk) (u_1,u_2,u_3,u_4):= _k=1^4F(u_k). 6.1 P is convex and connected The set [m,M)4⊂ℝ4[m,M)^4 ^4 is convex. The affine hyperplane H:=u∈ℝ4:∑uk=2πH:=\u ^4:Σ u_k=2π\ is convex. An intersection of convex sets is convex, so P=[m,M)4∩HP=[m,M)^4∩ H is convex. Any convex subset of ℝnR^n is path-connected (connect points by line segment), hence connected. 6.2 Ψ(P)⊂ℝ (P) is an interval Ψ is continuous because F is continuous on [m,M)[m,M) (all terms in (4.1) are continuous there since t(u)>0t(u)>0). Continuous image of a connected set in ℝR is connected; connected subsets of ℝR are intervals. Hence Ψ(P) (P) is an interval. By Lemma 1: λnonreal eigenvalue of some A(α,β,γ,δ)⇔P≠∅ and 0∈Ψ(P).λ\ nonreal eigenvalue of some A(α,β,γ,δ) P≠ \ and \ 0∈ (P). 7) Necessarily 0≤a≤10≤ a≤ 1 (and in fact a<1a<1 when b≠0b≠ 0) Assume b>0b>0 and P≠∅P≠ . Pick (u1,…,u4)∈P(u_1,…,u_4)∈ P. Then their average is u1+⋯+u44=2π4=π2 u_1+·s+u_44= 2π4= π2. Since each uk∈[m,M)u_k∈[m,M), the interval [m,M)[m,M) must contain π/2π/2. That is: m≤π2<M.m≤ π2<M. Now compute what m≤π/2m≤π/2 means. Recall m=arg(λ)=arg(a+ib),b>0m= (λ)= (a+ib),\ b>0. For b>0b>0, arg(a+ib)≤π/2 (a+ib)≤π/2 holds exactly when a≥0a≥ 0 (points in upper half-plane with angle at most π/2π/2 are those with nonnegative real part). Hence: m≤π2⇔a≥0.m≤ π2 a≥ 0. Next, M>π/2M>π/2 where M=arg(z)=arg((a−1)+ib)M= (z)= ((a-1)+ib). Since b>0b>0, we have arg((a−1)+ib)>π/2 ((a-1)+ib)>π/2 exactly when a−1<0a-1<0, i.e., a<1a<1. Thus: M>π2⇔a<1.M> π2 a<1. Combining (7.1)–(7.3), P≠∅⇒0≤a<1P≠ 0≤ a<1. In particular 0≤a≤10≤ a≤ 1. This proves item (1). 8) Strict convexity of F We show F′(u)>0F (u)>0 on (0,π)(0,π), hence F is strictly convex on [m,M)⊂(0,π)[m,M)⊂(0,π). Recall F(u)=log(ycscu)−log(ycotu−x),y>0F(u)= (y u)- (y u-x),\ y>0. Differentiate. First term: dulog(ycscu)=du(logy+logcscu)=dulogcscu. ddu (y u)= ddu( y+ u)= ddu u. Since ducscu=−cscucotu ddu u=- u u, we get dulogcscu=−cscucotucscu=−cotu ddu u= - u u u=- u. So dulog(ycscu)=−cotu. ddu (y u)=- u. Second term: define g(u)=ycotu−xg(u)=y u-x. Then dulogg(u)=g′(u)g(u) ddu g(u)= g (u)g(u). Since ducotu=−csc2u ddu u=- ^2u, we have g′(u)=−ycsc2ug (u)=-y ^2u. Therefore dulog(ycotu−x)=−ycsc2uycotu−x. ddu (y u-x)= -y ^2uy u-x. Thus F′(u)=−cotu+ycsc2uycotu−x.F (u)=- u+ y ^2uy u-x. (8.3) Differentiate again. Write h(u)=ycsc2uycotu−xh(u)= y ^2uy u-x. Then F′(u)=csc2u+h′(u)F (u)= ^2u+h (u) (8.4) because du(−cotu)=csc2u ddu(- u)= ^2u. Compute h′(u)h (u) via quotient rule. Let p(u)=ycsc2u,q(u)=ycotu−xp(u)=y ^2u,\ q(u)=y u-x. Then h=p/qh=p/q, so h′=p′q−pq′q2h = p q-pq q^2. Compute p′(u)=−2ycsc2ucotup (u)=-2y ^2u u. Also q′(u)=−ycsc2uq (u)=-y ^2u. h′=(−2ycsc2ucotu)(ycotu−x)−(ycsc2u)(−ycsc2u)(ycotu−x)2.h = (-2y ^2u u)(y u-x)-(y ^2u)(-y ^2u)(y u-x)^2. Factor ycsc2uy ^2u: h′=ycsc2u[−2cotu(ycotu−x)+ycsc2u](ycotu−x)2h = y ^2u [-2 u(y u-x)+y ^2u ](y u-x)^2. Use csc2u=1+cot2u ^2u=1+ ^2u: −2cotu(ycotu−x)+y(1+cot2u)=y−ycot2u+2xcotu.-2 u(y u-x)+y(1+ ^2u)=y-y ^2u+2x u. So h′=ycsc2u(y−ycot2u+2xcotu)(ycotu−x)2.h = y ^2u(y-y ^2u+2x u)(y u-x)^2. (8.5) Now combine with F′=csc2u+h′F = ^2u+h . Put over common denominator: F′(u)=csc2u(1+y(y−ycot2u+2xcotu)(ycotu−x)2)=csc2u⋅(ycotu−x)2+y(y−ycot2u+2xcotu)(ycotu−x)2.F (u)= ^2u (1+ y (y-y ^2u+2x u )(y u-x)^2 )= ^2u· (y u-x)^2+y(y-y ^2u+2x u)(y u-x)^2. Expand the numerator: (ycotu−x)2=y2cot2u−2xycotu+x2(y u-x)^2=y^2 ^2u-2xy u+x^2. And y(y−ycot2u+2xcotu)=y2−y2cot2u+2xycotuy(y-y ^2u+2x u)=y^2-y^2 ^2u+2xy u. Sum: (y2cot2u−2xycotu+x2)+(y2−y2cot2u+2xycotu)=x2+y2(y^2 ^2u-2xy u+x^2)+(y^2-y^2 ^2u+2xy u)=x^2+y^2. Therefore F′(u)=csc2u⋅x2+y2(ycotu−x)2=x2+y2(ycotu−x)2sin2u.F (u)= ^2u· x^2+y^2(y u-x)^2= x^2+y^2(y u-x)^2 ^2u. Since x2+y2>0x^2+y^2>0 (because y>0y>0) and the denominator is positive on (0,π)(0,π), we conclude F′(u)>0F (u)>0 for all u∈(0,π)u∈(0,π). So F is strictly convex. 9) Right boundary: necessity of a+b≤1a+b≤ 1, and attainment of CRCR Assume b>0b>0, P≠∅P≠ , and 0∈Ψ(P)0∈ (P). 9.1 Jensen inequality step (fully justified) Let (u1,…,u4)∈P(u_1,…,u_4)∈ P. Since F is convex on [m,M)[m,M), Jensen’s inequality gives 14∑F(uk)≥F(14∑uk) 14Σ F(u_k)≥ F ( 14Σ u_k ). Multiply by 4: ∑k=14F(uk)≥4F(2π4)=4F(π2). _k=1^4F(u_k)≥ 4F ( 2π4 )=4F ( π2 ). (9.1) If 0∈Ψ(P), then infPΨ≤0. But (9.1) shows Ψ≥4F(π/2) on P. Therefore 4F(π/2)≤0, i.e., F(π/2)≤0.If 0∈ (P), then _P ≤ 0. But (9.1) shows ≥ 4F(π/2) on P. Therefore 4F(π/2)≤ 0, i.e., F(π/2)≤ 0. 9.2 Compute F(π/2)F(π/2) and deduce a+b≤1a+b≤ 1 Using (4.1): F(π/2)=log(ycsc(π/2))−log(ycot(π/2)−x)=log(y)−log(−x)F(π/2)= (y (π/2))- (y (π/2)-x)= (y)- (-x). Since y=b,x=a−1y=b,x=a-1, so −x=1−a-x=1-a. Hence F(π/2)=log(b1−a)F(π/2)= ( b1-a ). So F(π/2)≤0F(π/2)≤ 0 is equivalent to b1−a≤1⇔b≤1−a⇔a+b≤1 b1-a≤ 1 b≤ 1-a a+b≤ 1. This proves item (2) for b>0b>0. By conjugation symmetry, for b<0b<0 it becomes a+|b|≤1a+|b|≤ 1. 9.3 Attainment of segment CRCR Let x∈(0,1]x∈(0,1] and choose α=β=γ=δ=1−x∈[0,1)α=β=γ=δ=1-x∈[0,1). Then (2.1) reads (z+x)4=x4(z+x)^4=x^4. So z+x=xeiπ2kz+x=xe^i π2k. The nonreal solutions correspond to k=1,3k=1,3: z+x=±ix⇒λ=1−x±ixz+x=± ix λ=1-x± ix. Thus the segment λ=1−x+ixλ=1-x+ix for x∈[0,1]x∈[0,1] is attained. 10) Two regimes for supPΨ _P : Lemmas 2 and 3 proved in full Lemma 10 (Unbounded supremum). If 3m+M≤2π3m+M≤ 2π, then supPΨ=+∞ _P =+∞. Proof. Fix u4∈[m,M)u_4∈[m,M). Define u1=u2=u3=2π−u43u_1=u_2=u_3= 2π-u_43. For u4u_4 sufficiently close to M, (u1,u2,u3,u4)∈P(u_1,u_2,u_3,u_4)∈ P. As u4↑Mu_4 M, t(u4)↓0t(u_4) 0, so F(u4)→+∞F(u_4)→+∞. Meanwhile u1→(2π−M)/3∈(m,M)u_1→(2π-M)/3∈(m,M), so F(u1)F(u_1) is bounded. Thus Ψ→+∞ →+∞. ∎ Lemma 11 (Finite maximum in the tight regime). If 3m+M>2π3m+M>2π and we define U:=2π−3mU:=2π-3m, then U<MU<M, and maxPΨ=3F(m)+F(U), _P =3F(m)+F(U), with equality iff (u1,u2,u3,u4)(u_1,u_2,u_3,u_4) is a permutation of (U,m,m,m)(U,m,m,m). Proof. 3m+M>2π⇒2π−3m<M3m+M>2π 2π-3m<M. So U∈[m,M)U∈[m,M). By symmetry, we rearrange (uk)(u_k) to nonincreasing v1≥v2≥v3≥v4v_1≥ v_2≥ v_3≥ v_4. Then v1≤2π−3m=Uv_1≤ 2π-3m=U. Every feasible ordered vector lies in a compact set majorized by x=(U,m,m,m)x=(U,m,m,m). By Karamata’s inequality and strict convexity of F, the sum is maximized at x. Thus maxPΨ=3F(m)+F(U) _P =3F(m)+F(U). ∎ 11) Necessity of G(a,b)≥0G(a,b)≥ 0 11.1 A geometry/algebra lemma: G≤0⇒3m+M>2πG≤ 0 3m+M>2π (Lemma 4) Lemma 12. Assume b>0b>0 and 0≤a≤10≤ a≤ 1. If G(a,b)≤0G(a,b)≤ 0, then 3m+M>2π3m+M>2π. Proof. Let s=b2s=b^2. G(a,b)=s2+s(2a2+2a−1)+(a2+a)2+2a2G(a,b)=s^2+s(2a^2+2a-1)+(a^2+a)^2+2a^2 (11.1). G≤0G≤ 0 implies s lies between roots of the quadratic. The discriminant Δ=−(2a+1)(6a−1) =-(2a+1)(6a-1). So Δ>0 >0 iff a<1/6a<1/6. Case a=0a=0: 3m+M=3(π/2)+π−arctan(b)>2π3m+M=3(π/2)+π- (b)>2π. Case a>0a>0: We show b2>3a2b^2>3a^2, which implies m>π/3m>π/3. Let tanm=b/a m=b/a. We want to show tan(3m)>b1−a (3m)> b1-a. Using triple-angle identity and substituting t=b/at=b/a, we find the sign of the difference depends on N(a,b)=4a3−3a2−4ab2+b2N(a,b)=4a^3-3a^2-4ab^2+b^2. Using s≥s−(a)s≥ s_-(a), algebraic expansion shows s−(a)>s0(a)s_-(a)>s_0(a) where s0s_0 is the root of N. Hence N(a,b)>0N(a,b)>0, which confirms 3m+M>2π3m+M>2π. ∎ 11.2 Tight regime: compute maxPΨ _P explicitly and relate it to G In the tight regime, maxPΨ=log(|λ|3bcscUbcotU+1−a) _P = ( |λ|^3\,b Ub U+1-a ) (11.25). Substituting trigonometric identities for cscU,cotU U, U in terms of a,ba,b yields |λ|6≥a(a2−3b2)+(1−a)(b2−3a2)|λ|^6≥ a(a^2-3b^2)+(1-a)(b^2-3a^2). This factors as |λ−1|2G(a,b)≥0|λ-1|^2G(a,b)≥ 0. Thus, maxPΨ≥0⇔G(a,b)≥0 _P ≥ 0 G(a,b)≥ 0. (11.39). 11.3 Finish necessity of G(a,b)≥0G(a,b)≥ 0 If λ is a nonreal eigenvalue, then supPΨ≥0 _P ≥ 0. If 3m+M≤2π3m+M≤ 2π, Lemma 4 contraposition implies G(a,b)>0G(a,b)>0. If 3m+M>2π3m+M>2π, the tight regime applies and G(a,b)≥0G(a,b)≥ 0. This proves item (3). 12) Left boundary CLCL: attainment by AL(α)A_L(α) and equivalence to G=0G=0 For AL(α)A_L(α), the characteristic equation is λ3(λ−α)=1−αλ^3(λ-α)=1-α. For nonreal λ, α=λ4−1λ3−1α= λ^4-1λ^3-1 (12.2). The imaginary part is ℑ(pq)=b|λ−1|2G(a,b)|λ3−1|2 ( pq)= b|λ-1|^2G(a,b)|λ^3-1|^2. Thus ℑ(α)=0⇔G(a,b)=0 (α)=0 G(a,b)=0. Factoring the equation as (λ−1)[λ3+(1−α)(λ2+λ+1)]=0(λ-1)[λ^3+(1-α)(λ^2+λ+1)]=0, one finds exactly one real root and a conjugate pair for α∈[0,1)α∈[0,1). At α=0,λ=iα=0,λ=i. As α→1,λ→0α→ 1,λ→ 0. 13) Every strict interior point is attained (constructive existence) Fix λ satisfying 0≤a≤1,a+b<1,G(a,b)>00≤ a≤ 1,a+b<1,G(a,b)>0. P is non-empty as π/2∈[m,M)π/2∈[m,M). Ψ(π/2,…,π/2)=4log(b1−a)<0 (π/2,…,π/2)=4 ( b1-a)<0 since a+b<1a+b<1. Conversely, supPΨ>0 _P >0 because G(a,b)>0G(a,b)>0 (using either regime). By the Intermediate Value Theorem on the convex set P, there exists a point where Ψ=0 =0, which corresponds to parameters α,β,γ,δ∈[0,1)α,β,γ,δ∈[0,1). Summary For any nonreal eigenvalue λ=a+ibλ=a+ib of some 4-cycle row-stochastic matrix, we have proved 0≤a≤10≤ a≤ 1, a+|b|≤1a+|b|≤ 1, and G(a,|b|)≥0G(a,|b|)≥ 0. Conversely, strict interior points are realized. The boundaries CRCR and CLCL are attained by specific matrix families. Appendix D Final version Theorem 23 (Spectral region for a 4-cycle row-stochastic matrix). Fix A(α,β,γ,δ)=(α1−α000β1−β000γ1−γ1−δ00δ),α,β,γ,δ∈[0,1).A(α,β,γ,δ)= pmatrixα&1-α&0&0\\ 0&β&1-β&0\\ 0&0&γ&1-γ\\ 1-δ&0&0&δ pmatrix, α,β,γ,δ∈[0,1). Let λ=a+ib∈σ(A(α,β,γ,δ))λ=a+ib∈σ(A(α,β,γ,δ)) with b≠0b≠ 0, and write b+=|b|b_+=|b|. Then: 1. 0≤a≤10≤ a≤ 1. 2. a+b+≤1a+b_+≤ 1. 3. For b≥0b≥ 0 define G(a,b):=(b2+a2+a)2+2a2−b2.G(a,b):=(b^2+a^2+a)^2+2a^2-b^2. Then G(a,b+)≥0G(a,b_+)≥ 0. Conversely, if b>0b>0 and 0≤a≤1,a+b<1,G(a,b)>0,0≤ a≤ 1, a+b<1, G(a,b)>0, then there exist α,β,γ,δ∈[0,1)α,β,γ,δ∈[0,1) such that λ∈σ(A(α,β,γ,δ))λ∈σ(A(α,β,γ,δ)). (The lower half-plane follows by conjugation.) Moreover: • The segment CR:λ=1−x+ix(x∈[0,1])CR:\ λ=1-x+ix\ (x∈[0,1]) is attained by α=β=γ=δ=1−xα=β=γ=δ=1-x. • The curve portion CLCL in the upper half-plane joining i to 0 defined by G(a,b)=0G(a,b)=0 is attained by the family AL(α)=(α1−α00001000011000),α∈[0,1).A_L(α)= pmatrixα&1-α&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0 pmatrix, α∈[0,1). • Every point strictly inside b>0, 0≤a≤1,a+b<1,G(a,b)>0\b>0,\ 0≤ a≤ 1,\ a+b<1,\ G(a,b)>0\ is attained. Proof. Because A is real, λ∈σ(A)⇒λ¯∈σ(A)λ∈σ(A) λ∈σ(A). Thus it suffices to prove the necessary conditions for b>0b>0 and then replace b by |b||b| in the final statements. The constructive (converse) direction will also be proved for b>0b>0. 1) Eigenvalue equation ⇒ a multiplicative constraint. Let v=(v1,v2,v3,v4)⊤≠0v=(v_1,v_2,v_3,v_4) ≠ 0 satisfy Av=λvAv=λ v. Writing the eigenvalue equation row-by-row: αv1+(1−α)v2 α v_1+(1-α)v_2 =λv1, =λ v_1, βv2+(1−β)v3 β v_2+(1-β)v_3 =λv2, =λ v_2, γv3+(1−γ)v4 γ v_3+(1-γ)v_4 =λv3, =λ v_3, (1−δ)v1+δv4 (1-δ)v_1+δ v_4 =λv4. =λ v_4. Rearrange: (λ−α)v1=(1−α)v2,(λ−β)v2=(1−β)v3,(λ−γ)v3=(1−γ)v4,(λ−δ)v4=(1−δ)v1.(λ-α)v_1=(1-α)v_2, (λ-β)v_2=(1-β)v_3, (λ-γ)v_3=(1-γ)v_4, (λ-δ)v_4=(1-δ)v_1. (1) Claim 1. If λ∉ℝλ and α,β,γ,δ∈[0,1)α,β,γ,δ∈[0,1) then v1v2v3v4≠0v_1v_2v_3v_4≠ 0 for every eigenvector v≠0v≠ 0. Proof. Because α,β,γ,δ∈[0,1)α,β,γ,δ∈[0,1) we have 1−α,1−β,1−γ,1−δ>01-α,1-β,1-γ,1-δ>0. If v1=0v_1=0 then the first equation in (1) gives (1−α)v2=0(1-α)v_2=0 hence v2=0v_2=0. Then the second gives v3=0v_3=0, the third gives v4=0v_4=0, contradicting v≠0v≠ 0. Cyclically the same holds for any coordinate, so no vkv_k can vanish. ∎ Multiply the four equalities in (1) and cancel v1v2v3v4≠0v_1v_2v_3v_4≠ 0 (Claim 1): (λ−α)(λ−β)(λ−γ)(λ−δ)=(1−α)(1−β)(1−γ)(1−δ).(λ-α)(λ-β)(λ-γ)(λ-δ)=(1-α)(1-β)(1-γ)(1-δ). (2) Conversely, if λ∉ℝλ and (2) holds, then λ≠α,β,γ,δλ≠α,β,γ,δ (since α,β,γ,δ∈ℝα,β,γ,δ ), and one can solve (1) recursively: choose v1≠0v_1≠ 0, set v2=λ−α1−αv1v_2= λ-α1-αv_1, then v3=λ−β1−βv2v_3= λ-β1-βv_2, v4=λ−γ1−γv3v_4= λ-γ1-γv_3; the final equation closes exactly when (2) holds. Thus for nonreal λ, (2) is equivalent to λ∈σ(A)λ∈σ(A) for some choice of parameters. 2) Substitute tk=1−parametert_k=1-parameter and shift z=λ−1z=λ-1. Define t1=1−α,t2=1−β,t3=1−γ,t4=1−δ.t_1=1-α, t_2=1-β, t_3=1-γ, t_4=1-δ. Then each tk∈(0,1]t_k∈(0,1]. Set z=λ−1z=λ-1. Then λ−(1−tk)=z+tk,(1−α)(1−β)(1−γ)(1−δ)=t1t2t3t4.λ-(1-t_k)=z+t_k, (1-α)(1-β)(1-γ)(1-δ)=t_1t_2t_3t_4. So (2) becomes (z+t1)(z+t2)(z+t3)(z+t4)=t1t2t3t4.(z+t_1)(z+t_2)(z+t_3)(z+t_4)=t_1t_2t_3t_4. (3) Write z=x+iyz=x+iy with x=Re(z)=a−1,y=Im(z)=b.x=Re(z)=a-1, y=Im(z)=b. In the necessity direction we assume b=y>0b=y>0. 3) Argument parametrization: t↔u=Arg(z+t)t u=Arg(z+t). Fix z=x+iyz=x+iy with y>0y>0. For any t>0t>0, the point z+t=(x+t)+iyz+t=(x+t)+iy lies in the open upper half-plane, so its argument u(t)=Arg(z+t)u(t)=Arg(z+t) is well-defined and belongs to (0,π)(0,π). 3.1) Inverse formulas. Given u∈(0,π)u∈(0,π) and z+t=(x+t)+iyz+t=(x+t)+iy with Arg(z+t)=uArg(z+t)=u, we have tanu=yx+t⟹x+t=ycotu⟹t=t(u):=ycotu−x. u= yx+t x+t=ycotu t=t(u):=ycotu-x. Also |z+t|2=(x+t)2+y2=(ycotu)2+y2=y2(cot2u+1)=y2csc2u,|z+t|^2=(x+t)^2+y^2=(ycotu)^2+y^2=y^2(cot^2u+1)=y^2 ^2u, hence |z+t(u)|=ycscu.|z+t(u)|=y u. 3.2) Endpoints and monotonicity. Define m:=Arg(z+1)=Arg(λ),M:=Arg(z)=Arg(λ−1).m:=Arg(z+1)=Arg(λ), M:=Arg(z)=Arg(λ-1). Because u(t)=arctan(yx+t)u(t)= \! ( yx+t ) and yx+t yx+t is strictly decreasing in t, the map t↦u(t)t u(t) is continuous and strictly decreasing on (0,∞)(0,∞). In particular, on t∈(0,1]t∈(0,1] it is a continuous strictly decreasing bijection onto u∈[m,M)u∈[m,M): t∈(0,1]⟺u∈[m,M).t∈(0,1] u∈[m,M). (4) (As t↑1t 1, u(t)↓mu(t) m; as t↓0t 0, u(t)↑Mu(t) M.) 4) The function F and reformulation. For u∈[m,M)u∈[m,M) define F(u):=log|z+t(u)|−logt(u)=log(ycscu)−log(ycotu−x).F(u):= |z+t(u)|- t(u)= (y u)- (ycotu-x). Note t(u)>0t(u)>0 on [m,M)[m,M) by construction. 5) Exact equivalence of eigenvalue condition in terms of (uk)(u_k). Lemma 13 (Exact equivalence). A nonreal λ satisfies (3) for some t1,t2,t3,t4∈(0,1]t_1,t_2,t_3,t_4∈(0,1] if and only if there exist u1,u2,u3,u4∈[m,M)u_1,u_2,u_3,u_4∈[m,M) such that u1+u2+u3+u4 u_1+u_2+u_3+u_4 =2π, =2π, (5) F(u1)+F(u2)+F(u3)+F(u4) F(u_1)+F(u_2)+F(u_3)+F(u_4) =0. =0. (6) Proof. (⇒ ) Assume (3) holds with tk∈(0,1]t_k∈(0,1]. Let uk=Arg(z+tk)∈(0,π)u_k=Arg(z+t_k)∈(0,π). Because tk∈(0,1]t_k∈(0,1], (4) gives uk∈[m,M)u_k∈[m,M). Write z+tk=|z+tk|eiukz+t_k=|z+t_k|e^iu_k. The left side of (3) has argument u1+⋯+u4u_1+·s+u_4 modulo 2π2π; the right side t1t2t3t4>0t_1t_2t_3t_4>0 has argument 0 modulo 2π2π. Hence u1+⋯+u4≡0(mod2π)u_1+·s+u_4≡ 0 2π. Since each uk∈(0,π)u_k∈(0,π), the sum lies in (0,4π)(0,4π), so the only possible multiple of 2π2π is 2π2π, proving (5). Taking moduli in (3) gives ∏|z+tk|=∏tkΠ|z+t_k|=Π t_k. Take logs: ∑k=14log|z+tk|=∑k=14logtk⟺∑k=14(log|z+tk|−logtk)=0. _k=1^4 |z+t_k|= _k=1^4 t_k _k=1^4 ( |z+t_k|- t_k )=0. But by definition F(uk)=log|z+tk|−logtkF(u_k)= |z+t_k|- t_k, so (6) holds. (⇐ ) Conversely, assume uk∈[m,M)u_k∈[m,M) satisfy (5)–(6). Define tk=t(uk)t_k=t(u_k). Then tk∈(0,1]t_k∈(0,1] by (4). Also z+tk=|z+tk|eiukz+t_k=|z+t_k|e^iu_k. Multiply: ∏k=14(z+tk)=(∏k=14|z+tk|)ei∑uk=(∏k=14|z+tk|)ei2π=∏k=14|z+tk|. _k=1^4(z+t_k)= ( _k=1^4|z+t_k| )e^iΣ u_k= ( _k=1^4|z+t_k| )e^i2π= _k=1^4|z+t_k|. Thus ∏(z+tk)Π(z+t_k) is a positive real number whose modulus is ∏|z+tk|Π|z+t_k|. Equation (6) means ∑log|z+tk|=∑logtkΣ |z+t_k|=Σ t_k, hence ∏|z+tk|=∏tkΠ|z+t_k|=Π t_k. Therefore ∏(z+tk)=∏tkΠ(z+t_k)=Π t_k, i.e. (3) holds. ∎ 6) The feasible set P and the function Ψ . Let P:=(u1,u2,u3,u4)∈[m,M)4:u1+u2+u3+u4=2π,Ψ(u1,…,u4):=∑k=14F(uk).P:= \(u_1,u_2,u_3,u_4)∈[m,M)^4:\ u_1+u_2+u_3+u_4=2π \, (u_1,…,u_4):= _k=1^4F(u_k). Then P is convex (intersection of a box with an affine hyperplane), hence path-connected, hence connected. Since F is continuous on [m,M)[m,M), Ψ is continuous on P. The continuous image of a connected set in ℝR is an interval, so Ψ(P)⊂ℝ (P) is an interval. By Lemma 13, λis a nonreal eigenvalue of someA(α,β,γ,δ)⟺P≠∅and 0∈Ψ(P).λ\ is a nonreal eigenvalue of some\ A(α,β,γ,δ) P≠ \ and\ 0∈ (P). (7) 7) Necessarily 0≤a≤10≤ a≤ 1 (indeed a<1a<1 if b≠0b≠ 0). Assume b>0b>0 and P≠∅P≠ . Choose (u1,…,u4)∈P(u_1,…,u_4)∈ P. Then u1+u2+u3+u44=2π4=π2. u_1+u_2+u_3+u_44= 2π4= π2. Since each uk∈[m,M)u_k∈[m,M), the interval [m,M)[m,M) must contain π/2π/2, i.e. m≤π2<M.m≤ π2<M. (8) Now m=Arg(λ)=Arg(a+ib)m=Arg(λ)=Arg(a+ib) with b>0b>0. In the open upper half-plane, Arg(a+ib)≤π/2Arg(a+ib)≤π/2 holds iff a≥0a≥ 0. Thus m≤π/2⇒a≥0m≤π/2 a≥ 0. Also M=Arg(λ−1)=Arg((a−1)+ib)M=Arg(λ-1)=Arg((a-1)+ib). Since b>0b>0, the argument is strictly bigger than π/2π/2 iff Re(λ−1)=a−1<0Re(λ-1)=a-1<0, i.e. a<1a<1. Thus π/2<M⇒a<1π/2<M a<1. Therefore P≠∅⇒0≤a<1P≠ 0≤ a<1, proving item (1) (and the sharper a<1a<1 for b≠0b≠ 0). 8) Strict convexity of F. We show F′(u)>0F (u)>0 on (0,π)(0,π). Recall F(u)=log(ycscu)−log(ycotu−x),y>0.F(u)= (y u)- (ycotu-x), y>0. Differentiate: dulog(ycscu)=dulogcscu=−cotu. ddu (y u)= ddu u=-cotu. Let g(u)=ycotu−xg(u)=ycotu-x. Then g′(u)=−ycsc2ug (u)=-y ^2u and dulog(ycotu−x)=g′(u)g(u)=−ycsc2uycotu−x. ddu (ycotu-x)= g (u)g(u)=- y ^2uycotu-x. Hence F′(u)=−cotu+ycsc2uycotu−x.F (u)=-cotu+ y ^2uycotu-x. A direct computation (as in the user’s draft) yields F′(u)=x2+y2(ycotu−x)2sin2u.F (u)= x^2+y^2(ycotu-x)^2 ^2u. Because y>0y>0 we have x2+y2>0x^2+y^2>0, and the denominator is positive for u∈(0,π)u∈(0,π) (since t(u)=ycotu−x>0t(u)=ycotu-x>0). Thus F′(u)>0F (u)>0 on (0,π)(0,π), so F is strictly convex on [m,M)⊂(0,π)[m,M)⊂(0,π). 9) Right boundary: necessity of a+b≤1a+b≤ 1, and attainment of CRCR. Assume b>0b>0, P≠∅P≠ , and 0∈Ψ(P)0∈ (P). 9.1) Jensen step. For any (u1,…,u4)∈P(u_1,…,u_4)∈ P, by convexity of F: 14∑k=14F(uk)≥F(14∑k=14uk)=F(π/2). 14 _k=1^4F(u_k)≥ F\! ( 14 _k=1^4u_k )=F(π/2). Thus Ψ(u1,…,u4)≥4F(π/2)for all (u1,…,u4)∈P. (u_1,…,u_4)≥ 4F(π/2) all (u_1,…,u_4)∈ P. (9) If 0∈Ψ(P)0∈ (P), then infPΨ≤0 _P ≤ 0, but (9) implies infPΨ≥4F(π/2) _P ≥ 4F(π/2), hence F(π/2)≤0F(π/2)≤ 0. 9.2) Compute F(π/2)F(π/2). Using the explicit F: F(π/2)=log(ycsc(π/2))−log(ycot(π/2)−x)=log(y)−log(−x).F(π/2)= (y (π/2))- (ycot(π/2)-x)= (y)- (-x). Because y=by=b and x=a−1x=a-1, we have −x=1−a-x=1-a, so F(π/2)=log(b1−a).F(π/2)= ( b1-a ). Thus F(π/2)≤0⇔b1−a≤1⇔b≤1−a⇔a+b≤1F(π/2)≤ 0 b1-a≤ 1 b≤ 1-a a+b≤ 1. This proves item (2) for b>0b>0; for b<0b<0 apply conjugation to obtain a+|b|≤1a+|b|≤ 1. 9.3) Attainment of segment CRCR. Fix x∈(0,1]x∈(0,1] and choose α=β=γ=δ=1−xα=β=γ=δ=1-x. Then t1=t2=t3=t4=xt_1=t_2=t_3=t_4=x and (3) becomes (z+x)4=x4.(z+x)^4=x^4. Hence z+x=xeikπ/2z+x=xe^ikπ/2 for k=0,1,2,3k=0,1,2,3. The nonreal solutions correspond to k=1,3k=1,3: z+x=±ixz+x=± ix, i.e. λ=1−x±ixλ=1-x± ix. Thus the upper segment λ=1−x+ixλ=1-x+ix, x∈[0,1]x∈[0,1], is attained. 10) Two regimes for supPΨ _P : Lemmas 2 and 3 with full details. Lemma 14 (Unbounded supremum). If 3m+M≤2π3m+M≤ 2π, then supPΨ=+∞ _P =+∞. Proof. Fix u4∈[m,M)u_4∈[m,M) and define u1=u2=u3=(2π−u4)/3u_1=u_2=u_3=(2π-u_4)/3. Then u1+u2+u3+u4=2πu_1+u_2+u_3+u_4=2π, so the only requirement to have (u1,…,u4)∈P(u_1,…,u_4)∈ P is u1∈[m,M)u_1∈[m,M). If u4u_4 is sufficiently close to M from below, then u4∈[m,M)u_4∈[m,M) and u1=2π−u43>2π−M3.u_1= 2π-u_43> 2π-M3. The hypothesis 3m+M≤2π3m+M≤ 2π implies (2π−M)/3≥m(2π-M)/3≥ m, hence u1≥mu_1≥ m for all such u4u_4. Also u1<Mu_1<M because u4>mu_4>m implies u1<(2π−m)/3≤(2π−m)/3<Mu_1<(2π-m)/3≤(2π-m)/3<M (since M>π/2M>π/2 and m>0m>0). Therefore (u1,u2,u3,u4)∈P(u_1,u_2,u_3,u_4)∈ P for u4u_4 close enough to M. Now as u4↑Mu_4 M, we have t(u4)↓0t(u_4) 0 (because u(t)u(t) decreases and u(t)↑Mu(t) M corresponds to t↓0t 0), so F(u4)=log|z+t(u4)|−logt(u4)→+∞(t(u4)↓0).F(u_4)= |z+t(u_4)|- t(u_4)→+∞ (t(u_4) 0). Meanwhile u1=(2π−u4)/3u_1=(2π-u_4)/3 converges to (2π−M)/3∈[m,M)(2π-M)/3∈[m,M), so F(u1)F(u_1) stays bounded. Thus Ψ=3F(u1)+F(u4)→+∞ =3F(u_1)+F(u_4)→+∞, proving supPΨ=+∞ _P =+∞. ∎ Lemma 15 (Finite maximum in the tight regime). Assume the tight regime 3m+M>2π3m+M>2π, and define U:=2π−3m.U:=2π-3m. Then U∈[m,M)U∈[m,M) (indeed U<MU<M), and maxPΨ=3F(m)+F(U), _P =3F(m)+F(U), with equality if and only if (u1,u2,u3,u4)(u_1,u_2,u_3,u_4) is a permutation of (U,m,m,m)(U,m,m,m). Proof. Step 1: U∈[m,M)U∈[m,M). Because 3m+M>2π3m+M>2π, we have U=2π−3m<MU=2π-3m<M. Also m≤π/2m≤π/2 (since a≥0a≥ 0 from Step 7), hence U=2π−3m≥2π−3(π/2)=π/2≥mU=2π-3m≥ 2π-3(π/2)=π/2≥ m. Therefore U∈[m,M)U∈[m,M). Step 2: reduce to ordered vectors and a compact domain. Define the ordered feasible set P↓:=(v1,v2,v3,v4)∈[m,M)4:v1≥v2≥v3≥v4,v1+v2+v3+v4=2π.P :=\(v_1,v_2,v_3,v_4)∈[m,M)^4:\ v_1≥ v_2≥ v_3≥ v_4,\ v_1+v_2+v_3+v_4=2π\. Because Ψ(u1,…,u4) (u_1,…,u_4) is symmetric in the coordinates, we have maxPΨ=maxP↓Ψ, _P = _P , since every point of P can be permuted into P↓P without changing Ψ . Next we prove that in the tight regime, every (v1,…,v4)∈P↓(v_1,…,v_4)∈ P actually lies in a closed box [m,U]4[m,U]^4, so that maxP↓Ψ _P is attained. Indeed, for any (v1,…,v4)∈P↓(v_1,…,v_4)∈ P we have v2,v3,v4≥mv_2,v_3,v_4≥ m, so v1=2π−(v2+v3+v4)≤2π−3m=U.v_1=2π-(v_2+v_3+v_4)≤ 2π-3m=U. Since v1≥v2≥v3≥v4v_1≥ v_2≥ v_3≥ v_4, this implies vk≤v1≤Uv_k≤ v_1≤ U for each k. Thus P↓⊂[m,U]4∩v1+v2+v3+v4=2π,v1≥v2≥v3≥v4.P ⊂[m,U]^4∩\v_1+v_2+v_3+v_4=2π,\ v_1≥ v_2≥ v_3≥ v_4\. Because U<MU<M, the upper endpoint M (open) is avoided: [m,U][m,U] is closed and bounded. Hence P↓P is a closed subset of a compact set, so P↓P is compact. Since Ψ is continuous, Ψ attains its maximum on P↓P . Step 3: prove the majorization inequalities explicitly. Let x:=(U,m,m,m).x:=(U,m,m,m). We claim that x majorizes every v=(v1,v2,v3,v4)∈P↓v=(v_1,v_2,v_3,v_4)∈ P , i.e. ∑j=1kvj≤∑j=1kxj(k=1,2,3),∑j=14vj=∑j=14xj=2π. _j=1^kv_j≤ _j=1^kx_j (k=1,2,3), _j=1^4v_j= _j=1^4x_j=2π. (10) We verify the partial sums: • For k=1k=1: as shown above, v1≤U=x1v_1≤ U=x_1. • For k=2k=2: since v3,v4≥mv_3,v_4≥ m, v1+v2=2π−(v3+v4)≤2π−2m=(2π−3m)+m=U+m=x1+x2.v_1+v_2=2π-(v_3+v_4)≤ 2π-2m=(2π-3m)+m=U+m=x_1+x_2. • For k=3k=3: since v4≥mv_4≥ m, v1+v2+v3=2π−v4≤2π−m=(2π−3m)+2m=U+2m=x1+x2+x3.v_1+v_2+v_3=2π-v_4≤ 2π-m=(2π-3m)+2m=U+2m=x_1+x_2+x_3. • For k=4k=4: both sums equal 2π2π by definition of P↓P and U: x1+x2+x3+x4=U+3m=(2π−3m)+3m=2π=∑j=14vj.x_1+x_2+x_3+x_4=U+3m=(2π-3m)+3m=2π= _j=1^4v_j. Thus (10) holds, i.e. x majorizes v. Step 4: apply Karamata (with strictness). Karamata’s inequality states: if f is convex on an interval and x majorizes v, then ∑f(xi)≥∑f(vi)Σ f(x_i)≥Σ f(v_i). Here F is convex (indeed strictly convex) on [m,M)[m,M), hence on [m,U]⊂[m,M)[m,U]⊂[m,M). Therefore Ψ(v1,v2,v3,v4)=∑k=14F(vk)≤∑k=14F(xk)=3F(m)+F(U). (v_1,v_2,v_3,v_4)= _k=1^4F(v_k)≤ _k=1^4F(x_k)=3F(m)+F(U). So maxP↓Ψ≤3F(m)+F(U) _P ≤ 3F(m)+F(U). On the other hand, the point (U,m,m,m)(U,m,m,m) itself belongs to P (since U∈[m,M)U∈[m,M) and U+3m=2πU+3m=2π), hence to P↓P (since U≥mU≥ m). Thus maxP↓Ψ≥Ψ(U,m,m,m)=3F(m)+F(U). _P ≥ (U,m,m,m)=3F(m)+F(U). So equality holds and maxPΨ=maxP↓Ψ=3F(m)+F(U). _P = _P =3F(m)+F(U). Finally, because F is strictly convex, the equality case in Karamata implies that if x majorizes v and ∑F(xi)=∑F(vi)Σ F(x_i)=Σ F(v_i), then v is a permutation of x. (Indeed, strict convexity forces equality only when the majorization is trivial in the sense that the vectors agree up to permutation.) Hence equality occurs if and only if (u1,u2,u3,u4)(u_1,u_2,u_3,u_4) is a permutation of (U,m,m,m)(U,m,m,m). ∎ 11) Necessity of G(a,b)≥0G(a,b)≥ 0. 11.1) A geometry/algebra lemma: G≤0⇒G≤ 0 tight regime. Lemma 16. Assume b>0b>0 and 0≤a≤10≤ a≤ 1. If G(a,b)≤0G(a,b)≤ 0 then 3m+M>2π3m+M>2π. Proof. Throughout, m=Arg(λ)=Arg(a+ib)∈(0,π/2]m=Arg(λ)=Arg(a+ib)∈(0,π/2] and M=Arg(λ−1)=Arg((a−1)+ib)∈(π/2,π)M=Arg(λ-1)=Arg((a-1)+ib)∈(π/2,π). Step 1: rewrite G as a quadratic in s=b2s=b^2. Let s=b2≥0s=b^2≥ 0. Expand: G(a,b)=(s+a2+a)2+2a2−s=s2+s(2a2+2a−1)+(a2+a)2+2a2.G(a,b)=(s+a^2+a)^2+2a^2-s=s^2+s(2a^2+2a-1)+(a^2+a)^2+2a^2. Thus G(a,b)≤0G(a,b)≤ 0 means s2+s(2a2+2a−1)+(a2+a)2+2a2≤0.s^2+s(2a^2+2a-1)+(a^2+a)^2+2a^2≤ 0. (11) Compute the discriminant: Δ =(2a2+2a−1)2−4((a2+a)2+2a2) =(2a^2+2a-1)^2-4 ((a^2+a)^2+2a^2 ) =(4a4+8a3+4a2−4a2−4a+1)−4(a4+2a3+a2+2a2) =(4a^4+8a^3+4a^2-4a^2-4a+1)-4(a^4+2a^3+a^2+2a^2) =(4a4+8a3−4a+1)−(4a4+8a3+12a2) =(4a^4+8a^3-4a+1)-(4a^4+8a^3+12a^2) =1−4a−12a2 =1-4a-12a^2 =−(2a+1)(6a−1). =-(2a+1)(6a-1). So Δ>0 >0 iff a<1/6a<1/6 (since 2a+1>02a+1>0). If a≥1/6a≥ 1/6 then Δ≤0 ≤ 0 and the quadratic (11) is everywhere >0>0 (it is monic), so G(a,b)≤0G(a,b)≤ 0 is impossible. Hence, under G(a,b)≤0G(a,b)≤ 0, we must have 0≤a<16.0≤ a< 16. (12) Step 2: handle a=0a=0 separately. If a=0a=0 then λ=ibλ=ib with b>0b>0, so m=π/2m=π/2. Also λ−1=−1+ibλ-1=-1+ib so M=π−arctan(b)M=π- (b). Therefore 3m+M=3π2+π−arctan(b)=5π2−arctan(b)>2π.3m+M= 3π2+π- (b)= 5π2- (b)>2π. So the claim holds when a=0a=0. Henceforth assume a>0a>0 (still with a<1/6a<1/6 by (12)). Step 3: show b2>3a2b^2>3a^2, hence m>π/3m>π/3, hence 3m>π3m>π. Let s=b2s=b^2. Since G(a,b)≤0G(a,b)≤ 0 and the quadratic is monic, s lies between its real roots. Let s−(a)s_-(a) be the smaller root: s−(a)=−(2a2+2a−1)−Δ2=1−2a−2a2−(2a+1)(1−6a)2.s_-(a)= -(2a^2+2a-1)- 2= 1-2a-2a^2- (2a+1)(1-6a)2. Then G(a,b)≤0G(a,b)≤ 0 implies s=b2≥s−(a).s=b^2≥ s_-(a). (13) We now prove the strict inequality s−(a)>3a2for all a∈(0,1/6).s_-(a)>3a^2 all a∈(0,1/6). (14) Indeed, (14) is equivalent to 1−2a−2a2−(2a+1)(1−6a)2>3a2⟺1−2a−8a2>(2a+1)(1−6a). 1-2a-2a^2- (2a+1)(1-6a)2>3a^2 1-2a-8a^2> (2a+1)(1-6a). For a∈(0,1/6)a∈(0,1/6), the left side is positive (check at a=1/6a=1/6: 1−1/3−8/36=4/9>01-1/3-8/36=4/9>0), so we can square both sides without changing the inequality: (1−2a−8a2)2>(2a+1)(1−6a).(1-2a-8a^2)^2>(2a+1)(1-6a). Expand the left side: (1−2a−8a2)2=1−4a−12a2+32a3+64a4.(1-2a-8a^2)^2=1-4a-12a^2+32a^3+64a^4. Expand the right side: (2a+1)(1−6a)=1−4a−12a2.(2a+1)(1-6a)=1-4a-12a^2. Thus the inequality becomes 1−4a−12a2+32a3+64a4>1−4a−12a2⟺32a3+64a4>0,1-4a-12a^2+32a^3+64a^4>1-4a-12a^2 32a^3+64a^4>0, which holds strictly for all a>0a>0. This proves (14). Combining with (13) gives b2=s≥s−(a)>3a2⟹ba>3.b^2=s≥ s_-(a)>3a^2 ba> 3. Since m=Arg(a+ib)=arctan(b/a)m=Arg(a+ib)= (b/a) for a>0a>0, we get m>arctan(3)=π3.m> ( 3)= π3. Therefore 3m>π3m>π. In particular 3m∈(π,3π/2)3m∈(π,3π/2) because m<π/2m<π/2. Step 4: rewrite 3m+M>2π3m+M>2π as a tangent inequality. Because a<1a<1 and b>0b>0, the point λ−1=(a−1)+ibλ-1=(a-1)+ib lies in quadrant I, so M=Arg(λ−1)=π−arctan(b1−a).M=Arg(λ-1)=π- \! ( b1-a ). Thus 3m+M>2π 3m+M>2π ⟺3m+π−arctan(b1−a)>2π 3m+π- \! ( b1-a )>2π ⟺3m>π+arctan(b1−a). 3m>π+ \! ( b1-a ). (15) Set θ:=arctan(b1−a)∈(0,π/2)θ:= \! ( b1-a )∈(0,π/2) (since b>0b>0 and a<1a<1). Then π+θ∈(π,3π/2)π+θ∈(π,3π/2). We already know 3m∈(π,3π/2)3m∈(π,3π/2) from Step 3. On (π,3π/2)(π,3π/2) the tangent function is continuous and strictly increasing. Therefore (15) is equivalent to tan(3m)>tan(π+θ)=tanθ=b1−a. (3m)> (π+θ)= θ= b1-a. (16) Step 5: express tan(3m) (3m) and reduce (16) to N(a,b)>0N(a,b)>0. Since tanm=b/a m=b/a (with a>0a>0), the triple-angle identity gives tan(3m)=3tanm−tan3m1−3tan2m=3(b/a)−(b/a)31−3(b/a)2=3a2b−b3a(a2−3b2). (3m)= 3 m- ^3m1-3 ^2m= 3(b/a)-(b/a)^31-3(b/a)^2= 3a^2b-b^3a(a^2-3b^2). Equivalently (multiplying numerator and denominator by −1-1), tan(3m)=b(b2−3a2)a(3b2−a2). (3m)= b(b^2-3a^2)a(3b^2-a^2). (17) By Step 3, b2>3a2b^2>3a^2 so all factors in (17) are positive, and tan(3m)>0 (3m)>0. Plug (17) into (16) and cancel b>0b>0: b2−3a2a(3b2−a2)>11−a. b^2-3a^2a(3b^2-a^2)> 11-a. All denominators are positive (since a∈(0,1)a∈(0,1) and 3b2−a2>03b^2-a^2>0), so we can cross-multiply: (1−a)(b2−3a2) (1-a)(b^2-3a^2) >a(3b2−a2) >a(3b^2-a^2) b2−ab2−3a2+3a3 b^2-ab^2-3a^2+3a^3 >3ab2−a3 >3ab^2-a^3 b2−4ab2+4a3−3a2 b^2-4ab^2+4a^3-3a^2 >0. >0. Define N(a,b):=4a3−3a2−4ab2+b2=(1−4a)b2+(4a3−3a2).N(a,b):=4a^3-3a^2-4ab^2+b^2=(1-4a)b^2+(4a^3-3a^2). (18) Then (16) is equivalent to N(a,b)>0.N(a,b)>0. (19) Step 6: show G(a,b)≤0⇒N(a,b)>0G(a,b)≤ 0 N(a,b)>0 by comparing roots in s=b2s=b^2. Fix a∈(0,1/6)a∈(0,1/6) and view N as an affine function of s=b2s=b^2: N(a,b)=(1−4a)s+(4a3−3a2).N(a,b)=(1-4a)s+(4a^3-3a^2). Because a<1/6<1/4a<1/6<1/4, we have 1−4a>01-4a>0, so N is strictly increasing in s. Therefore, under the constraint s≥s−(a)s≥ s_-(a) from (13), it suffices to prove s−(a)>s0(a),s_-(a)>s_0(a), (20) where s0(a)s_0(a) is the unique root of N in s: N(a,b)=0⇔(1−4a)s=3a2−4a3⇔s=s0(a):=a2(3−4a)1−4a.N(a,b)=0 (1-4a)s=3a^2-4a^3 s=s_0(a):= a^2(3-4a)1-4a. (21) Indeed, if (20) holds then s≥s−(a)s≥ s_-(a) implies s>s0(a)s>s_0(a), and since N increases in s this implies N(a,b)>0N(a,b)>0. We now prove (20). Starting from s−(a)=1−2a−2a2−(2a+1)(1−6a)2s_-(a)= 1-2a-2a^2- (2a+1)(1-6a)2, we compute: s−(a)>s0(a) s_-(a)>s_0(a) ⟺1−2a−2a2−(2a+1)(1−6a)2>a2(3−4a)1−4a 1-2a-2a^2- (2a+1)(1-6a)2> a^2(3-4a)1-4a ⟺(1−4a)(1−2a−2a2−(2a+1)(1−6a))>2a2(3−4a). (1-4a) (1-2a-2a^2- (2a+1)(1-6a) )>2a^2(3-4a). Expand (1−4a)(1−2a−2a2)(1-4a)(1-2a-2a^2): (1−4a)(1−2a−2a2)=1−6a+6a2+8a3.(1-4a)(1-2a-2a^2)=1-6a+6a^2+8a^3. So the inequality becomes 1−6a+6a2+8a3−(1−4a)(2a+1)(1−6a)>6a2−8a3,1-6a+6a^2+8a^3-(1-4a) (2a+1)(1-6a)>6a^2-8a^3, i.e. 1−6a+16a3>(1−4a)(2a+1)(1−6a).1-6a+16a^3>(1-4a) (2a+1)(1-6a). For a∈(0,1/6)a∈(0,1/6) both sides are positive, so we may square: (1−6a+16a3)2>(1−4a)2(2a+1)(1−6a).(1-6a+16a^3)^2>(1-4a)^2(2a+1)(1-6a). (22) Compute (2a+1)(1−6a)=1−4a−12a2(2a+1)(1-6a)=1-4a-12a^2. Then the right side is (1−4a)2(1−4a−12a2)=(1−8a+16a2)(1−4a−12a2).(1-4a)^2(1-4a-12a^2)=(1-8a+16a^2)(1-4a-12a^2). Expand: (1−8a+16a2)(1−4a−12a2)=1−12a+36a2+32a3−192a4.(1-8a+16a^2)(1-4a-12a^2)=1-12a+36a^2+32a^3-192a^4. Expand the left side: (1−6a+16a3)2=1−12a+36a2+32a3−192a4+256a6.(1-6a+16a^3)^2=1-12a+36a^2+32a^3-192a^4+256a^6. Thus the left side equals the right side plus 256a6256a^6, so (22) holds strictly for a>0a>0. Hence s−(a)>s0(a)s_-(a)>s_0(a), and therefore N(a,b)>0N(a,b)>0. Step 7: conclude 3m+M>2π3m+M>2π. We have proved G(a,b)≤0⇒N(a,b)>0⇒G(a,b)≤ 0 N(a,b)>0 (16) holds, which implies (15), which is equivalent to 3m+M>2π3m+M>2π. ∎ 11.2) Tight regime: compute maxPΨ _P explicitly and factor to G. Assume b>0b>0, 0≤a<10≤ a<1, and we are in the tight regime 3m+M>2π3m+M>2π. Then by Lemma 15, maxPΨ=3F(m)+F(U),U=2π−3m. _P =3F(m)+F(U), U=2π-3m. Step 1: compute F(m)F(m) exactly. By definition m=Arg(z+1)m=Arg(z+1), so the associated t is t=1t=1. Equivalently, using t(u)=ycotu−xt(u)=ycotu-x and (x,y)=(a−1,b)(x,y)=(a-1,b): t(m)=bcotm−(a−1)=1.t(m)=bcotm-(a-1)=1. Then F(m)=log|z+1|−log1=log|λ|.F(m)= |z+1|- 1= |λ|. Step 2: compute F(U)F(U) exactly. By definition, F(U)=log(bcscU)−log(bcotU−(a−1))=log(bcscU)−log(bcotU+1−a).F(U)= (b U)- (bcotU-(a-1))= (b U)- (bcotU+1-a). Step 3: obtain the announced closed form. Combine: maxPΨ _P =3log|λ|+log(bcscU)−log(bcotU+1−a) =3 |λ|+ (b U)- (bcotU+1-a) =log(|λ|3bcscUbcotU+1−a). = ( |λ|^3\,b UbcotU+1-a ). (23) This is the detailed derivation of the formula announced as (11.25) in the draft. Step 4: rewrite cscU U and cotUcotU algebraically in terms of (a,b)(a,b). We use U=2π−3mU=2π-3m. Because U∈(0,π)U∈(0,π) in the tight regime (indeed U<M<πU<M<π), we have sinU>0 U>0. We have the trigonometric identities sin(2π−θ)=−sinθ,cos(2π−θ)=cosθ, (2π-θ)=- θ, (2π-θ)= θ, hence sinU=−sin(3m),cotU=cosUsinU=cos(3m)−sin(3m)=−cot(3m),cscU=1sinU=−csc(3m). U=- (3m), = U U= (3m)- (3m)=-cot(3m), U= 1 U=- (3m). Now express sinm m and cosm m using λ=a+ibλ=a+ib: |λ|=a2+b2,cosm=a|λ|,sinm=b|λ|.|λ|= a^2+b^2, m= a|λ|, m= b|λ|. Use triple-angle formulas: sin(3m)=3sinm−4sin3m,cos(3m)=4cos3m−3cosm. (3m)=3 m-4 ^3m, (3m)=4 ^3m-3 m. Compute: sin(3m) (3m) =3b|λ|−4(b|λ|)3=3b|λ|2−4b3|λ|3=b(3(a2+b2)−4b2)|λ|3=b(3a2−b2)|λ|3, =3 b|λ|-4 ( b|λ| )^3= 3b|λ|^2-4b^3|λ|^3= b(3(a^2+b^2)-4b^2)|λ|^3= b(3a^2-b^2)|λ|^3, cos(3m) (3m) =4(a|λ|)3−3a|λ|=4a3−3a|λ|2|λ|3=a(4a2−3(a2+b2))|λ|3=a(a2−3b2)|λ|3. =4 ( a|λ| )^3-3 a|λ|= 4a^3-3a|λ|^2|λ|^3= a(4a^2-3(a^2+b^2))|λ|^3= a(a^2-3b^2)|λ|^3. Therefore cscU=−1sin(3m)=−|λ|3b(3a2−b2),cotU=−cos(3m)sin(3m)=−a(a2−3b2)b(3a2−b2). U=- 1 (3m)=- |λ|^3b(3a^2-b^2), =- (3m) (3m)=- a(a^2-3b^2)b(3a^2-b^2). (24) Step 5: substitute into (23) and reduce to an algebraic inequality. First compute bcscUb U using (24): bcscU=−|λ|33a2−b2=|λ|3b2−3a2.b U=- |λ|^33a^2-b^2= |λ|^3b^2-3a^2. Next compute the denominator bcotU+1−abcotU+1-a: bcotU+1−a=−a(a2−3b2)3a2−b2+1−a.bcotU+1-a=- a(a^2-3b^2)3a^2-b^2+1-a. Put over the common denominator (3a2−b2)(3a^2-b^2): bcotU+1−a bcotU+1-a =−a(a2−3b2)3a2−b2+(1−a)(3a2−b2)3a2−b2 = -a(a^2-3b^2)3a^2-b^2+ (1-a)(3a^2-b^2)3a^2-b^2 =−(a3−3ab2)+3a2−b2−3a3+ab23a2−b2 = -(a^3-3ab^2)+3a^2-b^2-3a^3+ab^23a^2-b^2 =−4a3+3a2+4ab2−b23a2−b2. = -4a^3+3a^2+4ab^2-b^23a^2-b^2. Define N(a,b)N(a,b) as in (18): N(a,b)=4a3−3a2−4ab2+b2.N(a,b)=4a^3-3a^2-4ab^2+b^2. Then the last numerator is exactly −N(a,b)-N(a,b), so bcotU+1−a=−N(a,b)3a2−b2=N(a,b)b2−3a2.bcotU+1-a= -N(a,b)3a^2-b^2= N(a,b)b^2-3a^2. Consequently, the positive quantity inside the logarithm in (23) is |λ|3bcscUbcotU+1−a |λ|^3\,b UbcotU+1-a =|λ|3⋅|λ|3b2−3a2N(a,b)b2−3a2=|λ|6N(a,b). = |λ|^3· |λ|^3b^2-3a^2 N(a,b)b^2-3a^2= |λ|^6N(a,b). (25) Thus maxPΨ=log(|λ|6N(a,b)). _P = ( |λ|^6N(a,b) ). (26) Since the left side is defined, the argument of the logarithm is positive; equivalently N(a,b)>0N(a,b)>0 in the tight regime. From (26), we have maxPΨ≥0⟺|λ|6N(a,b)≥1⟺|λ|6≥N(a,b). _P ≥ 0 |λ|^6N(a,b)≥ 1 |λ|^6≥ N(a,b). Step 6: factor |λ|6−N(a,b)|λ|^6-N(a,b) as |λ−1|2G(a,b)|λ-1|^2\,G(a,b). Let r:=|λ|2=a2+b2r:=|λ|^2=a^2+b^2. Then |λ|6=r3|λ|^6=r^3. Also |λ−1|2=(a−1)2+b2=r+1−2a|λ-1|^2=(a-1)^2+b^2=r+1-2a. Rewrite G in terms of r: G(a,b)=(b2+a2+a)2+2a2−b2=(r+a)2+2a2−(r−a2)=r2+(2a−1)r+4a2.G(a,b)=(b^2+a^2+a)^2+2a^2-b^2=(r+a)^2+2a^2-(r-a^2)=r^2+(2a-1)r+4a^2. Now multiply: |λ−1|2G(a,b) |λ-1|^2\,G(a,b) =(r+1−2a)(r2+(2a−1)r+4a2) =(r+1-2a) (r^2+(2a-1)r+4a^2 ) =r3+(4a−1)r+4a2−8a3(direct expansion; the r2 terms cancel) =r^3+(4a-1)r+4a^2-8a^3 (direct expansion; the $r^2$ terms cancel) =r3−(4a3−3a2−4ab2+b2) =r^3- (4a^3-3a^2-4ab^2+b^2 ) =|λ|6−N(a,b). =|λ|^6-N(a,b). Therefore |λ|6≥N(a,b)⟺|λ−1|2G(a,b)≥0.|λ|^6≥ N(a,b) |λ-1|^2\,G(a,b)≥ 0. (27) Because b>0b>0 implies |λ−1|2>0|λ-1|^2>0, (27) is equivalent to G(a,b)≥0.G(a,b)≥ 0. Combining with the previous equivalences yields the precise statement announced in the draft: maxPΨ≥0⟺G(a,b)≥0. _P ≥ 0 G(a,b)≥ 0. 11.3) Finish necessity of G(a,b)≥0G(a,b)≥ 0. Assume λ is a nonreal eigenvalue of some A(α,β,γ,δ)A(α,β,γ,δ). By (7), we have 0∈Ψ(P)0∈ (P), hence supPΨ≥0 _P ≥ 0. If 3m+M≤2π3m+M≤ 2π, then Lemma 16 by contrapositive implies G(a,b)>0G(a,b)>0. If 3m+M>2π3m+M>2π, then we are in the tight regime and by the computation in §11.2, supPΨ=maxPΨ≥0⇔G(a,b)≥0 _P = _P ≥ 0 G(a,b)≥ 0. Thus always G(a,b)≥0G(a,b)≥ 0. Finally, by conjugation symmetry, replace b by |b||b| to obtain item (3) for all b≠0b≠ 0. 12) Left boundary CLCL: attainment by AL(α)A_L(α) and equivalence to G=0G=0. 12.1) Characteristic equation for AL(α)A_L(α). Let AL(α)=(α1−α00001000011000).A_L(α)= pmatrixα&1-α&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0 pmatrix. Let v≠0v≠ 0 satisfy ALv=λvA_Lv=λ v. Row-by-row: αv1+(1−α)v2=λv1,v3=λv2,v4=λv3,v1=λv4.α v_1+(1-α)v_2=λ v_1, v_3=λ v_2, v_4=λ v_3, v_1=λ v_4. From the last three equalities, v3=λv2,v4=λ2v2,v1=λ3v2.v_3=λ v_2, v_4=λ^2v_2, v_1=λ^3v_2. Substitute into the first: α(λ3v2)+(1−α)v2=λ(λ3v2)⟺αλ3+(1−α)=λ4.α(λ^3v_2)+(1-α)v_2=λ(λ^3v_2) αλ^3+(1-α)=λ^4. Equivalently, λ3(λ−α)=1−α.λ^3(λ-α)=1-α. (28) For λ∉ℝλ this is equivalent to λ∈σ(AL(α))λ∈σ(A_L(α)). Solving (28) for α (and using λ3≠1λ^3≠ 1 for nonreal λ) gives α=λ4−1λ3−1.α= λ^4-1λ^3-1. (29) 12.2) Compute ℑ(α) (α) and show ℑ(α)=0⇔G(a,b)=0 (α)=0 G(a,b)=0. Let p=λ4−1p=λ^4-1 and q=λ3−1q=λ^3-1. Then α=p/qα=p/q. For any complex numbers p,qp,q with q≠0q≠ 0, ℑ(pq)=ℑ(pq¯|q|2)=ℑ(pq¯)|q|2. ( pq )= ( p q|q|^2 )= (p q)|q|^2. Thus ℑ(α)=ℑ((λ4−1)(λ¯3−1))|λ3−1|2. (α)= ((λ^4-1)( λ^3-1) )|λ^3-1|^2. (30) Expand the numerator: (λ4−1)(λ¯3−1)=λ4λ¯3−λ4−λ¯3+1.(λ^4-1)( λ^3-1)=λ^4 λ^3-λ^4- λ^3+1. Take imaginary parts: ℑ(λ4λ¯3)−ℑ(λ4)−ℑ(λ¯3). (λ^4 λ^3)- (λ^4)- ( λ^3). Now λ4λ¯3=λ(λλ¯)3=λ|λ|6λ^4 λ^3=λ(λ λ)^3=λ|λ|^6, hence ℑ(λ4λ¯3)=ℑ(λ)|λ|6=b|λ|6. (λ^4 λ^3)= (λ)|λ|^6=b|λ|^6. Also ℑ(λ¯3)=−ℑ(λ3) ( λ^3)=- (λ^3), so the numerator of (30) equals b|λ|6−ℑ(λ4)+ℑ(λ3).b|λ|^6- (λ^4)+ (λ^3). (31) Compute ℑ(λ3) (λ^3) and ℑ(λ4) (λ^4) for λ=a+ibλ=a+ib: λ3=(a3−3ab2)+i(3a2b−b3)⇒ℑ(λ3)=b(3a2−b2),λ^3=(a^3-3ab^2)+i(3a^2b-b^3) (λ^3)=b(3a^2-b^2), and λ4=(a4−6a2b2+b4)+i(4a3b−4ab3)⇒ℑ(λ4)=4ab(a2−b2).λ^4=(a^4-6a^2b^2+b^4)+i(4a^3b-4ab^3) (λ^4)=4ab(a^2-b^2). Substitute into (31): ℑ((λ4−1)(λ¯3−1)) ((λ^4-1)( λ^3-1) ) =b|λ|6−4ab(a2−b2)+b(3a2−b2) =b|λ|^6-4ab(a^2-b^2)+b(3a^2-b^2) =b(|λ|6−(4a3−3a2−4ab2+b2)) =b (|λ|^6-(4a^3-3a^2-4ab^2+b^2) ) =b(|λ|6−N(a,b)). =b (|λ|^6-N(a,b) ). By the factorization proved in §11.2, |λ|6−N(a,b)=|λ−1|2G(a,b).|λ|^6-N(a,b)=|λ-1|^2\,G(a,b). Therefore ℑ(α)=b|λ−1|2G(a,b)|λ3−1|2. (α)= b\,|λ-1|^2\,G(a,b)|λ^3-1|^2. (32) Since b>0b>0 and both |λ−1|2|λ-1|^2 and |λ3−1|2|λ^3-1|^2 are strictly positive for nonreal λ, we conclude ℑ(α)=0⟺G(a,b)=0. (α)=0 G(a,b)=0. Thus the upper-half-plane boundary curve G(a,b)=0G(a,b)=0 is exactly traced by the nonreal eigenvalues of AL(α)A_L(α). 12.3) Existence of a conjugate pair and endpoints i and 0. From (28) we have the characteristic polynomial λ4−αλ3−(1−α)=0.λ^4-αλ^3-(1-α)=0. One checks λ=1λ=1 is always a root: 1−α−(1−α)=0.1-α-(1-α)=0. Indeed the polynomial factors as (λ−1)(λ3+(1−α)(λ2+λ+1))=0.(λ-1) (λ^3+(1-α)(λ^2+λ+1) )=0. (33) Let fα(λ):=λ3+(1−α)(λ2+λ+1),λ∈ℝ.f_α(λ):=λ^3+(1-α)(λ^2+λ+1), λ . Then fα′(λ)=3λ2+2(1−α)λ+(1−α).f _α(λ)=3λ^2+2(1-α)λ+(1-α). The discriminant of this quadratic is Δα=4(1−α)2−12(1−α)=4(1−α)((1−α)−3)<0(α∈[0,1)), _α=4(1-α)^2-12(1-α)=4(1-α) ((1-α)-3 )<0 (α∈[0,1)), so fα′(λ)>0f _α(λ)>0 for all real λ. Hence fαf_α is strictly increasing and has exactly one real root. Therefore the remaining two roots of fαf_α are nonreal and form a complex conjugate pair. One of them lies in the upper half-plane. At α=0α=0, (28) becomes λ4=1λ^4=1, whose upper-half-plane nonreal root is λ=iλ=i. As α↑1α 1, the coefficients in (33) vary continuously, so the roots vary continuously; and from fα(0)=(1−α)>0f_α(0)=(1-α)>0 and fα(−1)=−1+(1−α)(1−1+1)=−α<0f_α(-1)=-1+(1-α)(1-1+1)=-α<0, the unique real root of fαf_α lies in (−1,0)(-1,0), while the conjugate pair has modulus tending to 0 as α↑1α 1. Thus the upper branch connects i (at α=0α=0) to 0 (as α→1−α→ 1^-), as stated. 13) Every strict interior point is attained (constructive existence). Fix λ=a+ibλ=a+ib with b>0,0≤a≤1,a+b<1,G(a,b)>0.b>0, 0≤ a≤ 1, a+b<1, G(a,b)>0. We must construct α,β,γ,δ∈[0,1)α,β,γ,δ∈[0,1) with λ∈σ(A(α,β,γ,δ))λ∈σ(A(α,β,γ,δ)). Step 1: P≠∅P≠ and exhibit a point with Ψ<0 <0. Because 0≤a<10≤ a<1 and b>0b>0, we have m≤π/2<Mm≤π/2<M as in Step 7, hence π/2∈[m,M)π/2∈[m,M). Define u(0):=(π2,π2,π2,π2)∈P.u^(0):= ( π2, π2, π2, π2 )∈ P. Then Ψ(u(0))=4F(π/2)=4log(b1−a). (u^(0))=4F(π/2)=4 ( b1-a ). Since a+b<1a+b<1, we have b<1−ab<1-a, hence b1−a<1 b1-a<1, hence log(b1−a)<0 \! ( b1-a )<0. Thus Ψ(u(0))<0. (u^(0))<0. (34) Step 2: produce a point with Ψ>0 >0 using G(a,b)>0G(a,b)>0. We show ∃u(1)∈Psuch thatΨ(u(1))>0.∃\,u^(1)∈ P\ such that\ (u^(1))>0. (35) There are two cases: Case 1: 3m+M≤2π3m+M≤ 2π. Then Lemma 14 gives supPΨ=+∞ _P =+∞, so in particular there exists u(1)∈Pu^(1)∈ P with Ψ(u(1))>0 (u^(1))>0. Case 2: 3m+M>2π3m+M>2π (tight regime). Then Lemma 15 gives supPΨ=maxPΨ, _P = _P , and §11.2 gives maxPΨ≥0⇔G(a,b)≥0 _P ≥ 0 G(a,b)≥ 0. Since we assume G(a,b)>0G(a,b)>0, we get maxPΨ>0 _P >0, so there exists u(1)∈Pu^(1)∈ P with Ψ(u(1))>0 (u^(1))>0. For example, take u(1)=(U,m,m,m)u^(1)=(U,m,m,m) (in any order), which achieves the maximum. Thus (35) holds in all cases. Step 3: apply IVT along a line segment in P (fully explicit). Because P is convex, for t∈[0,1]t∈[0,1] the convex combination u(t):=(1−t)u(0)+tu(1)u(t):=(1-t)u^(0)+t\,u^(1) lies in P. Define the scalar function ϕ(t):=Ψ(u(t))=Ψ((1−t)u(0)+tu(1)).φ(t):= (u(t))= ((1-t)u^(0)+t\,u^(1) ). Because Ψ is continuous on P and u(t)u(t) is continuous in t, the composition ϕφ is continuous on [0,1][0,1]. We have ϕ(0)=Ψ(u(0))<0φ(0)= (u^(0))<0 by (34), and ϕ(1)=Ψ(u(1))>0φ(1)= (u^(1))>0 by (35). By the Intermediate Value Theorem, there exists t∗∈(0,1)t_*∈(0,1) such that ϕ(t∗)=0φ(t_*)=0. Set (u1,u2,u3,u4):=u(t∗)∈P,(u_1,u_2,u_3,u_4):=u(t_*)∈ P, so u1+u2+u3+u4=2π,Ψ(u1,u2,u3,u4)=0.u_1+u_2+u_3+u_4=2π, (u_1,u_2,u_3,u_4)=0. By Lemma 13, the corresponding tk=t(uk)∈(0,1]t_k=t(u_k)∈(0,1] satisfy (3), hence λ is an eigenvalue of the original matrix for parameters α=1−t1,β=1−t2,γ=1−t3,δ=1−t4.α=1-t_1, β=1-t_2, γ=1-t_3, δ=1-t_4. Because each tk∈(0,1]t_k∈(0,1], each parameter lies in [0,1)[0,1). This constructs α,β,γ,δα,β,γ,δ with λ∈σ(A(α,β,γ,δ))λ∈σ(A(α,β,γ,δ)), proving the converse for strict interior points. Combining Steps 7, 9, and 11 yields the necessary conditions 0≤a≤10≤ a≤ 1, a+|b|≤1a+|b|≤ 1, and G(a,|b|)≥0G(a,|b|)≥ 0 for all nonreal eigenvalues. The explicit families in Steps 9.3 and 12 attain the boundary segments CRCR and CLCL. Finally Step 13 proves every strict interior point is attained. ∎