{
  "index": "1949-A-2",
  "type": "GEO",
  "tag": [
    "GEO",
    "ALG",
    "ANA"
  ],
  "difficulty": "",
  "question": "2. We consider three vectors drawn from the same initial point \\( O \\), of lengths \\( a, b \\), and \\( c \\), respectively. Let \\( E \\) be the parallelepiped with vertex \\( O \\) of which the given vectors are the edges and \\( H \\) the parallelepiped with vertex \\( O \\) of which these vectors are the altitudes. Show that the product of the volumes of \\( E \\) and \\( H \\) equals ( \\( a b c)^{2} \\), and generalize the theorem, with proof, to \\( n \\) dimensions.",
  "solution": "Solution. We proceed at once to the general case. Let \\( \\mathbf{v}_{1}, \\mathbf{v}_{2}, \\ldots, \\mathbf{v}_{n} \\) be vectors in an \\( n \\)-dimensional space. To say that these vectors span a parallelepiped \\( P \\) means that they are linearly independent and that\n\\[\nP=\\left\\{\\Sigma \\lambda_{i} \\mathbf{v}_{i}: 0 \\leq \\lambda_{i} \\leq 1\\right\\} .\n\\]\n\nThe volume of the parallelepiped \\( P \\) is \\( |\\operatorname{det} A| \\) where \\( A \\) is the \\( n \\times n \\) matrix whose rows are the components of \\( \\mathbf{v}_{1}, \\mathbf{v}_{2}, \\ldots, \\mathbf{v}_{n} \\) in some Cartesian coordinate system.\n\nSince the \\( \\mathbf{v} \\) 's are linearly independent there are linear functionals \\( f_{1} \\), \\( \\ldots, f_{n} \\) such that\n\\[\nf_{j}\\left(\\mathbf{v}_{i}\\right)=\\delta_{i j}\\left\\|\\mathbf{v}_{i}\\right\\|^{2}\n\\]\nfor all \\( i, j \\), where \\( \\delta_{i j} \\) is the Kronecker delta ( \\( \\delta_{i j}=0 \\) if \\( i \\neq j \\), \\( \\delta_{i i}=1 \\) ). Since every linear functional is realizable with an inner product there are vectors \\( \\mathbf{w}_{1}, \\ldots, \\mathbf{w}_{n} \\) such that\n\\[\n\\left(\\mathbf{v}_{i}, \\mathbf{w}_{i}\\right)=\\delta_{i j}\\left\\|_{\\mathbf{v}_{i}}\\right\\|^{2}\n\\]\nfor all \\( i, j \\). Now the \\( \\mathbf{w} \\) 's are linearly independent, for if \\( \\Sigma \\alpha_{j} \\mathbf{w}_{j}=0 \\), then\n\\[\n\\left(\\mathbf{v}_{i}, \\Sigma \\alpha_{j} \\mathbf{w}_{j}\\right)=\\alpha_{i}\\left\\|\\mathbf{v}_{i}\\right\\|^{2}=0, \\quad i=1,2, \\ldots, n\n\\]\nwhence \\( \\alpha_{1}=\\alpha_{2}=\\cdots=\\alpha_{n}=0 \\).\nTherefore the vectors \\( w_{1}, w_{2}, \\ldots, w_{n} \\) span a parallelepiped \\( Q \\). The vector \\( \\mathbf{v}_{i} \\) is the altitude of \\( Q \\) on the face spanned by all the \\( \\mathbf{w} \\) 's except \\( \\mathbf{w}_{i} \\), since it is perpendicular to that face, and the projection of \\( \\mathbf{w}_{i} \\) in the direction of \\( \\mathbf{v}_{i} \\) has length \\( \\left\\|\\mathbf{v}_{i}\\right\\| \\) by (1).\n\nLet \\( B \\) be the \\( n \\times n \\) matrix whose rows are \\( \\mathrm{w}_{1}, \\mathrm{w}_{2}, \\ldots, \\mathrm{w}_{n} \\). Then \\( \\operatorname{vol} Q=|\\operatorname{det} B|=\\left|\\operatorname{det} B^{T}\\right| \\), where \\( B^{T} \\) is the transpose of \\( B \\).\n\nNow equation (1) shows that \\( A B^{T} \\) is the diagonal matrix\n\\[\n\\operatorname{diag}\\left(\\left\\|\\mathbf{v}_{1}\\right\\|^{2},\\left\\|\\mathbf{v}_{2}\\right\\|^{2}, \\ldots,\\left\\|\\mathbf{v}_{n}\\right\\|^{2}\\right)\n\\]\n\nHence\n\\[\n\\begin{aligned}\n(\\operatorname{vol} P)(\\operatorname{vol} Q) & =|\\operatorname{det} A| \\cdot\\left|\\operatorname{det} B^{T}\\right|=\\left|\\operatorname{det} A B^{T}\\right| \\\\\n& =\\left\\|\\mathbf{v}_{1}\\right\\|^{2} \\cdot\\left\\|\\mathbf{v}_{2}\\right\\|^{2} \\cdots \\cdot\\left\\|\\mathbf{v}_{n}\\right\\|^{2} .\n\\end{aligned}\n\\]\n\nIn the three-dimensional case, the problem calls the parallelepipeds \\( E \\) and \\( H \\) instead of \\( P \\) and \\( Q \\), and, since \\( \\left\\|\\mathrm{v}_{1}\\right\\|=a,\\left\\|\\mathrm{v}_{2}\\right\\|=b \\), and \\( \\left\\|\\mathrm{v}_{3}\\right\\|=c \\), the formula above is the required result for the three-dimensional case.",
  "vars": [
    "O",
    "P",
    "Q",
    "E",
    "H",
    "A",
    "B",
    "v_1",
    "v_2",
    "v_3",
    "v_i",
    "v_n",
    "w_1",
    "w_2",
    "w_i",
    "w_n",
    "f_j",
    "i",
    "j"
  ],
  "params": [
    "a",
    "b",
    "c",
    "n",
    "\\\\delta_ij"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "O": "originpoint",
        "P": "parallelepipedp",
        "Q": "parallelepipedq",
        "E": "parallelepipede",
        "H": "parallelepipedh",
        "A": "matrixa",
        "B": "matrixb",
        "v_1": "vectorone",
        "v_2": "vectortwo",
        "v_3": "vectorthree",
        "v_i": "vectori",
        "v_n": "vectorn",
        "w_1": "dualvectorone",
        "w_2": "dualvectortwo",
        "w_i": "dualvectori",
        "w_n": "dualvectorn",
        "f_j": "functionalj",
        "i": "indexi",
        "j": "indexj",
        "a": "lengtha",
        "b": "lengthb",
        "c": "lengthc",
        "n": "dimensionn",
        "\\\\delta_ij": "kroneckerdelta"
      },
      "question": "2. We consider three vectors drawn from the same initial point \\( originpoint \\), of lengths \\( lengtha, lengthb \\), and \\( lengthc \\), respectively. Let \\( parallelepipede \\) be the parallelepiped with vertex \\( originpoint \\) of which the given vectors are the edges and \\( parallelepipedh \\) the parallelepiped with vertex \\( originpoint \\) of which these vectors are the altitudes. Show that the product of the volumes of \\( parallelepipede \\) and \\( parallelepipedh \\) equals \\( ( lengtha \\; lengthb \\; lengthc )^{2} \\), and generalize the theorem, with proof, to \\( dimensionn \\) dimensions.",
      "solution": "Solution. We proceed at once to the general case. Let \\( \\mathbf{vectorone}, \\mathbf{vectortwo}, \\ldots, \\mathbf{vectorn} \\) be vectors in an \\( dimensionn \\)-dimensional space. To say that these vectors span a parallelepiped \\( parallelepipedp \\) means that they are linearly independent and that\n\\[\nparallelepipedp = \\{ \\Sigma \\lambda_{indexi} \\, \\mathbf{vectori} : 0 \\le \\lambda_{indexi} \\le 1 \\} .\n\\]\n\nThe volume of the parallelepiped \\( parallelepipedp \\) is \\( |\\operatorname{det} matrixa| \\) where \\( matrixa \\) is the \\( dimensionn \\times dimensionn \\) matrix whose rows are the components of \\( \\mathbf{vectorone}, \\mathbf{vectortwo}, \\ldots, \\mathbf{vectorn} \\) in some Cartesian coordinate system.\n\nSince the \\( \\mathbf{vectori} \\)'s are linearly independent there are linear functionals \\( functionalj \\), \\ldots , \\( functionalj \\) such that\n\\[\nfunctionalj(\\mathbf{vectori}) = kroneckerdelta \\, \\| \\mathbf{vectori} \\|^{2}\n\\]\nfor all \\( indexi, indexj \\), where \\( kroneckerdelta \\) is the Kronecker delta (\\( kroneckerdelta = 0 \\) if \\( indexi \\ne indexj \\), \\( kroneckerdelta = 1 \\)). Because every linear functional can be represented by an inner product, there are vectors \\( \\mathbf{dualvectorone}, \\ldots, \\mathbf{dualvectorn} \\) such that\n\\[\n( \\mathbf{vectori}, \\mathbf{dualvectori} ) = kroneckerdelta \\, \\| \\mathbf{vectori} \\|^{2}\n\\]\nfor all \\( indexi, indexj \\). The vectors \\( \\mathbf{dualvectori} \\) are linearly independent, for if \\( \\Sigma \\alpha_{indexj} \\mathbf{dualvectorj} = 0 \\), then\n\\[\n( \\mathbf{vectori}, \\Sigma \\alpha_{indexj} \\mathbf{dualvectorj} ) = \\alpha_{indexi} \\, \\| \\mathbf{vectori} \\|^{2} = 0,\\qquad indexi = 1,2, \\ldots , dimensionn,\n\\]\nwhence \\( \\alpha_{1} = \\alpha_{2} = \\cdots = \\alpha_{dimensionn} = 0 \\).\nTherefore the vectors \\( dualvectorone, dualvectortwo, \\ldots , dualvectorn \\) span a parallelepiped \\( parallelepipedq \\).  The vector \\( \\mathbf{vectori} \\) is the altitude of \\( parallelepipedq \\) on the face spanned by all the \\( \\mathbf{dualvectori} \\)'s except \\( \\mathbf{dualvectori} \\), since it is perpendicular to that face, and the projection of \\( \\mathbf{dualvectori} \\) in the direction of \\( \\mathbf{vectori} \\) has length \\( \\| \\mathbf{vectori} \\| \\).\n\nLet \\( matrixb \\) be the \\( dimensionn \\times dimensionn \\) matrix whose rows are \\( \\mathbf{dualvectorone}, \\mathbf{dualvectortwo}, \\ldots, \\mathbf{dualvectorn} \\). Then \\( \\operatorname{vol} parallelepipedq = |\\operatorname{det} matrixb| = |\\operatorname{det} matrixb^{T}| \\), where \\( matrixb^{T} \\) is the transpose of \\( matrixb \\).\n\nEquation (1) shows that \\( matrixa \\, matrixb^{T} \\) is the diagonal matrix\n\\[\n\\operatorname{diag}( \\| \\mathbf{vectorone} \\|^{2}, \\| \\mathbf{vectortwo} \\|^{2}, \\ldots , \\| \\mathbf{vectorn} \\|^{2} ).\n\\]\nHence\n\\[\n\\begin{aligned}\n( \\operatorname{vol} parallelepipedp )( \\operatorname{vol} parallelepipedq )\n&= |\\operatorname{det} matrixa| \\cdot |\\operatorname{det} matrixb^{T}| = |\\operatorname{det}( matrixa \\, matrixb^{T} )| \\\\ \n&= \\| \\mathbf{vectorone} \\|^{2} \\cdot \\| \\mathbf{vectortwo} \\|^{2} \\cdots \\| \\mathbf{vectorn} \\|^{2}.\n\\end{aligned}\n\\]\n\nIn the three-dimensional case, the problem labels the parallelepipeds \\( parallelepipede \\) and \\( parallelepipedh \\) instead of \\( parallelepipedp \\) and \\( parallelepipedq \\). Since \\( \\| \\mathbf{vectorone} \\| = lengtha, \\| \\mathbf{vectortwo} \\| = lengthb \\), and \\( \\| \\mathbf{vectorthree} \\| = lengthc \\), the above formula gives the required result for three dimensions."
    },
    "descriptive_long_confusing": {
      "map": {
        "O": "blueberry",
        "P": "rainforest",
        "Q": "woodpecker",
        "E": "marshmallow",
        "H": "tortoise",
        "A": "chandelier",
        "B": "stormcloud",
        "v_1": "pineapple",
        "v_2": "hummingbird",
        "v_3": "whirligig",
        "v_i": "sailboat",
        "v_n": "blacksmith",
        "w_1": "grasshopper",
        "w_2": "partridge",
        "w_i": "buttercup",
        "w_n": "kingfisher",
        "f_j": "arrowhead",
        "i": "peppermint",
        "j": "scarecrow",
        "a": "lemonade",
        "b": "toothbrush",
        "c": "horseshoe",
        "n": "dandelion",
        "\\\\delta_ij": "moonlight"
      },
      "question": "2. We consider three vectors drawn from the same initial point \\( blueberry \\), of lengths \\( lemonade, toothbrush \\), and \\( horseshoe \\), respectively. Let \\( marshmallow \\) be the parallelepiped with vertex \\( blueberry \\) of which the given vectors are the edges and \\( tortoise \\) the parallelepiped with vertex \\( blueberry \\) of which these vectors are the altitudes. Show that the product of the volumes of \\( marshmallow \\) and \\( tortoise \\) equals ( \\( lemonade toothbrush horseshoe)^{2} \\), and generalize the theorem, with proof, to \\( dandelion \\) dimensions.",
      "solution": "Solution. We proceed at once to the general case. Let \\( \\mathbf{pineapple}, \\mathbf{hummingbird}, \\ldots, \\mathbf{blacksmith} \\) be vectors in an \\( dandelion \\)-dimensional space. To say that these vectors span a parallelepiped \\( rainforest \\) means that they are linearly independent and that\\n\\[\\nrainforest=\\left\\{\\Sigma \\lambda_{peppermint} \\mathbf{sailboat}: 0 \\leq \\lambda_{peppermint} \\leq 1\\right\\} .\\n\\]\\nThe volume of the parallelepiped \\( rainforest \\) is \\( |\\operatorname{det} chandelier| \\) where \\( chandelier \\) is the \\( dandelion \\times dandelion \\) matrix whose rows are the components of \\( \\mathbf{pineapple}, \\mathbf{hummingbird}, \\ldots, \\mathbf{blacksmith} \\) in some Cartesian coordinate system.\\n\\nSince the \\( \\mathbf{v} \\) 's are linearly independent there are linear functionals \\( f_{1}, \\ldots, f_{dandelion} \\) such that\\n\\[\\nf_{scarecrow}\\left(\\mathbf{sailboat}\\right)=moonlight\\left\\|\\mathbf{sailboat}\\right\\|^{2}\\n\\]for all \\( peppermint, scarecrow \\), where \\( moonlight \\) is the Kronecker delta ( \\( moonlight=0 \\) if \\( peppermint \\neq scarecrow \\), \\( moonlight=1 \\) ). Since every linear functional is realizable with an inner product there are vectors \\( \\mathbf{grasshopper}, \\ldots, \\mathbf{kingfisher} \\) such that\\n\\[\\n\\left(\\mathbf{sailboat}, \\mathbf{buttercup}\\right)=moonlight\\left\\|\\mathbf{sailboat}\\right\\|^{2}\\n\\]for all \\( peppermint, scarecrow \\). Now the \\( \\mathbf{w} \\) 's are linearly independent, for if \\( \\Sigma \\alpha_{scarecrow} \\mathbf{w}_{scarecrow}=0 \\), then\\n\\[\\n\\left(\\mathbf{sailboat}, \\Sigma \\alpha_{scarecrow} \\mathbf{w}_{scarecrow}\\right)=\\alpha_{peppermint}\\left\\|\\mathbf{sailboat}\\right\\|^{2}=0, \\quad peppermint=1,2, \\ldots, dandelion\\n\\]whence \\( \\alpha_{1}=\\alpha_{2}=\\cdots=\\alpha_{dandelion}=0 \\). Therefore the vectors grasshopper, partridge, \\ldots, kingfisher span a parallelepiped \\( woodpecker \\). The vector \\( \\mathbf{sailboat} \\) is the altitude of \\( woodpecker \\) on the face spanned by all the \\( \\mathbf{w} \\) 's except \\( \\mathbf{buttercup} \\), since it is perpendicular to that face, and the projection of \\( \\mathbf{buttercup} \\) in the direction of \\( \\mathbf{sailboat} \\) has length \\( \\left\\|\\mathbf{sailboat}\\right\\| \\) by (1).\\n\\nLet \\( stormcloud \\) be the \\( dandelion \\times dandelion \\) matrix whose rows are \\( \\mathrm{grasshopper}, \\mathrm{partridge}, \\ldots, \\mathrm{kingfisher} \\). Then \\( \\operatorname{vol} woodpecker=|\\operatorname{det} stormcloud|=\\left|\\operatorname{det} stormcloud^{T}\\right| \\), where \\( stormcloud^{T} \\) is the transpose of \\( stormcloud \\).\\n\\nNow equation (1) shows that \\( chandelier\\,stormcloud^{T} \\) is the diagonal matrix\\n\\[\\n\\operatorname{diag}\\left(\\left\\|\\mathbf{pineapple}\\right\\|^{2},\\left\\|\\mathbf{hummingbird}\\right\\|^{2}, \\ldots, \\left\\|\\mathbf{blacksmith}\\right\\|^{2}\\right)\\n\\]\\nHence\\n\\[\\n\\begin{aligned}\\n(\\operatorname{vol} rainforest)(\\operatorname{vol} woodpecker) & =|\\operatorname{det} chandelier| \\cdot \\left|\\operatorname{det} stormcloud^{T}\\right| = \\left|\\operatorname{det} (chandelier\\,stormcloud^{T})\\right| \\\\ & =\\left\\|\\mathbf{pineapple}\\right\\|^{2} \\cdot \\left\\|\\mathbf{hummingbird}\\right\\|^{2} \\cdots \\cdot \\left\\|\\mathbf{blacksmith}\\right\\|^{2} .\\n\\end{aligned}\\n\\]\\nIn the three-dimensional case, the problem calls the parallelepipeds \\( marshmallow \\) and \\( tortoise \\) instead of \\( rainforest \\) and \\( woodpecker \\), and, since \\( \\left\\|\\mathrm{pineapple}\\right\\|=lemonade, \\left\\|\\mathrm{hummingbird}\\right\\|=toothbrush \\), and \\( \\left\\|\\mathrm{whirligig}\\right\\|=horseshoe \\), the formula above is the required result for the three-dimensional case."
    },
    "descriptive_long_misleading": {
      "map": {
        "O": "abysspoint",
        "P": "vacuumshape",
        "Q": "voidsolid",
        "E": "emptybox",
        "H": "hollowshell",
        "A": "antiarray",
        "B": "blankmatrix",
        "v_1": "stillvector",
        "v_2": "calmvector",
        "v_3": "hushvector",
        "v_i": "idlevector",
        "v_n": "inertvector",
        "w_1": "quietvector",
        "w_2": "mutevector",
        "w_i": "numbvector",
        "w_n": "nullvector",
        "f_j": "silentfunction",
        "i": "constantix",
        "j": "unchanging",
        "a": "shortside",
        "b": "briefside",
        "c": "smallside",
        "n": "singledim",
        "\\delta_ij": "nonkronecker"
      },
      "question": "2. We consider three vectors drawn from the same initial point \\( abysspoint \\), of lengths \\( shortside, briefside \\), and \\( smallside \\), respectively. Let \\( emptybox \\) be the parallelepiped with vertex \\( abysspoint \\) of which the given vectors are the edges and \\( hollowshell \\) the parallelepiped with vertex \\( abysspoint \\) of which these vectors are the altitudes. Show that the product of the volumes of \\( emptybox \\) and \\( hollowshell \\) equals ( \\( shortside briefside smallside)^{2} \\), and generalize the theorem, with proof, to \\( singledim \\) dimensions.",
      "solution": "Solution. We proceed at once to the general case. Let \\( \\mathbf{stillvector}_{1}, \\mathbf{calmvector}_{2}, \\ldots, \\mathbf{inertvector}_{singledim} \\) be vectors in an \\( singledim \\)-dimensional space. To say that these vectors span a parallelepiped \\( vacuumshape \\) means that they are linearly independent and that\n\\[\nvacuumshape=\\left\\{\\Sigma \\lambda_{constantix} \\mathbf{idlevector}_{constantix}: 0 \\leq \\lambda_{constantix} \\leq 1\\right\\} .\n\\]\n\nThe volume of the parallelepiped \\( vacuumshape \\) is \\( |\\operatorname{det} antiarray| \\) where \\( antiarray \\) is the \\( singledim \\times singledim \\) matrix whose rows are the components of \\( \\mathbf{stillvector}_{1}, \\mathbf{calmvector}_{2}, \\ldots, \\mathbf{inertvector}_{singledim} \\) in some Cartesian coordinate system.\n\nSince the \\( \\mathbf{v} \\) 's are linearly independent there are linear functionals \\( silentfunction_{1}, \\ldots, silentfunction_{singledim} \\) such that\n\\[\nsilentfunction_{unchanging}\\left(\\mathbf{idlevector}_{constantix}\\right)=nonkronecker_{constantix\\ unchanging}\\left\\|\\mathbf{idlevector}_{constantix}\\right\\|^{2}\n\\]\nfor all \\( constantix, unchanging \\), where \\( nonkronecker_{constantix\\ unchanging} \\) is the Kronecker delta ( \\( nonkronecker_{constantix\\ unchanging}=0 \\) if \\( constantix \\neq unchanging \\), \\( nonkronecker_{constantix\\ constantix}=1 \\) ). Since every linear functional is realizable with an inner product there are vectors \\( \\mathbf{quietvector}_{1}, \\ldots, \\mathbf{nullvector}_{singledim} \\) such that\n\\[\n\\left(\\mathbf{idlevector}_{constantix}, \\mathbf{numbvector}_{constantix}\\right)=nonkronecker_{constantix\\ unchanging}\\left\\|\\mathbf{idlevector}_{constantix}\\right\\|^{2}\n\\]\nfor all \\( constantix, unchanging \\). Now the \\( \\mathbf{w} \\) 's are linearly independent, for if \\( \\Sigma \\alpha_{unchanging} \\mathbf{numbvector}_{unchanging}=0 \\), then\n\\[\n\\left(\\mathbf{idlevector}_{constantix}, \\Sigma \\alpha_{unchanging} \\mathbf{numbvector}_{unchanging}\\right)=\\alpha_{constantix}\\left\\|\\mathbf{idlevector}_{constantix}\\right\\|^{2}=0, \\quad constantix=1,2, \\ldots, singledim\n\\]\nwhence \\( \\alpha_{1}=\\alpha_{2}=\\cdots=\\alpha_{singledim}=0 \\).\nTherefore the vectors \\( quietvector_{1}, quietvector_{2}, \\ldots, quietvector_{singledim} \\) span a parallelepiped \\( voidsolid \\). The vector \\( \\mathbf{idlevector}_{constantix} \\) is the altitude of \\( voidsolid \\) on the face spanned by all the \\( \\mathbf{numbvector} \\) 's except \\( \\mathbf{numbvector}_{constantix} \\), since it is perpendicular to that face, and the projection of \\( \\mathbf{numbvector}_{constantix} \\) in the direction of \\( \\mathbf{idlevector}_{constantix} \\) has length \\( \\left\\|\\mathbf{idlevector}_{constantix}\\right\\| \\) by (1).\n\nLet \\( blankmatrix \\) be the \\( singledim \\times singledim \\) matrix whose rows are \\( \\mathrm{quietvector}_{1}, \\mathrm{mutevector}_{2}, \\ldots, \\mathrm{nullvector}_{singledim} \\). Then \\( \\operatorname{vol} voidsolid=|\\operatorname{det} blankmatrix|=\\left|\\operatorname{det} blankmatrix^{T}\\right| \\), where \\( blankmatrix^{T} \\) is the transpose of \\( blankmatrix \\).\n\nNow equation (1) shows that \\( antiarray blankmatrix^{T} \\) is the diagonal matrix\n\\[\n\\operatorname{diag}\\left(\\left\\|\\mathbf{stillvector}_{1}\\right\\|^{2},\\left\\|\\mathbf{calmvector}_{2}\\right\\|^{2}, \\ldots,\\left\\|\\mathbf{inertvector}_{singledim}\\right\\|^{2}\\right)\n\\]\n\nHence\n\\[\n\\begin{aligned}\n(\\operatorname{vol} vacuumshape)(\\operatorname{vol} voidsolid) & =|\\operatorname{det} antiarray| \\cdot\\left|\\operatorname{det} blankmatrix^{T}\\right|=\\left|\\operatorname{det} antiarray blankmatrix^{T}\\right| \\\\\n& =\\left\\|\\mathbf{stillvector}_{1}\\right\\|^{2} \\cdot\\left\\|\\mathbf{calmvector}_{2}\\right\\|^{2} \\cdots \\cdot\\left\\|\\mathbf{inertvector}_{singledim}\\right\\|^{2} .\n\\end{aligned}\n\\]\n\nIn the three-dimensional case, the problem calls the parallelepipeds \\( emptybox \\) and \\( hollowshell \\) instead of \\( vacuumshape \\) and \\( voidsolid \\), and, since \\( \\left\\|\\mathrm{stillvector}_{1}\\right\\|=shortside,\\left\\|\\mathrm{calmvector}_{2}\\right\\|=briefside \\), and \\( \\left\\|\\mathrm{hushvector}_{3}\\right\\|=smallside \\), the formula above is the required result for the three-dimensional case."
    },
    "garbled_string": {
      "map": {
        "O": "mavncytq",
        "P": "zlbyrken",
        "Q": "rksvumeq",
        "E": "jahyxdeo",
        "H": "qopnezal",
        "A": "tpqirfux",
        "B": "xevmanor",
        "v_1": "exybdhom",
        "v_2": "krqadlen",
        "v_3": "phormdaz",
        "v_i": "bjraxuln",
        "v_n": "kwhumelo",
        "w_1": "vchoymsd",
        "w_2": "gtwlepor",
        "w_i": "yzqfanid",
        "w_n": "swnekplo",
        "f_j": "dlygirca",
        "i": "hrqstlva",
        "j": "mpowzcre",
        "a": "cvtebqsl",
        "b": "znofkjla",
        "c": "wexparcd",
        "n": "lycaspem",
        "\\\\delta_ij": "zlwqmsnf"
      },
      "question": "2. We consider three vectors drawn from the same initial point \\( mavncytq \\), of lengths \\( cvtebqsl, znofkjla \\), and \\( wexparcd \\), respectively. Let \\( jahyxdeo \\) be the parallelepiped with vertex \\( mavncytq \\) of which the given vectors are the edges and \\( qopnezal \\) the parallelepiped with vertex \\( mavncytq \\) of which these vectors are the altitudes. Show that the product of the volumes of \\( jahyxdeo \\) and \\( qopnezal \\) equals ( \\( cvtebqsl znofkjla wexparcd)^{2} \\), and generalize the theorem, with proof, to \\( lycaspem \\) dimensions.",
      "solution": "Solution. We proceed at once to the general case. Let \\( \\mathbf{exybdhom}, \\mathbf{krqadlen}, \\ldots, \\mathbf{kwhumelo} \\) be vectors in an \\( lycaspem \\)-dimensional space. To say that these vectors span a parallelepiped \\( zlbyrken \\) means that they are linearly independent and that\n\\[\nzlbyrken=\\left\\{\\Sigma \\lambda_{hrqstlva} \\mathbf{bjraxuln}: 0 \\leq \\lambda_{hrqstlva} \\leq 1\\right\\} .\n\\]\n\nThe volume of the parallelepiped \\( zlbyrken \\) is \\( |\\operatorname{det} tpqirfux| \\) where \\( tpqirfux \\) is the \\( lycaspem \\times lycaspem \\) matrix whose rows are the components of \\( \\mathbf{exybdhom}, \\mathbf{krqadlen}, \\ldots, \\mathbf{kwhumelo} \\) in some Cartesian coordinate system.\n\nSince the \\( \\mathbf{v} \\) 's are linearly independent there are linear functionals \\( dlygirca_{1} \\), \\( \\ldots, dlygirca_{lycaspem} \\) such that\n\\[\ndlygirca_{mpowzcre}\\left(\\mathbf{bjraxuln}\\right)=zlwqmsnf\\left\\|\\mathbf{bjraxuln}\\right\\|^{2}\n\\]\nfor all \\( hrqstlva, mpowzcre \\), where \\( zlwqmsnf \\) is the Kronecker delta ( \\( zlwqmsnf=0 \\) if \\( hrqstlva \\neq mpowzcre \\), \\( zlwqmsnf=1 \\) ). Since every linear functional is realizable with an inner product there are vectors \\( \\mathbf{vchoymsd}, \\ldots, \\mathbf{swnekplo} \\) such that\n\\[\n\\left(\\mathbf{bjraxuln}, \\mathbf{yzqfanid}\\right)=zlwqmsnf\\left\\|_{\\mathbf{bjraxuln}}\\right\\|^{2}\n\\]\nfor all \\( hrqstlva, mpowzcre \\). Now the \\( \\mathbf{w} \\) 's are linearly independent, for if \\( \\Sigma \\alpha_{mpowzcre} \\mathbf{yzqfanid}=0 \\), then\n\\[\n\\left(\\mathbf{bjraxuln}, \\Sigma \\alpha_{mpowzcre} \\mathbf{yzqfanid}\\right)=\\alpha_{hrqstlva}\\left\\|\\mathbf{bjraxuln}\\right\\|^{2}=0, \\quad hrqstlva=1,2, \\ldots, lycaspem\n\\]\nwhence \\( \\alpha_{1}=\\alpha_{2}=\\cdots=\\alpha_{lycaspem}=0 \\).\nTherefore the vectors \\( vchoymsd, gtwlepor, \\ldots, swnekplo \\) span a parallelepiped \\( rksvumeq \\). The vector \\( \\mathbf{bjraxuln} \\) is the altitude of \\( rksvumeq \\) on the face spanned by all the \\( \\mathbf{w} \\) 's except \\( \\mathbf{yzqfanid} \\), since it is perpendicular to that face, and the projection of \\( \\mathbf{yzqfanid} \\) in the direction of \\( \\mathbf{bjraxuln} \\) has length \\( \\left\\|\\mathbf{bjraxuln}\\right\\| \\) by (1).\n\nLet \\( xevmanor \\) be the \\( lycaspem \\times lycaspem \\) matrix whose rows are \\( \\mathrm{vchoymsd}, \\mathrm{gtwlepor}, \\ldots, \\mathrm{swnekplo} \\). Then \\( \\operatorname{vol} rksvumeq=|\\operatorname{det} xevmanor|=\\left|\\operatorname{det} xevmanor^{T}\\right| \\), where \\( xevmanor^{T} \\) is the transpose of \\( xevmanor \\).\n\nNow equation (1) shows that \\( tpqirfux xevmanor^{T} \\) is the diagonal matrix\n\\[\n\\operatorname{diag}\\left(\\left\\|\\mathbf{exybdhom}\\right\\|^{2},\\left\\|\\mathbf{krqadlen}\\right\\|^{2}, \\ldots,\\left\\|\\mathbf{kwhumelo}\\right\\|^{2}\\right)\n\\]\n\nHence\n\\[\n\\begin{aligned}\n(\\operatorname{vol} zlbyrken)(\\operatorname{vol} rksvumeq) & =|\\operatorname{det} tpqirfux| \\cdot\\left|\\operatorname{det} xevmanor^{T}\\right|=\\left|\\operatorname{det} tpqirfux xevmanor^{T}\\right| \\\\\n& =\\left\\|\\mathbf{exybdhom}\\right\\|^{2} \\cdot\\left\\|\\mathbf{krqadlen}\\right\\|^{2} \\cdots \\cdot\\left\\|\\mathbf{kwhumelo}\\right\\|^{2} .\n\\end{aligned}\n\\]\n\nIn the three-dimensional case, the problem calls the parallelepipeds \\( jahyxdeo \\) and \\( qopnezal \\) instead of \\( zlbyrken \\) and \\( rksvumeq \\), and, since \\( \\left\\|\\mathrm{exybdhom}\\right\\|=cvtebqsl,\\left\\|\\mathrm{krqadlen}\\right\\|=znofkjla \\), and \\( \\left\\|\\mathrm{phormdaz}\\right\\|=wexparcd \\), the formula above is the required result for the three-dimensional case."
    },
    "kernel_variant": {
      "question": "Fix once and for all  \n\n• a real Euclidean vector space (V,⟨·,·⟩) of dimension n ≥ 2,  \n• the origin O ∈ V, and  \n• a positively oriented orthonormal basis 𝔈 = (e₁,…,e_n).\n\nRow convention.  \nFor an ordered basis (frame)  \n  F = (v₁,…,v_n)  (v_i≠0)  \nwrite the vectors as ROWS of the matrix  \n  A = Mat_𝔈(F) ∈ GL_n(ℝ),  A = (v₁^T;…;v_n^T).  \n\nIts row Gram matrix is G := A A^T, and the squared edge–lengths form the diagonal matrix  \n  D := diag(ℓ₁²,…,ℓ_n²) (with ℓ_i := ‖v_i‖>0).\n\nDefinition (altitude operator).  \nGiven A ∈ GL_n(ℝ) let Alt(A)=:B be the unique matrix whose i-th ROW w_i^T satisfies  \n (1) w_i ⊥ span{v_j : j≠i},  (2) ⟨v_i,w_i⟩ = ℓ_i².\n\n(The w_i are the altitudes from O of the parallelotope with edges v_i.)\n\nProblems.\n\n(a)  Prove the matrix identity Alt(A)=D A^{-T}.  \n\n(b)  Denoting by K(F) the edge parallelotope with edges v_i and by L(F) the altitude parallelotope with edges Alt(F), show  \n  Vol_n(K(F))·Vol_n(L(F)) = (ℓ₁ℓ₂⋯ℓ_n)²   (remember that Vol_n(P)=|det Mat_𝔈(P)|).  \n\n(c)  Put W:=Alt(A)=D A^{-T}.  For m_i:=‖w_i‖ set E:=diag(m₁²,…,m_n²).\n\n(i)  Show Alt²(A)=E W^{-T}=S(A)·A with  \n  S(A):=E D^{-1}=diag(s₁,…,s_n), s_i=ℓ_i²·(G^{-1})_{ii}>0.\n\n(ii)  Prove det S(A)=∏ s_i=(∏ m_i²)/(∏ ℓ_i²).  \n\n(iii)  Deduce det Alt²(A)=det S(A)·det A.\n\n(d)  (Shape–scalar frames)  \nShow the equivalence of  \n (i) Alt²(A)=λ(A)·A for some λ(A)>0;  \n (ii) S(A)=λ(A)·I_n (i.e. s₁=⋯=s_n);  \n (iii) G_{ii}(G^{-1})_{ii} is independent of i.  \n\nMoreover λ(A)=s_i, hence λ(A)ⁿ=det S(A).\n\n(e)  Projectivised frame space.  \nLet ℙ(GL_n):=GL_n/ℝ^× (right action by non-zero scalars) and define the rational map  \n  ẐAlt : ℙ(GL_n) ⇢ ℙ(GL_n), [A]↦[Alt(A)].  \n\n(i)  Prove that ẐAlt is dominant.  \n\n(ii)  Show that a projective class [A] is fixed by ẐAlt  \n    iff Alt(A)=A  \n    iff S(A)=I_n  iff A A^T=D.  \nThus the fixed locus consists precisely of the classes of frames whose vectors are pairwise orthogonal (no restriction on their lengths).\n\n(f)  Specialise to n = 3.  \nClassify all ordered bases F for which Alt(F)=R·F with a proper rotation R ∈ SO(3).  \nShow that this occurs iff the basis vectors are pairwise orthogonal, i.e. A A^T=D, in which case necessarily Alt(F)=F and R=I_3.",
      "solution": "Notation is taken from the statement: A=(v₁^T;…;v_n^T), G=AA^T, D=diag(ℓ_i²) and Alt(A)=(w₁^T;…;w_n^T).\n\n(a)  Altitude matrix.  \nConditions (1)–(2) are equivalent to the single matrix equation  \n  A · Alt(A)^T = D. (∗)\n\nBecause A ∈ GL_n(ℝ), multiplying (∗) from the left by A^{-1} gives Alt(A)=D A^{-T}.  \nThus Alt is a rational self-map of GL_n.\n\n(b)  Product of volumes.  \nVol_n(K(F)) = |det A| and, by (a), Alt(A)=D A^{-T}; hence  \n det Alt(A)=det(D A^{-T})=(ℓ₁²⋯ℓ_n²)/(det A).  \nTherefore  \n Vol_n(K(F))·Vol_n(L(F))  \n = |det A|·|det Alt(A)|  \n = |det A|·|(ℓ₁²⋯ℓ_n²)/(det A)|  \n = (ℓ₁ℓ₂⋯ℓ_n)².\n\n(c-i)  Second altitude.  \nSet W = Alt(A) = D A^{-T}.  Re-applying (a) to W gives  \n Alt²(A)=Alt(W)=E W^{-T}.  \nFor every i  \n w_i^T = ℓ_i² e_i^T A^{-T},  \nso  \n m_i² = ℓ_i⁴·(A^{-T}A^{-1})_{ii}=ℓ_i⁴·(G^{-1})_{ii}.  \nHence  \n S(A):=E D^{-1}=diag(s_i) with s_i=ℓ_i²·(G^{-1})_{ii}>0,  \nand Alt²(A)=S(A)·A.\n\n(c-ii)  Since S(A) is diagonal, det S(A)=∏ s_i=(∏ m_i²)/(∏ ℓ_i²).\n\n(c-iii)  From Alt²(A)=S(A)A we get det Alt²(A)=det S(A)·det A.\n\n(d)  Shape–scalar frames.  \n(i)⇒(ii) Comparing Alt²(A)=λA with Alt²(A)=S(A)A yields S(A)=λ I_n.  \n(ii)⇒(iii) Equality of the diagonal entries of S(A) means s_i is constant, i.e. G_{ii}(G^{-1})_{ii} is independent of i.  \n(iii)⇒(i) If G_{ii}(G^{-1})_{ii}=c then s_i=c and S(A)=c I_n, so Alt²(A)=c A.  \nFinally λ(A)=s_i and λ(A)^n = det S(A) by (c-ii).\n\n(e)  Projective altitude map.  \n(i)  Dominance.  \nCompute the differential of Alt at the identity I.  \nLet H ∈ 𝔤𝔩_n(ℝ) and put A(ε)=I+εH.  Up to O(ε²) one finds  \n Alt(A(ε)) = I + ε( diag(H+H^T) − H^T ) + O(ε²).  \nDefine L(H):=diag(H+H^T) − H^T.  \nSolving L(H)=Y:  \n • If i≠j, Y_{ij}=−H_{ji}  ⇒ H_{ji}=−Y_{ij}.  \n • If i=j, Y_{ii}=H_{ii}.  \nThus L is bijective, its rank is n² and Alt is a submersion at I.  \nConsequently its projectivisation ẐAlt is dominant.\n\n(ii)  Fixed points.  \nIf [A] is fixed, Alt(A)=c A for some c∈ℝ^×.  Plugging into (∗) gives D A^{-T}=c A ⇒ D=c G.  \nTaking diagonal entries yields ℓ_i²=c ℓ_i² ⇒ c=1.  Hence Alt(A)=A and G=D.  \nConversely, G=D ⇒ Alt(A)=A and then S(A)=I_n.  \nThus the three conditions are equivalent, and the fixed locus is exactly the set of projective classes of pairwise-orthogonal frames.\n\n(f)  The three–dimensional case.  \nAssume Alt(F)=R F with some R∈SO(3); write A for the frame matrix.  Using (a):  \n D A^{-T}=R A ⇒ D = R G. (1)\n\n(i)  Orthogonality of the basis via Hadamard.  \nBoth D and G are symmetric positive-definite and have the same diagonal because diag G = diag D = (ℓ₁²,ℓ₂²,ℓ₃²).  \nHadamard’s inequality says det G ≤ ∏ G_{ii} = det D, with equality iff G is diagonal.  \nTaking determinants in (1) gives det D = det G, so equality holds in Hadamard’s inequality and therefore G is diagonal.  But a positive-definite diagonal matrix with the same diagonal as D must equal D; hence G=D and the vectors v_i are pairwise orthogonal.\n\n(ii)  Identification of R.  \nWith G=D equation (1) becomes D = R D.  Because D is invertible we can multiply by D^{-1} on the right to obtain R = I_3 outright; no further diagonalisation considerations are needed.  Thus Alt(F)=F.\n\nConversely, if the vectors are pairwise orthogonal (i.e. G=D) then Alt(F)=F, so Alt(F)=R F with R=I_3, which is indeed a proper rotation.\n\nTherefore Alt(F)=R F with R∈SO(3) occurs precisely when the basis vectors are pairwise orthogonal, and in that situation R=I_3.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.423458",
        "was_fixed": false,
        "difficulty_analysis": "•  Additional structures.  The problem introduces the \\emph{altitude operator} as a self-map of \\(\\mathrm{GL}_{n}(\\mathbb R)\\) and asks for its algebraic properties, creating a bridge between geometry of parallelotopes and linear algebra on the full general linear group.\n\n•  Multiple interacting concepts.  Parts (a)–(e) combine geometric reasoning (orthogonality, volumes), exterior algebra (Gram determinants), linear algebra (matrix inverses and transposes), group actions (projectivisation, rotations) and dynamical aspects (iteration of a nonlinear operator).\n\n•  Deeper theoretical requirements.  One has to manipulate Gram matrices, determinants, involutive rational maps, and perform a fixed–point classification—none of which appears in the original or first kernel variant.\n\n•  Higher technical load.  Whereas the original statement stops after a single volume identity, the enhanced variant demands (i) an explicit matrix formula, (ii) derivation of a second-iterate formula, (iii) projective-geometric interpretation, and (iv) a classification problem in dimension three.\n\n•  Non-trivial proofs.  Parts (c)–(e) cannot be settled by a direct determinant computation alone; they require recognising hidden symmetries, using the polar decomposition of matrices, and engaging with the structure of \\(\\mathrm{SO}(3)\\).\n\nHence the enhanced kernel variant is significantly more sophisticated and longer than both the original problem and the previous kernel version."
      }
    },
    "original_kernel_variant": {
      "question": "Fix once and for all  \n\n• a real Euclidean vector space (V,⟨·,·⟩) of dimension n ≥ 2,  \n• the origin O ∈ V, and  \n• a positively oriented orthonormal basis 𝔈 = (e₁,…,e_n).\n\nRow convention.  \nFor an ordered basis (frame)  \n  F = (v₁,…,v_n)  (v_i≠0)  \nwrite the vectors as ROWS of the matrix  \n  A = Mat_𝔈(F) ∈ GL_n(ℝ),  A = (v₁^T;…;v_n^T).  \n\nIts row Gram matrix is G := A A^T, and the squared edge–lengths form the diagonal matrix  \n  D := diag(ℓ₁²,…,ℓ_n²) (with ℓ_i := ‖v_i‖>0).\n\nDefinition (altitude operator).  \nGiven A ∈ GL_n(ℝ) let Alt(A)=:B be the unique matrix whose i-th ROW w_i^T satisfies  \n (1) w_i ⊥ span{v_j : j≠i},  (2) ⟨v_i,w_i⟩ = ℓ_i².\n\n(The w_i are the altitudes from O of the parallelotope with edges v_i.)\n\nProblems.\n\n(a)  Prove the matrix identity Alt(A)=D A^{-T}.  \n\n(b)  Denoting by K(F) the edge parallelotope with edges v_i and by L(F) the altitude parallelotope with edges Alt(F), show  \n  Vol_n(K(F))·Vol_n(L(F)) = (ℓ₁ℓ₂⋯ℓ_n)²   (remember that Vol_n(P)=|det Mat_𝔈(P)|).  \n\n(c)  Put W:=Alt(A)=D A^{-T}.  For m_i:=‖w_i‖ set E:=diag(m₁²,…,m_n²).\n\n(i)  Show Alt²(A)=E W^{-T}=S(A)·A with  \n  S(A):=E D^{-1}=diag(s₁,…,s_n), s_i=ℓ_i²·(G^{-1})_{ii}>0.\n\n(ii)  Prove det S(A)=∏ s_i=(∏ m_i²)/(∏ ℓ_i²).  \n\n(iii)  Deduce det Alt²(A)=det S(A)·det A.\n\n(d)  (Shape–scalar frames)  \nShow the equivalence of  \n (i) Alt²(A)=λ(A)·A for some λ(A)>0;  \n (ii) S(A)=λ(A)·I_n (i.e. s₁=⋯=s_n);  \n (iii) G_{ii}(G^{-1})_{ii} is independent of i.  \n\nMoreover λ(A)=s_i, hence λ(A)ⁿ=det S(A).\n\n(e)  Projectivised frame space.  \nLet ℙ(GL_n):=GL_n/ℝ^× (right action by non-zero scalars) and define the rational map  \n  ẐAlt : ℙ(GL_n) ⇢ ℙ(GL_n), [A]↦[Alt(A)].  \n\n(i)  Prove that ẐAlt is dominant.  \n\n(ii)  Show that a projective class [A] is fixed by ẐAlt  \n    iff Alt(A)=A  \n    iff S(A)=I_n  iff A A^T=D.  \nThus the fixed locus consists precisely of the classes of frames whose vectors are pairwise orthogonal (no restriction on their lengths).\n\n(f)  Specialise to n = 3.  \nClassify all ordered bases F for which Alt(F)=R·F with a proper rotation R ∈ SO(3).  \nShow that this occurs iff the basis vectors are pairwise orthogonal, i.e. A A^T=D, in which case necessarily Alt(F)=F and R=I_3.",
      "solution": "Notation is taken from the statement: A=(v₁^T;…;v_n^T), G=AA^T, D=diag(ℓ_i²) and Alt(A)=(w₁^T;…;w_n^T).\n\n(a)  Altitude matrix.  \nConditions (1)–(2) are equivalent to the single matrix equation  \n  A · Alt(A)^T = D. (∗)\n\nBecause A ∈ GL_n(ℝ), multiplying (∗) from the left by A^{-1} gives Alt(A)=D A^{-T}.  \nThus Alt is a rational self-map of GL_n.\n\n(b)  Product of volumes.  \nVol_n(K(F)) = |det A| and, by (a), Alt(A)=D A^{-T}; hence  \n det Alt(A)=det(D A^{-T})=(ℓ₁²⋯ℓ_n²)/(det A).  \nTherefore  \n Vol_n(K(F))·Vol_n(L(F))  \n = |det A|·|det Alt(A)|  \n = |det A|·|(ℓ₁²⋯ℓ_n²)/(det A)|  \n = (ℓ₁ℓ₂⋯ℓ_n)².\n\n(c-i)  Second altitude.  \nSet W = Alt(A) = D A^{-T}.  Re-applying (a) to W gives  \n Alt²(A)=Alt(W)=E W^{-T}.  \nFor every i  \n w_i^T = ℓ_i² e_i^T A^{-T},  \nso  \n m_i² = ℓ_i⁴·(A^{-T}A^{-1})_{ii}=ℓ_i⁴·(G^{-1})_{ii}.  \nHence  \n S(A):=E D^{-1}=diag(s_i) with s_i=ℓ_i²·(G^{-1})_{ii}>0,  \nand Alt²(A)=S(A)·A.\n\n(c-ii)  Since S(A) is diagonal, det S(A)=∏ s_i=(∏ m_i²)/(∏ ℓ_i²).\n\n(c-iii)  From Alt²(A)=S(A)A we get det Alt²(A)=det S(A)·det A.\n\n(d)  Shape–scalar frames.  \n(i)⇒(ii) Comparing Alt²(A)=λA with Alt²(A)=S(A)A yields S(A)=λ I_n.  \n(ii)⇒(iii) Equality of the diagonal entries of S(A) means s_i is constant, i.e. G_{ii}(G^{-1})_{ii} is independent of i.  \n(iii)⇒(i) If G_{ii}(G^{-1})_{ii}=c then s_i=c and S(A)=c I_n, so Alt²(A)=c A.  \nFinally λ(A)=s_i and λ(A)^n = det S(A) by (c-ii).\n\n(e)  Projective altitude map.  \n(i)  Dominance.  \nCompute the differential of Alt at the identity I.  \nLet H ∈ 𝔤𝔩_n(ℝ) and put A(ε)=I+εH.  Up to O(ε²) one finds  \n Alt(A(ε)) = I + ε( diag(H+H^T) − H^T ) + O(ε²).  \nDefine L(H):=diag(H+H^T) − H^T.  \nSolving L(H)=Y:  \n • If i≠j, Y_{ij}=−H_{ji}  ⇒ H_{ji}=−Y_{ij}.  \n • If i=j, Y_{ii}=H_{ii}.  \nThus L is bijective, its rank is n² and Alt is a submersion at I.  \nConsequently its projectivisation ẐAlt is dominant.\n\n(ii)  Fixed points.  \nIf [A] is fixed, Alt(A)=c A for some c∈ℝ^×.  Plugging into (∗) gives D A^{-T}=c A ⇒ D=c G.  \nTaking diagonal entries yields ℓ_i²=c ℓ_i² ⇒ c=1.  Hence Alt(A)=A and G=D.  \nConversely, G=D ⇒ Alt(A)=A and then S(A)=I_n.  \nThus the three conditions are equivalent, and the fixed locus is exactly the set of projective classes of pairwise-orthogonal frames.\n\n(f)  The three–dimensional case.  \nAssume Alt(F)=R F with some R∈SO(3); write A for the frame matrix.  Using (a):  \n D A^{-T}=R A ⇒ D = R G. (1)\n\n(i)  Orthogonality of the basis via Hadamard.  \nBoth D and G are symmetric positive-definite and have the same diagonal because diag G = diag D = (ℓ₁²,ℓ₂²,ℓ₃²).  \nHadamard’s inequality says det G ≤ ∏ G_{ii} = det D, with equality iff G is diagonal.  \nTaking determinants in (1) gives det D = det G, so equality holds in Hadamard’s inequality and therefore G is diagonal.  But a positive-definite diagonal matrix with the same diagonal as D must equal D; hence G=D and the vectors v_i are pairwise orthogonal.\n\n(ii)  Identification of R.  \nWith G=D equation (1) becomes D = R D.  Because D is invertible we can multiply by D^{-1} on the right to obtain R = I_3 outright; no further diagonalisation considerations are needed.  Thus Alt(F)=F.\n\nConversely, if the vectors are pairwise orthogonal (i.e. G=D) then Alt(F)=F, so Alt(F)=R F with R=I_3, which is indeed a proper rotation.\n\nTherefore Alt(F)=R F with R∈SO(3) occurs precisely when the basis vectors are pairwise orthogonal, and in that situation R=I_3.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.367678",
        "was_fixed": false,
        "difficulty_analysis": "•  Additional structures.  The problem introduces the \\emph{altitude operator} as a self-map of \\(\\mathrm{GL}_{n}(\\mathbb R)\\) and asks for its algebraic properties, creating a bridge between geometry of parallelotopes and linear algebra on the full general linear group.\n\n•  Multiple interacting concepts.  Parts (a)–(e) combine geometric reasoning (orthogonality, volumes), exterior algebra (Gram determinants), linear algebra (matrix inverses and transposes), group actions (projectivisation, rotations) and dynamical aspects (iteration of a nonlinear operator).\n\n•  Deeper theoretical requirements.  One has to manipulate Gram matrices, determinants, involutive rational maps, and perform a fixed–point classification—none of which appears in the original or first kernel variant.\n\n•  Higher technical load.  Whereas the original statement stops after a single volume identity, the enhanced variant demands (i) an explicit matrix formula, (ii) derivation of a second-iterate formula, (iii) projective-geometric interpretation, and (iv) a classification problem in dimension three.\n\n•  Non-trivial proofs.  Parts (c)–(e) cannot be settled by a direct determinant computation alone; they require recognising hidden symmetries, using the polar decomposition of matrices, and engaging with the structure of \\(\\mathrm{SO}(3)\\).\n\nHence the enhanced kernel variant is significantly more sophisticated and longer than both the original problem and the previous kernel version."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}