{
  "index": "2009-A-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "Let $d_n$ be the determinant of the $n \\times n$ matrix whose entries, from\nleft to right and then from top to bottom, are $\\cos 1, \\cos 2, \\dots, \\cos\nn^2$. (For example,\n\\[\n d_3 = \\left| \\begin{matrix} \\cos 1 & \\cos 2 & \\cos 3 \\\\\n               \\cos 4 & \\cos 5 & \\cos 6 \\\\\n\\cos 7 & \\cos 8 & \\cos 9\n              \\end{matrix} \\right|.\n\\]\nThe argument of $\\cos$ is always in radians, not degrees.) Evaluate\n$\\lim_{n\\to\\infty} d_n$.",
  "solution": "The limit is $0$; we will show this by checking that $d_n = 0$ for all $n \\geq 3$.\nStarting from the given matrix, add the third column to the first column; this does not change the\ndeterminant. However, thanks to the identity\n$\\cos x + \\cos y = 2 \\cos \\frac{x+y}{2} \\cos \\frac{x-y}{2}$,\nthe resulting matrix has the form\n\\[\n \\begin{pmatrix} 2 \\cos 2 \\cos 1 & \\cos 2 & \\cdots \\\\\n  2 \\cos (n+2) \\cos 1 & \\cos (n+2) & \\cdots \\\\\n  2 \\cos (2n+2) \\cos 1 & 2 \\cos (2n+2) & \\cdots \\\\\n\\vdots & \\vdots & \\ddots\n \\end{pmatrix}\n\\]\nwith the first column being a multiple of the second. Hence $d_n = 0$.\n\n\\textbf{Remark.}\nAnother way to draw the same conclusion is to observe that the given matrix is the sum\nof the two rank 1 matrices $A_{jk} = \\cos (j-1)n \\cos k$ and $B_{jk} = -\\sin (j-1)n \\sin k$,\nand so has rank at most 2. One can also use the matrices\n$A_{jk} = e^{i((j-1)n+k)}$, $B_{jk} = e^{-i(j-1)n+k}$.",
  "vars": [
    "A_jk",
    "B_jk",
    "d_n",
    "d_3",
    "j",
    "k",
    "n",
    "x",
    "y"
  ],
  "params": [],
  "sci_consts": [
    "e",
    "i"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "A_jk": "matrixaelem",
        "B_jk": "matrixbelem",
        "d_n": "detsequence",
        "d_3": "detthree",
        "j": "rowindex",
        "k": "colindex",
        "n": "sizeindex",
        "x": "angleone",
        "y": "angletwo"
      },
      "question": "Let $detsequence$ be the determinant of the $sizeindex \\times sizeindex$ matrix whose entries, from\nleft to right and then from top to bottom, are $\\cos 1, \\cos 2, \\dots, \\cos sizeindex^2$. (For example,\n\\[\n detthree = \\left| \\begin{matrix} \\cos 1 & \\cos 2 & \\cos 3 \\\\\n               \\cos 4 & \\cos 5 & \\cos 6 \\\\\n\\cos 7 & \\cos 8 & \\cos 9\n              \\end{matrix} \\right|.\n\\]\nThe argument of $\\cos$ is always in radians, not degrees.) Evaluate $\\lim_{sizeindex\\to\\infty} detsequence$.",
      "solution": "The limit is $0$; we will show this by checking that $detsequence = 0$ for all $sizeindex \\geq 3$.\nStarting from the given matrix, add the third column to the first column; this does not change the determinant. However, thanks to the identity\n$\\cos angleone + \\cos angletwo = 2 \\cos \\frac{angleone+angletwo}{2} \\cos \\frac{angleone-angletwo}{2}$,\nthe resulting matrix has the form\n\\[\n \\begin{pmatrix} 2 \\cos 2 \\cos 1 & \\cos 2 & \\cdots \\\\\n  2 \\cos (sizeindex+2) \\cos 1 & \\cos (sizeindex+2) & \\cdots \\\\\n  2 \\cos (2sizeindex+2) \\cos 1 & 2 \\cos (2sizeindex+2) & \\cdots \\\\\n\\vdots & \\vdots & \\ddots\n \\end{pmatrix}\n\\]\nwith the first column being a multiple of the second. Hence $detsequence = 0$.\n\n\\textbf{Remark.}\nAnother way to draw the same conclusion is to observe that the given matrix is the sum of the two rank 1 matrices $matrixaelem = \\cos ((rowindex-1)sizeindex) \\cos colindex$ and $matrixbelem = -\\sin ((rowindex-1)sizeindex) \\sin colindex$, and so has rank at most 2. One can also use the matrices $matrixaelem = e^{i((rowindex-1)sizeindex+colindex)}$, $matrixbelem = e^{-i((rowindex-1)sizeindex+colindex)}$."
    },
    "descriptive_long_confusing": {
      "map": {
        "A_jk": "daffodil",
        "B_jk": "chrysalis",
        "d_n": "hummingbird",
        "d_3": "woodpecker",
        "j": "raincloud",
        "k": "windchime",
        "n": "mountains",
        "x": "riverbank",
        "y": "sailboat"
      },
      "question": "Let $hummingbird$ be the determinant of the $mountains \\times mountains$ matrix whose entries, from\nleft to right and then from top to bottom, are $\\cos 1, \\cos 2, \\dots, \\cos\nmountains^2$. (For example,\n\\[\n woodpecker = \\left| \\begin{matrix} \\cos 1 & \\cos 2 & \\cos 3 \\\\\n               \\cos 4 & \\cos 5 & \\cos 6 \\\\\n\\cos 7 & \\cos 8 & \\cos 9\n              \\end{matrix} \\right|.\n\\]\nThe argument of $\\cos$ is always in radians, not degrees.) Evaluate\n$\\lim_{mountains\\to\\infty} hummingbird$.",
      "solution": "The limit is $0$; we will show this by checking that $hummingbird = 0$ for all $mountains \\geq 3$.\nStarting from the given matrix, add the third column to the first column; this does not change the\ndeterminant. However, thanks to the identity\n$\\cos riverbank + \\cos sailboat = 2 \\cos \\frac{riverbank+sailboat}{2} \\cos \\frac{riverbank-sailboat}{2}$,\nthe resulting matrix has the form\n\\[\n \\begin{pmatrix} 2 \\cos 2 \\cos 1 & \\cos 2 & \\cdots \\\\\n  2 \\cos (mountains+2) \\cos 1 & \\cos (mountains+2) & \\cdots \\\\\n  2 \\cos (2mountains+2) \\cos 1 & 2 \\cos (2mountains+2) & \\cdots \\\\\n\\vdots & \\vdots & \\ddots\n \\end{pmatrix}\n\\]\nwith the first column being a multiple of the second. Hence $hummingbird = 0$.\n\n\\textbf{Remark.}\nAnother way to draw the same conclusion is to observe that the given matrix is the sum\nof the two rank 1 matrices $daffodil_{raincloudwindchime} = \\cos (raincloud-1)mountains \\cos windchime$ and $chrysalis_{raincloudwindchime} = -\\sin (raincloud-1)mountains \\sin windchime$,\nand so has rank at most 2. One can also use the matrices\n$daffodil_{raincloudwindchime} = e^{i((raincloud-1)mountains+windchime)}$, $chrysalis_{raincloudwindchime} = e^{-i(raincloud-1)mountains+windchime}$.}",
      "confidence": "0.17"
    },
    "descriptive_long_misleading": {
      "map": {
        "A_jk": "singularmatrix",
        "B_jk": "fullrankmatrix",
        "d_n": "tracevalue",
        "d_3": "tracevaluethree",
        "j": "columnindex",
        "k": "rowindex",
        "n": "singleton",
        "x": "outputvalue",
        "y": "inputvalue"
      },
      "question": "Let $tracevalue$ be the determinant of the $singleton \\times singleton$ matrix whose entries, from left to right and then from top to bottom, are $\\cos 1, \\cos 2, \\dots, \\cos singleton^2$. (For example,\n\\[\n tracevaluethree = \\left| \\begin{matrix} \\cos 1 & \\cos 2 & \\cos 3 \\\\\n               \\cos 4 & \\cos 5 & \\cos 6 \\\\\n\\cos 7 & \\cos 8 & \\cos 9\n              \\end{matrix} \\right|.\n\\]\nThe argument of $\\cos$ is always in radians, not degrees.) Evaluate $\\lim_{singleton\\to\\infty} tracevalue$.",
      "solution": "The limit is $0$; we will show this by checking that $tracevalue = 0$ for all $singleton \\geq 3$.\nStarting from the given matrix, add the third column to the first column; this does not change the determinant. However, thanks to the identity\n$\\cos outputvalue + \\cos inputvalue = 2 \\cos \\frac{outputvalue+inputvalue}{2} \\cos \\frac{outputvalue-inputvalue}{2}$,\nthe resulting matrix has the form\n\\[\n \\begin{pmatrix} 2 \\cos 2 \\cos 1 & \\cos 2 & \\cdots \\\\\n  2 \\cos (singleton+2) \\cos 1 & \\cos (singleton+2) & \\cdots \\\\\n  2 \\cos (2singleton+2) \\cos 1 & 2 \\cos (2singleton+2) & \\cdots \\\\\n\\vdots & \\vdots & \\ddots\n \\end{pmatrix}\n\\]\nwith the first column being a multiple of the second. Hence $tracevalue = 0$.\n\n\\textbf{Remark.}\nAnother way to draw the same conclusion is to observe that the given matrix is the sum of the two rank 1 matrices $singularmatrix = \\cos (columnindex-1)singleton \\cos rowindex$ and $fullrankmatrix = -\\sin (columnindex-1)singleton \\sin rowindex$, and so has rank at most 2. One can also use the matrices $singularmatrix = e^{i((columnindex-1)singleton+rowindex)}$, $fullrankmatrix = e^{-i(columnindex-1)singleton+rowindex}$.}",
      "confidence": "0.14"
    },
    "garbled_string": {
      "map": {
        "A_jk": "qzxwvtnp",
        "B_jk": "hjgrksla",
        "d_n": "bxmcqrad",
        "d_3": "wyphlseo",
        "j": "cnhuskle",
        "k": "mvatspow",
        "n": "lrgdqfze",
        "x": "uosbekwr",
        "y": "kfdyhlam"
      },
      "question": "Let $bxmcqrad$ be the determinant of the $lrgdqfze \\times lrgdqfze$ matrix whose entries, from\nleft to right and then from top to bottom, are $\\cos 1, \\cos 2, \\dots, \\cos\nlrgdqfze^2$. (For example,\n\\[\n wyphlseo = \\left| \\begin{matrix} \\cos 1 & \\cos 2 & \\cos 3 \\\\\n               \\cos 4 & \\cos 5 & \\cos 6 \\\\\n\\cos 7 & \\cos 8 & \\cos 9\n              \\end{matrix} \\right|.\n\\]\nThe argument of $\\cos$ is always in radians, not degrees.) Evaluate\n$\\lim_{lrgdqfze\\to\\infty} bxmcqrad$.",
      "solution": "The limit is $0$; we will show this by checking that $bxmcqrad = 0$ for all $lrgdqfze \\geq 3$.\nStarting from the given matrix, add the third column to the first column; this does not change the\ndeterminant. However, thanks to the identity\n$\\cos uosbekwr + \\cos kfdyhlam = 2 \\cos \\frac{uosbekwr+kfdyhlam}{2} \\cos \\frac{uosbekwr-kfdyhlam}{2}$,\nthe resulting matrix has the form\n\\[\n \\begin{pmatrix} 2 \\cos 2 \\cos 1 & \\cos 2 & \\cdots \\\\\n  2 \\cos (lrgdqfze+2) \\cos 1 & \\cos (lrgdqfze+2) & \\cdots \\\\\n  2 \\cos (2lrgdqfze+2) \\cos 1 & 2 \\cos (2lrgdqfze+2) & \\cdots \\\\\n\\vdots & \\vdots & \\ddots\n \\end{pmatrix}\n\\]\nwith the first column being a multiple of the second. Hence $bxmcqrad = 0$.\n\n\\textbf{Remark.}\nAnother way to draw the same conclusion is to observe that the given matrix is the sum\nof the two rank 1 matrices $qzxwvtnp = \\cos (cnhuskle-1)lrgdqfze \\cos mvatspow$ and $hjgrksla = -\\sin (cnhuskle-1)lrgdqfze \\sin mvatspow$,\nand so has rank at most 2. One can also use the matrices\n$qzxwvtnp = e^{i((cnhuskle-1)lrgdqfze+mvatspow)}$, $hjgrksla = e^{-i(cnhuskle-1)lrgdqfze+mvatspow}$. "
    },
    "kernel_variant": {
      "question": "Let  \n  α = π/7 ,  β = √2 ,  γ = e  \n\nand, for every positive integer n, form the n × n real matrix  \n\n  Mₙ = (m_{jk})_{1≤j,k≤n}  with entries  \n\n  m_{jk} = cos(αj + βk) + cos(βj + γk) + sin(γj + αk)  \n        + cos(2αj + 3βk) + sin(3βj + 2γk).     (★)\n\n(Angles are measured in radians.)  Let Δₙ = det Mₙ.\n\na)  Prove that rank Mₙ ≤ 10 for every n.  \nb)  Prove that rank Mₙ = 10 for every n ≥ 10.  \nc)  Deduce that Δₙ = 0 for all n ≥ 11 and therefore  \n\n    lim_{n→∞} Δₙ = 0.\n\nYou may use, without proof, the following two facts.\n\nF1.  The twelve real numbers  \n  ±α, ±β, ±γ, ±2α, ±3β, ±2γ  \n  are pairwise incongruent modulo 2π.\n\nF2.  (Vandermonde independence)  \n  If t₁,…,t_m are distinct real numbers modulo 2π, then the complex sequences  \n  (e^{it_r j})_{j≥1} (1 ≤ r ≤ m) are linearly independent over ℂ.  \n  Equivalently, every m × m matrix (e^{it_r j})_{1≤j,r≤m} has non–zero determinant.",
      "solution": "Throughout write  \n\n u^{c}_{t}(j)=cos(tj), u^{s}_{t}(j)=sin(tj) (j=1,2,…),  \n v^{c}_{t}(k)=cos(tk), v^{s}_{t}(k)=sin(tk) (k=1,2,…).\n\n────────────────────────────────────────────────────────\n1. Separation of variables.  \nThe addition and double–angle identities give  \n\ncos(αj+βk)=u^{c}_{α}(j)v^{c}_{β}(k)−u^{s}_{α}(j)v^{s}_{β}(k)  \ncos(βj+γk)=u^{c}_{β}(j)v^{c}_{γ}(k)−u^{s}_{β}(j)v^{s}_{γ}(k)  \nsin(γj+αk)=u^{s}_{γ}(j)v^{c}_{α}(k)+u^{c}_{γ}(j)v^{s}_{α}(k)  \ncos(2αj+3βk)=u^{c}_{2α}(j)v^{c}_{3β}(k)−u^{s}_{2α}(j)v^{s}_{3β}(k)  \nsin(3βj+2γk)=u^{s}_{3β}(j)v^{c}_{2γ}(k)+u^{c}_{3β}(j)v^{s}_{2γ}(k).\n\nHence every entry of Mₙ is a linear combination of the ten rank-1 matrices  \n\n X₁=u^{c}_{α}⊗v^{c}_{β},   X₂=−u^{s}_{α}⊗v^{s}_{β},  \n X₃=u^{c}_{β}⊗v^{c}_{γ},   X4=−u^{s}_{β}⊗v^{s}_{γ},  \n X₅=u^{s}_{γ}⊗v^{c}_{α},   X₆=u^{c}_{γ}⊗v^{s}_{α},  \n X₇=u^{c}_{2α}⊗v^{c}_{3β},  X₈=−u^{s}_{2α}⊗v^{s}_{3β},  \n X₉=u^{s}_{3β}⊗v^{c}_{2γ},  X₁₀=u^{c}_{3β}⊗v^{s}_{2γ}. (1)\n\n────────────────────────────────────────────────────────\n2. Universal rank bound (part a).  \nEach X_r has rank 1, so  \n\n rank Mₙ ≤ Σ_{r=1}^{10} rank X_r = 10 for every n. ∎\n\n────────────────────────────────────────────────────────\n3. Independence of the j–factor sequences.  \nRetain the ten j-sequences appearing in (1):\n\n U₁=u^{c}_{α}, U₂=u^{s}_{α}, U₃=u^{c}_{β}, U₄=u^{s}_{β},  \n U₅=u^{s}_{γ}, U₆=u^{c}_{γ}, U₇=u^{c}_{2α}, U₈=u^{s}_{2α},  \n U₉=u^{s}_{3β}, U₁₀=u^{c}_{3β}. (2)\n\nUsing cos tj = ½(e^{itj}+e^{−itj}) and sin tj = ½i(e^{−itj}−e^{itj}), each U_r\nis a ℂ-linear combination of the exponentials  \n\n E_{±α}, E_{±β}, E_{±γ}, E_{±2α}, E_{±3β}, where E_{t}(j)=e^{itj}. (3)\n\nBecause the ten frequencies in (3) are pairwise distinct by F1, Fact F2 yields the ℂ-independence of the exponentials and hence  \n\n dim span_{ℝ}{U₁,…,U₁₀}=10. (4)\n\n────────────────────────────────────────────────────────\n4. Independence of the k–factor sequences.  \nDefine  \n\n V₁=v^{c}_{β}, V₂=v^{s}_{β}, V₃=v^{c}_{γ}, V₄=v^{s}_{γ},  \n V₅=v^{c}_{α}, V₆=v^{s}_{α}, V₇=v^{c}_{3β}, V₈=v^{s}_{3β},  \n V₉=v^{c}_{2γ}, V₁₀=v^{s}_{2γ}. (5)\n\nThese are ℝ-linear combinations of the exponentials  \n\n E_{±β}, E_{±γ}, E_{±α}, E_{±3β}, E_{±2γ}.  \n\nSince the ten frequencies are again distinct mod 2π (F1), F2 gives  \n\n dim span_{ℝ}{V₁,…,V₁₀}=10. (6)\n\n────────────────────────────────────────────────────────\n5. Truncations and factor matrices.  \nFix n ≥ 10 and truncate the sequences:\n\n U_r^{(n)}=(U_r(1),…,U_r(n))∈ℝⁿ, V_r^{(n)}=(V_r(1),…,V_r(n))∈ℝⁿ.\n\nSet  \n\n Pₙ = [U_1^{(n)} … U_{10}^{(n)}]  (n×10),  \n Qₙ = [V_1^{(n)} … V_{10}^{(n)}]  (n×10).\n\nTo keep the original signs, introduce the diagonal matrix  \n\n D = diag(1,−1,1,−1,1,1,1,−1,1,1). (7)\n\nWith the ordering chosen in (1) we have the exact factorisation  \n\n Mₙ = Pₙ D Qₙᵀ. (8)\n\n────────────────────────────────────────────────────────\n5.1 Rank of Pₙ.  \nSuppose Σ_{r=1}^{10} c_r U_r^{(n)} = 0.  \nThis means Σ_{r=1}^{10} c_r U_r(j)=0 for j=1,…,n.  \nReplacing each U_r by its exponential representation produces\n\n Σ_{s=1}^{10} d_s e^{it_s j}=0 (j=1,…,n),  \n\nwith distinct t_s in {±α,±β,±γ,±2α,±3β}.  \nBecause n≥10, the square sub-matrix W=(e^{it_s j})_{1≤j,s≤10} has det W≠0 by F2, forcing d_s=0 and therefore c_r=0.  Thus  \n\n rank Pₙ = 10. (9)\n\n────────────────────────────────────────────────────────\n5.2 Rank of Qₙ.  \nThe same argument using the ten frequencies {±β,±γ,±α,±3β,±2γ} gives  \n\n rank Qₙ = 10. (10)\n\n────────────────────────────────────────────────────────\n6. Exact rank of Mₙ (part b).  \nFrom (8) and Sylvester’s inequality,\n\n rank Mₙ ≥ rank (Pₙ D) + rank Qₙ − 10  \n      = rank Pₙ + rank Qₙ − 10 = 10.\n\nBut part (a) gave rank Mₙ ≤ 10, so  \n\n rank Mₙ = 10 for every n ≥ 10. ∎\n\n────────────────────────────────────────────────────────\n7. Singularity for n ≥ 11 (part c).  \nIf n ≥ 11, the n×n matrix Mₙ has rank 10 < n, hence det Mₙ = 0.  Therefore  \n\n Δₙ = 0 for all n ≥ 11. ∎\n\n────────────────────────────────────────────────────────\n8. Limit of Δₙ.  \nBecause Δₙ vanishes from n = 11 onward,  \n\n lim_{n→∞} Δₙ = 0. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.812932",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original problem (and its existing kernel variant)\nthe enhanced version is decisively more intricate.\n\n1.  Technical proliferation.  Five different trigonometric expressions\ninvolving three incommensurable constants appear in each entry instead\nof a single elementary sine.  The entries therefore look opaque and\npreclude routine “add-one-column-to-another’’ tricks.\n\n2.  Higher rank bookkeeping.  One has to recognise and count all\nseparable terms created when each trigonometric function is expanded,\nthen argue abstractly with rank rather than with explicit numerical\nrelationships.  The argument must track sixteen distinct outer products\ninstead of just two.\n\n3.  Multiple interacting concepts.  The solution blends trigonometric\naddition formulas, low-rank factorizations, and elementary linear-algebraic\nrank considerations; none of these suffices in isolation.\n\n4.  Deeper theoretical requirement.  The crux is to see that summing\nrank-1 matrices keeps the overall rank small—an idea that is\nstraightforward in principle but tricky to spot amid the dense\ntrigonometric clutter.\n\n5.  Subtlety of incommensurability.  Choosing α, β, γ pairwise\nincommensurable forbids accidental periodicity, blocking shortcuts\nsuch as combining columns by constant factors; one must truly rely on\nthe rank argument.\n\nBecause of these layers of complexity, a solver cannot dispatch the\nproblem with the single “rank-two’’ observation that kills the\noriginal exercise; instead she must unravel a much thicker algebraic\nfabric before the same conceptual hammer—low rank—can finally fall."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n  α = π/7 ,  β = √2 ,  γ = e  \n\nand, for every positive integer n, form the n × n real matrix  \n\n  Mₙ = (m_{jk})_{1≤j,k≤n}  with entries  \n\n  m_{jk} = cos(αj + βk) + cos(βj + γk) + sin(γj + αk)  \n        + cos(2αj + 3βk) + sin(3βj + 2γk).     (★)\n\n(Angles are measured in radians.)  Let Δₙ = det Mₙ.\n\na)  Prove that rank Mₙ ≤ 10 for every n.  \nb)  Prove that rank Mₙ = 10 for every n ≥ 10.  \nc)  Deduce that Δₙ = 0 for all n ≥ 11 and therefore  \n\n    lim_{n→∞} Δₙ = 0.\n\nYou may use, without proof, the following two facts.\n\nF1.  The twelve real numbers  \n  ±α, ±β, ±γ, ±2α, ±3β, ±2γ  \n  are pairwise incongruent modulo 2π.\n\nF2.  (Vandermonde independence)  \n  If t₁,…,t_m are distinct real numbers modulo 2π, then the complex sequences  \n  (e^{it_r j})_{j≥1} (1 ≤ r ≤ m) are linearly independent over ℂ.  \n  Equivalently, every m × m matrix (e^{it_r j})_{1≤j,r≤m} has non–zero determinant.",
      "solution": "Throughout write  \n\n u^{c}_{t}(j)=cos(tj), u^{s}_{t}(j)=sin(tj) (j=1,2,…),  \n v^{c}_{t}(k)=cos(tk), v^{s}_{t}(k)=sin(tk) (k=1,2,…).\n\n────────────────────────────────────────────────────────\n1. Separation of variables.  \nThe addition and double–angle identities give  \n\ncos(αj+βk)=u^{c}_{α}(j)v^{c}_{β}(k)−u^{s}_{α}(j)v^{s}_{β}(k)  \ncos(βj+γk)=u^{c}_{β}(j)v^{c}_{γ}(k)−u^{s}_{β}(j)v^{s}_{γ}(k)  \nsin(γj+αk)=u^{s}_{γ}(j)v^{c}_{α}(k)+u^{c}_{γ}(j)v^{s}_{α}(k)  \ncos(2αj+3βk)=u^{c}_{2α}(j)v^{c}_{3β}(k)−u^{s}_{2α}(j)v^{s}_{3β}(k)  \nsin(3βj+2γk)=u^{s}_{3β}(j)v^{c}_{2γ}(k)+u^{c}_{3β}(j)v^{s}_{2γ}(k).\n\nHence every entry of Mₙ is a linear combination of the ten rank-1 matrices  \n\n X₁=u^{c}_{α}⊗v^{c}_{β},   X₂=−u^{s}_{α}⊗v^{s}_{β},  \n X₃=u^{c}_{β}⊗v^{c}_{γ},   X4=−u^{s}_{β}⊗v^{s}_{γ},  \n X₅=u^{s}_{γ}⊗v^{c}_{α},   X₆=u^{c}_{γ}⊗v^{s}_{α},  \n X₇=u^{c}_{2α}⊗v^{c}_{3β},  X₈=−u^{s}_{2α}⊗v^{s}_{3β},  \n X₉=u^{s}_{3β}⊗v^{c}_{2γ},  X₁₀=u^{c}_{3β}⊗v^{s}_{2γ}. (1)\n\n────────────────────────────────────────────────────────\n2. Universal rank bound (part a).  \nEach X_r has rank 1, so  \n\n rank Mₙ ≤ Σ_{r=1}^{10} rank X_r = 10 for every n. ∎\n\n────────────────────────────────────────────────────────\n3. Independence of the j–factor sequences.  \nRetain the ten j-sequences appearing in (1):\n\n U₁=u^{c}_{α}, U₂=u^{s}_{α}, U₃=u^{c}_{β}, U₄=u^{s}_{β},  \n U₅=u^{s}_{γ}, U₆=u^{c}_{γ}, U₇=u^{c}_{2α}, U₈=u^{s}_{2α},  \n U₉=u^{s}_{3β}, U₁₀=u^{c}_{3β}. (2)\n\nUsing cos tj = ½(e^{itj}+e^{−itj}) and sin tj = ½i(e^{−itj}−e^{itj}), each U_r\nis a ℂ-linear combination of the exponentials  \n\n E_{±α}, E_{±β}, E_{±γ}, E_{±2α}, E_{±3β}, where E_{t}(j)=e^{itj}. (3)\n\nBecause the ten frequencies in (3) are pairwise distinct by F1, Fact F2 yields the ℂ-independence of the exponentials and hence  \n\n dim span_{ℝ}{U₁,…,U₁₀}=10. (4)\n\n────────────────────────────────────────────────────────\n4. Independence of the k–factor sequences.  \nDefine  \n\n V₁=v^{c}_{β}, V₂=v^{s}_{β}, V₃=v^{c}_{γ}, V₄=v^{s}_{γ},  \n V₅=v^{c}_{α}, V₆=v^{s}_{α}, V₇=v^{c}_{3β}, V₈=v^{s}_{3β},  \n V₉=v^{c}_{2γ}, V₁₀=v^{s}_{2γ}. (5)\n\nThese are ℝ-linear combinations of the exponentials  \n\n E_{±β}, E_{±γ}, E_{±α}, E_{±3β}, E_{±2γ}.  \n\nSince the ten frequencies are again distinct mod 2π (F1), F2 gives  \n\n dim span_{ℝ}{V₁,…,V₁₀}=10. (6)\n\n────────────────────────────────────────────────────────\n5. Truncations and factor matrices.  \nFix n ≥ 10 and truncate the sequences:\n\n U_r^{(n)}=(U_r(1),…,U_r(n))∈ℝⁿ, V_r^{(n)}=(V_r(1),…,V_r(n))∈ℝⁿ.\n\nSet  \n\n Pₙ = [U_1^{(n)} … U_{10}^{(n)}]  (n×10),  \n Qₙ = [V_1^{(n)} … V_{10}^{(n)}]  (n×10).\n\nTo keep the original signs, introduce the diagonal matrix  \n\n D = diag(1,−1,1,−1,1,1,1,−1,1,1). (7)\n\nWith the ordering chosen in (1) we have the exact factorisation  \n\n Mₙ = Pₙ D Qₙᵀ. (8)\n\n────────────────────────────────────────────────────────\n5.1 Rank of Pₙ.  \nSuppose Σ_{r=1}^{10} c_r U_r^{(n)} = 0.  \nThis means Σ_{r=1}^{10} c_r U_r(j)=0 for j=1,…,n.  \nReplacing each U_r by its exponential representation produces\n\n Σ_{s=1}^{10} d_s e^{it_s j}=0 (j=1,…,n),  \n\nwith distinct t_s in {±α,±β,±γ,±2α,±3β}.  \nBecause n≥10, the square sub-matrix W=(e^{it_s j})_{1≤j,s≤10} has det W≠0 by F2, forcing d_s=0 and therefore c_r=0.  Thus  \n\n rank Pₙ = 10. (9)\n\n────────────────────────────────────────────────────────\n5.2 Rank of Qₙ.  \nThe same argument using the ten frequencies {±β,±γ,±α,±3β,±2γ} gives  \n\n rank Qₙ = 10. (10)\n\n────────────────────────────────────────────────────────\n6. Exact rank of Mₙ (part b).  \nFrom (8) and Sylvester’s inequality,\n\n rank Mₙ ≥ rank (Pₙ D) + rank Qₙ − 10  \n      = rank Pₙ + rank Qₙ − 10 = 10.\n\nBut part (a) gave rank Mₙ ≤ 10, so  \n\n rank Mₙ = 10 for every n ≥ 10. ∎\n\n────────────────────────────────────────────────────────\n7. Singularity for n ≥ 11 (part c).  \nIf n ≥ 11, the n×n matrix Mₙ has rank 10 < n, hence det Mₙ = 0.  Therefore  \n\n Δₙ = 0 for all n ≥ 11. ∎\n\n────────────────────────────────────────────────────────\n8. Limit of Δₙ.  \nBecause Δₙ vanishes from n = 11 onward,  \n\n lim_{n→∞} Δₙ = 0. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.622204",
        "was_fixed": false,
        "difficulty_analysis": "Compared with the original problem (and its existing kernel variant)\nthe enhanced version is decisively more intricate.\n\n1.  Technical proliferation.  Five different trigonometric expressions\ninvolving three incommensurable constants appear in each entry instead\nof a single elementary sine.  The entries therefore look opaque and\npreclude routine “add-one-column-to-another’’ tricks.\n\n2.  Higher rank bookkeeping.  One has to recognise and count all\nseparable terms created when each trigonometric function is expanded,\nthen argue abstractly with rank rather than with explicit numerical\nrelationships.  The argument must track sixteen distinct outer products\ninstead of just two.\n\n3.  Multiple interacting concepts.  The solution blends trigonometric\naddition formulas, low-rank factorizations, and elementary linear-algebraic\nrank considerations; none of these suffices in isolation.\n\n4.  Deeper theoretical requirement.  The crux is to see that summing\nrank-1 matrices keeps the overall rank small—an idea that is\nstraightforward in principle but tricky to spot amid the dense\ntrigonometric clutter.\n\n5.  Subtlety of incommensurability.  Choosing α, β, γ pairwise\nincommensurable forbids accidental periodicity, blocking shortcuts\nsuch as combining columns by constant factors; one must truly rely on\nthe rank argument.\n\nBecause of these layers of complexity, a solver cannot dispatch the\nproblem with the single “rank-two’’ observation that kills the\noriginal exercise; instead she must unravel a much thicker algebraic\nfabric before the same conceptual hammer—low rank—can finally fall."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}