{
  "index": "1995-A-5",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "(real-valued) functions of a single variable $f$ which satisfy\n\\begin{align*}\n\\frac{dx_{1}}{dt} &= a_{11}x_{1} + a_{12}x_{2} + \\cdots +\na_{1n}x_{n} \\\\\n\\frac{dx_{2}}{dt} &= a_{21}x_{1} + a_{22}x_{2} + \\cdots +\na_{2n}x_{n} \\\\\n\\vdots && \\vdots \\\\\n\\frac{dx_{n}}{dt} &= a_{n1}x_{1} + a_{n2}x_{2} + \\cdots +\na_{nn}x_{n}\n\\end{align*}\nfor some constants $a_{ij}>0$. Suppose that for all $i$, $x_{i}(t)\n\\to 0$ as $t \\to \\infty$. Are the functions $x_{1},x_{2},\\dots,x_{n}$\nnecessarily linearly dependent?",
  "solution": "Everyone (presumably) knows that the set of solutions of a system of\nlinear first-order differential equations with constant coefficients\nis $n$-dimensional, with basis vectors of the form $f_{i}(t)\n\\vec{v}_{i}$ (i.e.\\ a function times a constant vector), where the\n$\\vec{v}_{i}$ are linearly independent. In\nparticular, our solution $\\vec{x}(t)$ can be written as $\\sum_{i=1}^{n}\nc_{i}f_{i}(t) \\vec{v}_{1}$.\n\nChoose a vector $\\vec{w}$ orthogonal to $\\vec{v}_{2}, \\dots,\n\\vec{v}_{n}$ but not to $\\vec{v}_1$. Since $\\vec{x}(t) \\to 0$ as $t\n\\to \\infty$, the same is true of $\\vec{w} \\cdot \\vec{x}$; but that is\nsimply $(\\vec{w} \\cdot \\vec{v}_{1}) c_{1} f_{1}(t)$. In other words,\nif $c_{i} \\neq 0$, then $f_{i}(t)$ must also go to 0.\n\nHowever, it is easy to exhibit a solution which does not go to 0. The\nsum of the eigenvalues of the matrix $A = (a_{ij})$, also known as the\ntrace of $A$, being the sum of the diagonal entries of $A$, is\nnonnegative, so $A$ has an eigenvalue $\\lambda$ with nonnegative real\npart, and a corresponding eigenvector $\\vec{v}$. Then $e^{\\lambda t}\n\\vec{v}$ is a solution that does not go to 0. (If $\\lambda$ is not\nreal, add this solution to its complex conjugate to get a real\nsolution, which still doesn't go to 0.)\n\nHence one of the $c_{i}$, say $c_{1}$, is zero, in which case\n$\\vec{x}(t) \\cdot \\vec{w} = 0$ for all $t$.",
  "vars": [
    "x_1",
    "x_2",
    "x_n",
    "x_i",
    "f",
    "f_i",
    "t"
  ],
  "params": [
    "a_11",
    "a_12",
    "a_1n",
    "a_21",
    "a_22",
    "a_2n",
    "a_n1",
    "a_nn",
    "a_ij",
    "n",
    "c_i",
    "c_1",
    "A",
    "w",
    "v_i",
    "v_1",
    "i",
    "j",
    "\\\\lambda"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x_1": "firstvar",
        "x_2": "secondvar",
        "x_n": "lastxvar",
        "x_i": "xindexvar",
        "f": "funcvar",
        "f_i": "funcindex",
        "t": "timevar",
        "a_11": "coefoneone",
        "a_12": "coefonetwo",
        "a_1n": "coefonelast",
        "a_21": "coeftwoone",
        "a_22": "coeftwotwo",
        "a_2n": "coeftwolast",
        "a_n1": "coeflastone",
        "a_nn": "coeflastlast",
        "a_ij": "coefgenij",
        "n": "dimension",
        "c_i": "constindex",
        "c_1": "constone",
        "A": "matrixa",
        "w": "vectorw",
        "v_i": "vectorindex",
        "v_1": "vectorone",
        "i": "indexi",
        "j": "indexj",
        "\\lambda": "eigenlambda"
      },
      "question": "(real-valued) functions of a single variable $funcvar$ which satisfy\n\\begin{align*}\n\\frac{d firstvar}{d timevar} &= coefoneone firstvar + coefonetwo secondvar + \\cdots + coefonelast lastxvar \\\\\n\\frac{d secondvar}{d timevar} &= coeftwoone firstvar + coeftwotwo secondvar + \\cdots + coeftwolast lastxvar \\\\\n\\vdots && \\vdots \\\\\n\\frac{d lastxvar}{d timevar} &= coeflastone firstvar + a_{n2} secondvar + \\cdots + coeflastlast lastxvar\n\\end{align*}\nfor some constants $coefgenij>0$. Suppose that for all $indexi$, $xindexvar(timevar) \\to 0$ as $timevar \\to \\infty$. Are the functions $firstvar, secondvar,\\dots,lastxvar$ necessarily linearly dependent?",
      "solution": "Everyone (presumably) knows that the set of solutions of a system of linear first-order differential equations with constant coefficients is $dimension$-dimensional, with basis vectors of the form $funcindex(timevar) \\vec{vectorindex}$ (i.e.\\ a function times a constant vector), where the $\\vec{vectorindex}$ are linearly independent. In particular, our solution $\\vec{x}(timevar)$ can be written as $\\sum_{indexi=1}^{dimension} constindex funcindex(timevar) \\vec{vectorone}$.\n\nChoose a vector $\\vec{vectorw}$ orthogonal to $\\vec{v}_{2}, \\dots, \\vec{v}_{dimension}$ but not to $\\vec{vectorone}$. Since $\\vec{x}(timevar) \\to 0$ as $timevar \\to \\infty$, the same is true of $\\vec{vectorw} \\cdot \\vec{x}$; but that is simply $(\\vec{vectorw} \\cdot \\vec{vectorone}) constone funcindex(timevar)$. In other words, if $constindex \\neq 0$, then $funcindex(timevar)$ must also go to 0.\n\nHowever, it is easy to exhibit a solution which does not go to 0. The sum of the eigenvalues of the matrix $matrixa = (coefgenij)$, also known as the trace of $matrixa$, being the sum of the diagonal entries of $matrixa$, is nonnegative, so $matrixa$ has an eigenvalue $eigenlambda$ with nonnegative real part, and a corresponding eigenvector $\\vec{v}$. Then $e^{eigenlambda timevar} \\vec{v}$ is a solution that does not go to 0. (If $eigenlambda$ is not real, add this solution to its complex conjugate to get a real solution, which still doesn't go to 0.)\n\nHence one of the $constindex$, say $constone$, is zero, in which case $\\vec{x}(timevar) \\cdot \\vec{vectorw} = 0$ for all $timevar$."
    },
    "descriptive_long_confusing": {
      "map": {
        "x_1": "marigolds",
        "x_2": "pinecones",
        "x_n": "lampstand",
        "x_i": "drumstick",
        "f": "ploughman",
        "f_i": "sugarcane",
        "t": "paperclip",
        "a_11": "raincloud",
        "a_12": "peppermint",
        "a_1n": "toothpick",
        "a_21": "binoculars",
        "a_22": "beachball",
        "a_2n": "gooseberry",
        "a_n1": "firebrick",
        "a_nn": "thumbtack",
        "a_ij": "strawberries",
        "n": "shoelaces",
        "c_i": "snowflake",
        "c_1": "honeycomb",
        "A": "masterplan",
        "w": "hairbrush",
        "v_i": "candlestick",
        "v_1": "goldfinch",
        "i": "paintwork",
        "j": "sandpiper",
        "\\lambda": "buttercup"
      },
      "question": "(real-valued) functions of a single variable $ploughman$ which satisfy\n\\begin{align*}\n\\frac{dmarigolds}{dpaperclip} &= raincloud marigolds + peppermint pinecones + \\cdots +\ntoothpick lampstand \\\\\n\\frac{dpinecones}{dpaperclip} &= binoculars marigolds + beachball pinecones + \\cdots +\ngooseberry lampstand \\\\\n\\vdots && \\vdots \\\\\n\\frac{dlampstand}{dpaperclip} &= firebrick marigolds + a_{n2} pinecones + \\cdots +\nthumbtack lampstand\n\\end{align*}\nfor some constants $strawberries>0$. Suppose that for all $paintwork$, $drumstick(paperclip)\n\\to 0$ as $paperclip \\to \\infty$. Are the functions $marigolds,pinecones,\\dots,lampstand$\nnecessarily linearly dependent?",
      "solution": "Everyone (presumably) knows that the set of solutions of a system of\nlinear first-order differential equations with constant coefficients\nis $shoelaces$-dimensional, with basis vectors of the form $sugarcane(paperclip)\n\\vec{candlestick}$ (i.e.\\ a function times a constant vector), where the\n$\\vec{candlestick}$ are linearly independent. In\nparticular, our solution $\\vec{x}(paperclip)$ can be written as $\\sum_{paintwork=1}^{shoelaces}\nsnowflake sugarcane(paperclip) \\vec{goldfinch}$.\n\nChoose a vector $\\vec{hairbrush}$ orthogonal to $\\vec{v}_{2}, \\dots,\n\\vec{v}_{n}$ but not to $\\vec{goldfinch}$. Since $\\vec{x}(paperclip) \\to 0$ as $paperclip\n\\to \\infty$, the same is true of $\\vec{hairbrush} \\cdot \\vec{x}$; but that is\nsimply $(\\vec{hairbrush} \\cdot \\vec{goldfinch}) honeycomb ploughman_{1}(paperclip)$. In other words,\nif $snowflake \\neq 0$, then $sugarcane(paperclip)$ must also go to 0.\n\nHowever, it is easy to exhibit a solution which does not go to 0. The\nsum of the eigenvalues of the matrix $masterplan = (strawberries)$, also known as the\ntrace of $masterplan$, being the sum of the diagonal entries of $masterplan$, is\nnonnegative, so $masterplan$ has an eigenvalue $buttercup$ with nonnegative real\npart, and a corresponding eigenvector $\\vec{v}$. Then $e^{buttercup paperclip}\n\\vec{v}$ is a solution that does not go to 0. (If $buttercup$ is not\nreal, add this solution to its complex conjugate to get a real\nsolution, which still doesn't go to 0.)\n\nHence one of the $snowflake$, say $honeycomb$, is zero, in which case\n$\\vec{x}(paperclip) \\cdot \\vec{hairbrush} = 0$ for all $paperclip$. "
    },
    "descriptive_long_misleading": {
      "map": {
        "x_1": "constantone",
        "x_2": "constanttwo",
        "x_n": "constantend",
        "x_i": "constantindex",
        "f": "fixedfunc",
        "f_i": "fixedindex",
        "t": "timeless",
        "a_11": "negativeone",
        "a_12": "negativetwo",
        "a_1n": "negativeend",
        "a_21": "negativealpha",
        "a_22": "negativebeta",
        "a_2n": "negativegamma",
        "a_n1": "negativedelta",
        "a_nn": "negativetheta",
        "a_ij": "negativepair",
        "n": "singularnum",
        "c_i": "nullindex",
        "c_1": "nullsingle",
        "A": "voidmatrix",
        "w": "emptiness",
        "v_i": "scalarindex",
        "v_1": "scalarsingle",
        "i": "outsider",
        "j": "bystander",
        "\\lambda": "positiveroot"
      },
      "question": "(real-valued) functions of a single variable fixedfunc which satisfy\n\\begin{align*}\n\\frac{dconstantone}{dtimeless} &= negativeone constantone + negativetwo constanttwo + \\cdots +\nnegativeend constantend \\\\\n\\frac{dconstanttwo}{dtimeless} &= negativealpha constantone + negativebeta constanttwo + \\cdots +\nnegativegamma constantend \\\\\n\\vdots && \\vdots \\\\\n\\frac{dconstantend}{dtimeless} &= negativedelta constantone + a_{n2} constanttwo + \\cdots +\nnegativetheta constantend\n\\end{align*}\nfor some constants negativepair>0. Suppose that for all outsider, constantindex(timeless)\n\\to 0 as timeless \\to \\infty. Are the functions constantone, constanttwo, \\dots, constantend\nnecessarily linearly dependent?",
      "solution": "Everyone (presumably) knows that the set of solutions of a system of\nlinear first-order differential equations with constant coefficients\nis singularnum-dimensional, with basis vectors of the form fixedindex(timeless)\n\\vec{scalarindex} (i.e.\\ a function times a constant vector), where the\n\\vec{scalarindex} are linearly independent. In\nparticular, our solution \\vec{x}(timeless) can be written as \\sum_{outsider=1}^{singularnum}\n nullindex fixedindex(timeless) \\vec{scalarsingle}.\n\nChoose a vector \\vec{emptiness} orthogonal to \\vec{v}_{2}, \\dots,\n\\vec{v}_{n} but not to \\vec{scalarsingle}. Since \\vec{x}(timeless) \\to 0 as timeless\n\\to \\infty, the same is true of \\vec{emptiness} \\cdot \\vec{x}; but that is\nsimply (\\vec{emptiness} \\cdot \\vec{scalarsingle}) nullsingle fixedfunc(timeless). In other words,\nif nullindex \\neq 0, then fixedindex(timeless) must also go to 0.\n\nHowever, it is easy to exhibit a solution which does not go to 0. The\nsum of the eigenvalues of the matrix voidmatrix = (negativepair), also known as the\ntrace of voidmatrix, being the sum of the diagonal entries of voidmatrix, is\nnonnegative, so voidmatrix has an eigenvalue positiveroot with nonnegative real\npart, and a corresponding eigenvector \\vec{v}. Then e^{positiveroot timeless}\n\\vec{v} is a solution that does not go to 0. (If positiveroot is not\nreal, add this solution to its complex conjugate to get a real\nsolution, which still doesn't go to 0.)\n\nHence one of the nullindex, say nullsingle, is zero, in which case\n\\vec{x}(timeless) \\cdot \\vec{emptiness} = 0 for all timeless."
    },
    "garbled_string": {
      "map": {
        "x_1": "qzxwvtnp",
        "x_2": "hjgrksla",
        "x_n": "mbcxdfeo",
        "x_i": "klmnrsop",
        "f": "sdqwejkl",
        "f_i": "plokmjnh",
        "t": "zxcvbnma",
        "a_11": "lkjhgfds",
        "a_12": "poiuytre",
        "a_1n": "qazwsxed",
        "a_21": "edcrfvtg",
        "a_22": "yhnujmki",
        "a_2n": "ujmnhygt",
        "a_n1": "ikmjuynh",
        "a_nn": "mnbvcxzq",
        "a_ij": "rfvgytbh",
        "n": "oikjuhtg",
        "c_i": "tgbyhnuj",
        "c_1": "vfrtgbhy",
        "A": "bgtfrved",
        "w": "njuhtgfr",
        "v_i": "cdewsxza",
        "v_1": "plmkoijn",
        "i": "lpoikmnj",
        "j": "mkjiolpn",
        "\\\\lambda": "asdfghjk"
      },
      "question": "(real-valued) functions of a single variable $sdqwejkl$ which satisfy\n\\begin{align*}\n\\frac{d qzxwvtnp}{d zxcvbnma} &= lkjhgfds qzxwvtnp + poiuytre hjgrksla + \\cdots +\nqazwsxed mbcxdfeo \\\\\n\\frac{d hjgrksla}{d zxcvbnma} &= edcrfvtg qzxwvtnp + yhnujmki hjgrksla + \\cdots +\nujmnhygt mbcxdfeo \\\\\n\\vdots && \\vdots \\\\\n\\frac{d mbcxdfeo}{d zxcvbnma} &= ikmjuynh qzxwvtnp + a_{n2} hjgrksla + \\cdots +\nmnbvcxzq mbcxdfeo\n\\end{align*}\nfor some constants $rfvgytbh>0$. Suppose that for all $lpoikmnj$, $klmnrsop(zxcvbnma)\n\\to 0$ as $zxcvbnma \\to \\infty$. Are the functions $qzxwvtnp,hjgrksla,\\dots,mbcxdfeo$\nnecessarily linearly dependent?",
      "solution": "Everyone (presumably) knows that the set of solutions of a system of\nlinear first-order differential equations with constant coefficients\nis $oikjuhtg$-dimensional, with basis vectors of the form $plokmjnh(zxcvbnma)\n\\vec{cdewsxza}$ (i.e.\\ a function times a constant vector), where the\n$\\vec{cdewsxza}$ are linearly independent. In\nparticular, our solution $\\vec{x}(zxcvbnma)$ can be written as $\\sum_{lpoikmnj=1}^{oikjuhtg}\ntgbyhnuj plokmjnh(zxcvbnma) \\vec{plmkoijn}$.\n\nChoose a vector $\\vec{njuhtgfr}$ orthogonal to $\\vec{cdewsxza}, \\dots,\n\\vec{v}_{n}$ but not to $\\vec{plmkoijn}$. Since $\\vec{x}(zxcvbnma) \\to 0$ as $zxcvbnma\n\\to \\infty$, the same is true of $\\vec{njuhtgfr} \\cdot \\vec{x}$; but that is\nsimply $(\\vec{njuhtgfr} \\cdot \\vec{plmkoijn}) vfrtgbhy plokmjnh(zxcvbnma)$. In other words,\nif $tgbyhnuj \\neq 0$, then $plokmjnh(zxcvbnma)$ must also go to 0.\n\nHowever, it is easy to exhibit a solution which does not go to 0. The\nsum of the eigenvalues of the matrix $bgtfrved = (rfvgytbh)$, also known as the\ntrace of $bgtfrved$, being the sum of the diagonal entries of $bgtfrved$, is\nnonnegative, so $bgtfrved$ has an eigenvalue $asdfghjk$ with nonnegative real\npart, and a corresponding eigenvector $\\vec{v}$. Then $e^{asdfghjk zxcvbnma}\n\\vec{v}$ is a solution that does not go to 0. (If $asdfghjk$ is not\nreal, add this solution to its complex conjugate to get a real\nsolution, which still doesn't go to 0.)\n\nHence one of the $tgbyhnuj$, say $vfrtgbhy$, is zero, in which case\n$\\vec{x}(zxcvbnma) \\cdot \\vec{njuhtgfr} = 0$ for all $zxcvbnma$. "
    },
    "kernel_variant": {
      "question": "Let n ≥ 2 and let  \n  A : ℝ ⟶ Mₙ(ℂ)  \nbe a 2π–periodic C¹ matrix–valued map such that every pair of matrices of the\nfamily commutes,  \n  A(s)A(t) = A(t)A(s)  for all s , t ∈ ℝ.  (∗)\n\nDenote by Φ(t) the principal fundamental matrix of the linear system  \n\n  x′(t) = A(t) x(t), t ≥ 0,  x(0)=x₀ ∈ ℂⁿ,  \n\nnormalised by Φ(0)=Iₙ, and put  \n\n  M := Φ(2π)                             (monodromy matrix).\n\nAssume that the spectral radius of M is strictly larger than one,  \n\n  ρ(M) = max{|μ| : μ ∈ σ(M)} > 1.  (1)\n\nLet x : [0,∞) ⟶ ℂⁿ be any C¹–solution whose coordinates all decay,  \n\n  lim_{t→∞} x_i(t) = 0  (i = 1,…,n).  (2)\n\nProve that the coordinate functions x₁,…,x_n are linearly dependent over ℂ; i.e. show that there exist complex constants c₁,…,c_n, not all zero, such that  \n\n  c₁ x₁(t) + ⋯ + c_n x_n(t) ≡ 0  for every t ≥ 0.\n\n(The commutativity hypothesis (∗) is automatic, for example, when A(t) is diagonal for every t, or when A(t)=f(t)B with a fixed matrix B and a scalar\nfunction f.)\n\n--------------------------------------------------------------------",
      "solution": "Step 0 – Consequences of the commutativity assumption.  \nBecause all A(t) commute, so do their integrals; hence\n\n  Φ(t)=exp (∫₀ᵗ A(τ)dτ)  for all t∈ℝ,             (3)\n\nand Φ(t) commutes with Φ(s) for every s,t.  In particular\n\n  Φ(t) M = M Φ(t)  for all t.                     (4)\n\nPut  \n\n  B := (1/2π) ∫₀^{2π} A(τ)dτ.  \n\nOwing to (3) we have  \n\n  M = Φ(2π) = exp(2π B).                           (5)\n\nStep 1 – Unstable Floquet multiplier.  \nFrom (1) choose μ∈σ(M) with |μ|>1 and fix a non–zero left eigenvector\n\n  w* M = μ w*.                                     (6)\n\nWrite w for the column-conjugate transpose of w*, i.e. regard w as a constant\nrow vector.\n\nStep 2 – A functional equation.  \nFor the given decaying solution set  \n\n  f(t):= w·x(t)=w* x(t).                           (7)\n\nBecause x(t)=Φ(t)c with c:=x₀, we obtain, using (4) and (6),\n\n  f(t+2π) = w* Φ(t+2π)c\n           = w* Φ(t)M c            (by (4))\n           = μ w* Φ(t)c\n           = μ f(t).                                (8)\n\nThus f satisfies the linear functional equation\n\n  f(t+2π)=μ f(t).                                 (9)\n\nStep 3 – Decay forces f≡0.  \nAssume, for a contradiction, that f is not identically zero; choose t₀≥0 with\nf(t₀)≠0.  Iterating (9) gives\n\n  f(t₀+2πk)=μ^{\\,k}f(t₀), k=0,1,2,….              (10)\n\nTaking absolute values and using |μ|>1 we find |f(t₀+2πk)|→∞ as k→∞,\ncontradicting the fact that each coordinate of x – and hence every constant\nlinear combination of them – tends to 0 by (2).  Therefore\n\n  f(t)≡0  for all t≥0.                            (11)\n\nStep 4 – Linear dependence of the coordinates.  \nWrite w=(c₁,…,c_n).  Then (11) is precisely\n\n  c₁ x₁(t)+⋯+c_n x_n(t) ≡ 0  (t≥0).               (12)\n\nBecause w* is a non-zero left eigenvector, the vector  \n(c₁,…,c_n) is not the zero vector, so the coordinate functions are linearly\ndependent, completing the proof. ∎\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.742390",
        "was_fixed": false,
        "difficulty_analysis": "1. Non-autonomous system: The coefficient matrix now depends on t, so\n   simple eigen-decomposition of a constant matrix no longer works.\n   One must invoke Floquet theory and handle a periodic fundamental\n   matrix decomposition.\n2. Spectral condition on the monodromy matrix replaces “trace > 0” and\n   requires understanding of Floquet multipliers and exponents—concepts\n   absent from the original exercise.\n3. The proof uses left eigenvectors of the monodromy matrix and a\n   functional equation f(t+T)=μ f(t) whose analysis is more delicate\n   than the exponential estimate for constant-coefficient systems.\n4. Because of the periodic factor P(t) in Φ(t), the naive attempt to\n   copy the original argument fails; a new idea (working with the\n   monodromy matrix and its eigenvectors) is essential.\n5. The learner must combine several advanced tools—Floquet theory,\n   spectral radius considerations, and functional iteration—to obtain\n   the desired linear dependence.  These additional layers of theory\n   and the subtler logical steps make the enhanced variant\n   significantly harder than both the original problem and the current\n   kernel version."
      }
    },
    "original_kernel_variant": {
      "question": "Let n ≥ 2 and let  \n  A : ℝ ⟶ Mₙ(ℂ)  \nbe a 2π–periodic C¹ matrix–valued map such that every pair of matrices of the\nfamily commutes,  \n  A(s)A(t) = A(t)A(s)  for all s , t ∈ ℝ.  (∗)\n\nDenote by Φ(t) the principal fundamental matrix of the linear system  \n\n  x′(t) = A(t) x(t), t ≥ 0,  x(0)=x₀ ∈ ℂⁿ,  \n\nnormalised by Φ(0)=Iₙ, and put  \n\n  M := Φ(2π)                             (monodromy matrix).\n\nAssume that the spectral radius of M is strictly larger than one,  \n\n  ρ(M) = max{|μ| : μ ∈ σ(M)} > 1.  (1)\n\nLet x : [0,∞) ⟶ ℂⁿ be any C¹–solution whose coordinates all decay,  \n\n  lim_{t→∞} x_i(t) = 0  (i = 1,…,n).  (2)\n\nProve that the coordinate functions x₁,…,x_n are linearly dependent over ℂ; i.e. show that there exist complex constants c₁,…,c_n, not all zero, such that  \n\n  c₁ x₁(t) + ⋯ + c_n x_n(t) ≡ 0  for every t ≥ 0.\n\n(The commutativity hypothesis (∗) is automatic, for example, when A(t) is diagonal for every t, or when A(t)=f(t)B with a fixed matrix B and a scalar\nfunction f.)\n\n--------------------------------------------------------------------",
      "solution": "Step 0 – Consequences of the commutativity assumption.  \nBecause all A(t) commute, so do their integrals; hence\n\n  Φ(t)=exp (∫₀ᵗ A(τ)dτ)  for all t∈ℝ,             (3)\n\nand Φ(t) commutes with Φ(s) for every s,t.  In particular\n\n  Φ(t) M = M Φ(t)  for all t.                     (4)\n\nPut  \n\n  B := (1/2π) ∫₀^{2π} A(τ)dτ.  \n\nOwing to (3) we have  \n\n  M = Φ(2π) = exp(2π B).                           (5)\n\nStep 1 – Unstable Floquet multiplier.  \nFrom (1) choose μ∈σ(M) with |μ|>1 and fix a non–zero left eigenvector\n\n  w* M = μ w*.                                     (6)\n\nWrite w for the column-conjugate transpose of w*, i.e. regard w as a constant\nrow vector.\n\nStep 2 – A functional equation.  \nFor the given decaying solution set  \n\n  f(t):= w·x(t)=w* x(t).                           (7)\n\nBecause x(t)=Φ(t)c with c:=x₀, we obtain, using (4) and (6),\n\n  f(t+2π) = w* Φ(t+2π)c\n           = w* Φ(t)M c            (by (4))\n           = μ w* Φ(t)c\n           = μ f(t).                                (8)\n\nThus f satisfies the linear functional equation\n\n  f(t+2π)=μ f(t).                                 (9)\n\nStep 3 – Decay forces f≡0.  \nAssume, for a contradiction, that f is not identically zero; choose t₀≥0 with\nf(t₀)≠0.  Iterating (9) gives\n\n  f(t₀+2πk)=μ^{\\,k}f(t₀), k=0,1,2,….              (10)\n\nTaking absolute values and using |μ|>1 we find |f(t₀+2πk)|→∞ as k→∞,\ncontradicting the fact that each coordinate of x – and hence every constant\nlinear combination of them – tends to 0 by (2).  Therefore\n\n  f(t)≡0  for all t≥0.                            (11)\n\nStep 4 – Linear dependence of the coordinates.  \nWrite w=(c₁,…,c_n).  Then (11) is precisely\n\n  c₁ x₁(t)+⋯+c_n x_n(t) ≡ 0  (t≥0).               (12)\n\nBecause w* is a non-zero left eigenvector, the vector  \n(c₁,…,c_n) is not the zero vector, so the coordinate functions are linearly\ndependent, completing the proof. ∎\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.573790",
        "was_fixed": false,
        "difficulty_analysis": "1. Non-autonomous system: The coefficient matrix now depends on t, so\n   simple eigen-decomposition of a constant matrix no longer works.\n   One must invoke Floquet theory and handle a periodic fundamental\n   matrix decomposition.\n2. Spectral condition on the monodromy matrix replaces “trace > 0” and\n   requires understanding of Floquet multipliers and exponents—concepts\n   absent from the original exercise.\n3. The proof uses left eigenvectors of the monodromy matrix and a\n   functional equation f(t+T)=μ f(t) whose analysis is more delicate\n   than the exponential estimate for constant-coefficient systems.\n4. Because of the periodic factor P(t) in Φ(t), the naive attempt to\n   copy the original argument fails; a new idea (working with the\n   monodromy matrix and its eigenvectors) is essential.\n5. The learner must combine several advanced tools—Floquet theory,\n   spectral radius considerations, and functional iteration—to obtain\n   the desired linear dependence.  These additional layers of theory\n   and the subtler logical steps make the enhanced variant\n   significantly harder than both the original problem and the current\n   kernel version."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}