{
  "index": "1940-B-6",
  "type": "ALG",
  "tag": [
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "14. Prove that\n\\[\n\\left(\\begin{array}{lllll}\na_{1}^{2}+k & a_{1} a_{2} & a_{1} a_{3} & \\ldots & a_{1} a_{n} \\\\\na_{2} a_{1} & a_{2}^{2}+k & a_{2} a_{3} & \\ldots & a_{2} a_{n} \\\\\n\\ldots & \\ldots & \\ldots & \\ldots & \\ldots \\\\\na_{n} a_{1} & a_{n} a_{2} & a_{n} a_{3} & \\ldots & a_{n}^{2}+k\n\\end{array}\\right)\n\\]\nis divisible by \\( \\boldsymbol{k}^{n-1} \\) and find its other factor.",
  "solution": "First Solution. Let \\( B \\) be the matrix\n\\[\n\\left(\\begin{array}{lllll}\na_{1}^{2} & a_{1} a_{2} & a_{1} a_{3} & \\cdots & a_{1} a_{n} \\\\\na_{2} a_{1} & a_{2}^{2} & a_{2} a_{3} & \\cdots & a_{2} a_{n} \\\\\n\\cdots & \\cdots & \\cdots & \\cdots & \\cdots \\\\\na_{n} a_{1} & a_{n} a_{2} & a_{n} a_{3} & \\cdots & a_{n}^{2}\n\\end{array}\\right)\n\\]\n\\( B \\) has rank at most one, since any two rows (or columns) are clearly dependent. So there are \\( (n-1) \\) zeros among the eigenvalues of \\( B \\). Therefore the characteristic polynomial of \\( B \\) is divisible by \\( x^{n-1} \\). Hence\n\\[\n\\begin{aligned}\n\\operatorname{det}(x \\cdot I-B) & =x^{n}-(\\operatorname{trace} B) x^{n-1} \\\\\n& =x^{n-1}\\left(x-a_{1}{ }^{2}-a_{2}{ }^{2}-\\cdots-a_{n}{ }^{2}\\right)\n\\end{aligned}\n\\]\nso\n\\[\n\\begin{aligned}\n\\operatorname{det}(B+k I) & =(-1)^{n} \\operatorname{det}(-k I-B) \\\\\n& =k^{n-1}\\left(k+a_{1}{ }^{2}+a_{2}{ }^{2}+\\cdots+a_{n}{ }^{2}\\right)\n\\end{aligned}\n\\]\nand the other factor is \\( \\left(k+a_{1}{ }^{2}+a_{2}{ }^{2}+\\cdots+a_{n}{ }^{2}\\right) \\).\nSecond Solution. Assume for a moment that none of the \\( a \\) 's are zero, and let\n\\[\nB_{n}=\\left(\\begin{array}{lllll}\na_{1}{ }^{2}+k & a_{1} a_{2} & a_{1} a_{3} & \\cdots & a_{1} a_{n} \\\\\na_{2} a_{1} & a_{2}{ }^{2}+k & a_{2} a_{3} & \\cdots & a_{2} a_{n} \\\\\n\\cdots & \\cdots & \\cdots & \\cdots & \\cdots \\\\\na_{n} a_{1} & a_{n} a_{2} & a_{n} a_{3} & \\cdots & a_{n}{ }^{2}+k\n\\end{array}\\right)\n\\]\n\nSince the determinant is linear in the last row, we find\n\nNow in the first of these new determinants, subtract multiples of the last rows from the others to get\n\\[\n\\operatorname{det}\\left(\\begin{array}{lll:l}\nk & & 0 & 0 \\\\\n0 & k & k & 0 \\\\\n\\hdashline a_{n} a_{1} & a_{n} a_{n-1} & a_{n}^{2}\n\\end{array}\\right) .\n\\]\n\nThen we have\n\\[\n\\operatorname{det} B_{n}=k^{n-1} a_{n}^{2}+k \\operatorname{det} B_{n-1}\n\\]\n\nSince \\( \\operatorname{det} B_{1}=k+a_{1}{ }^{2} \\), the relation\n\\[\n\\operatorname{det} B_{n}=k^{n-1}\\left(k+a_{1}^{2}+\\cdots+a_{n}^{2}\\right)\n\\]\nfollows easily by induction.\nAlthough this derivation depends on the assumption that the \\( a \\) 's are not zero, the result remains valid for the case where some of the \\( a \\) 's are zero, since \\( D_{n} \\) is evidently some polynomial in \\( k \\) and the \\( a \\) 's which agrees with \\( k^{n-1}\\left(k+a_{1}{ }^{2}+\\cdots+a_{n}{ }^{2}\\right) \\) as long as none of the \\( a \\) 's are zero. Therefore (1) must be a polynomial identity.\n\nAlternatively, we can regard the computation as taking place in the field \\( Q\\left(k, a_{1}, a_{2}, \\ldots, a_{n}\\right) \\) where \\( k \\) and the \\( a_{i} \\) are independent indeterminates. In this field the condition \\( a_{i} \\neq 0 \\) is satisfied.\n\nFor a discussion of such fields, see I. N. Herstein, Topics in Algebra, Blaisdell, Waltham, Mass., 1964.",
  "vars": [
    "x",
    "a",
    "a_1",
    "a_2",
    "a_3",
    "a_n-1",
    "a_n",
    "a_i"
  ],
  "params": [
    "k",
    "n",
    "B",
    "I",
    "B_n",
    "B_n-1",
    "D_n",
    "Q"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "charvar",
        "a": "genericco",
        "a_1": "coeffone",
        "a_2": "coefftwo",
        "a_3": "coeffthr",
        "a_n-1": "coeffnmin",
        "a_n": "coefflast",
        "a_i": "coeffindi",
        "k": "constk",
        "n": "sizenum",
        "B": "matrixb",
        "I": "identmx",
        "B_n": "matrixbn",
        "B_n-1": "matrixbnm",
        "D_n": "determin",
        "Q": "fieldrat"
      },
      "question": "14. Prove that\n\\[\n\\left(\\begin{array}{lllll}\ncoeffone^{2}+constk & coeffone coefftwo & coeffone coeffthr & \\ldots & coeffone coefflast \\\\\ncoefftwo coeffone & coefftwo^{2}+constk & coefftwo coeffthr & \\ldots & coefftwo coefflast \\\\\n\\ldots & \\ldots & \\ldots & \\ldots & \\ldots \\\\\ncoefflast coeffone & coefflast coefftwo & coefflast coeffthr & \\ldots & coefflast^{2}+constk\n\\end{array}\\right)\n\\]\nis divisible by \\( \\boldsymbol{constk}^{sizenum-1} \\) and find its other factor.",
      "solution": "First Solution. Let \\( matrixb \\) be the matrix\n\\[\n\\left(\\begin{array}{lllll}\ncoeffone^{2} & coeffone coefftwo & coeffone coeffthr & \\cdots & coeffone coefflast \\\\\ncoefftwo coeffone & coefftwo^{2} & coefftwo coeffthr & \\cdots & coefftwo coefflast \\\\\n\\cdots & \\cdots & \\cdots & \\cdots & \\cdots \\\\\ncoefflast coeffone & coefflast coefftwo & coefflast coeffthr & \\cdots & coefflast^{2}\n\\end{array}\\right)\n\\]\n\\( matrixb \\) has rank at most one, since any two rows (or columns) are clearly dependent. So there are \\( (sizenum-1) \\) zeros among the eigenvalues of \\( matrixb \\). Therefore the characteristic polynomial of \\( matrixb \\) is divisible by \\( charvar^{sizenum-1} \\). Hence\n\\[\n\\begin{aligned}\n\\operatorname{det}(charvar \\cdot identmx-matrixb) & =charvar^{sizenum}-(\\operatorname{trace} matrixb) charvar^{sizenum-1} \\\\\n& =charvar^{sizenum-1}\\left(charvar-coeffone^{2}-coefftwo^{2}-\\cdots-coefflast^{2}\\right)\n\\end{aligned}\n\\]\nso\n\\[\n\\begin{aligned}\n\\operatorname{det}(matrixb+constk\\,identmx) & =(-1)^{sizenum} \\, \\operatorname{det}(-constk\\,identmx-matrixb) \\\\\n& =constk^{sizenum-1}\\left(constk+coeffone^{2}+coefftwo^{2}+\\cdots+coefflast^{2}\\right)\n\\end{aligned}\n\\]\nand the other factor is \\( \\left(constk+coeffone^{2}+coefftwo^{2}+\\cdots+coefflast^{2}\\right) \\).\n\nSecond Solution. Assume for a moment that none of the \\( genericco \\) 's are zero, and let\n\\[\nmatrixbn=\\left(\\begin{array}{lllll}\ncoeffone^{2}+constk & coeffone coefftwo & coeffone coeffthr & \\cdots & coeffone coefflast \\\\\ncoefftwo coeffone & coefftwo^{2}+constk & coefftwo coeffthr & \\cdots & coefftwo coefflast \\\\\n\\cdots & \\cdots & \\cdots & \\cdots & \\cdots \\\\\ncoefflast coeffone & coefflast coefftwo & coefflast coeffthr & \\cdots & coefflast^{2}+constk\n\\end{array}\\right)\n\\]\nSince the determinant is linear in the last row, we find\n\nNow in the first of these new determinants, subtract multiples of the last rows from the others to get\n\\[\n\\operatorname{det}\\left(\\begin{array}{lll:l}\nconstk & & 0 & 0 \\\\\n0 & constk & constk & 0 \\\\\n\\hdashline coefflast coeffone & coefflast a_{n-1} & coefflast^{2}\n\\end{array}\\right) .\n\\]\nThen we have\n\\[\n\\operatorname{det} matrixbn = constk^{sizenum-1} coefflast^{2} + constk \\, \\operatorname{det} matrixbnm\n\\]\nSince \\( \\operatorname{det} matrixb_{1} = constk + coeffone^{2} \\), the relation\n\\[\n\\operatorname{det} matrixbn = constk^{sizenum-1}\\left(constk+coeffone^{2}+\\cdots+coefflast^{2}\\right)\n\\]\nfollows easily by induction.\n\nAlthough this derivation depends on the assumption that the \\( genericco \\) 's are not zero, the result remains valid for the case where some of the \\( genericco \\) 's are zero, since \\( determin \\) is evidently some polynomial in \\( constk \\) and the \\( genericco \\) 's which agrees with \\( constk^{sizenum-1}\\left(constk+coeffone^{2}+\\cdots+coefflast^{2}\\right) \\) as long as none of the \\( genericco \\) 's are zero. Therefore (1) must be a polynomial identity.\n\nAlternatively, we can regard the computation as taking place in the field \\( fieldrat\\left(constk, coeffone, coefftwo, \\ldots, coefflast\\right) \\) where \\( constk \\) and the \\( coeffindi \\) are independent indeterminates. In this field the condition \\( coeffindi \\neq 0 \\) is satisfied.\n\nFor a discussion of such fields, see I. N. Herstein, Topics in Algebra, Blaisdell, Waltham, Mass., 1964."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "latitude",
        "a": "caravanser",
        "a_1": "driftwood",
        "a_2": "moonshine",
        "a_3": "earthworm",
        "a_n-1": "sailcloth",
        "a_n": "breadcrumb",
        "a_i": "soapstone",
        "k": "peppermint",
        "n": "floodgate",
        "B": "shoelace",
        "I": "raincloud",
        "B_n": "crosswalk",
        "B_n-1": "overglaze",
        "D_n": "thumbtack",
        "Q": "marshland"
      },
      "question": "14. Prove that\n\\[\n\\left(\\begin{array}{lllll}\ndriftwood^{2}+peppermint & driftwood moonshine & driftwood earthworm & \\ldots & driftwood breadcrumb \\\\\nmoonshine driftwood & moonshine^{2}+peppermint & moonshine earthworm & \\ldots & moonshine breadcrumb \\\\\n\\ldots & \\ldots & \\ldots & \\ldots & \\ldots \\\\\nbreadcrumb driftwood & breadcrumb moonshine & breadcrumb earthworm & \\ldots & breadcrumb^{2}+peppermint\n\\end{array}\\right)\n\\]\nis divisible by \\( \\boldsymbol{peppermint}^{floodgate-1} \\) and find its other factor.",
      "solution": "First Solution. Let \\( shoelace \\) be the matrix\n\\[\n\\left(\\begin{array}{lllll}\ndriftwood^{2} & driftwood moonshine & driftwood earthworm & \\cdots & driftwood breadcrumb \\\\\nmoonshine driftwood & moonshine^{2} & moonshine earthworm & \\cdots & moonshine breadcrumb \\\\\n\\cdots & \\cdots & \\cdots & \\cdots & \\cdots \\\\\nbreadcrumb driftwood & breadcrumb moonshine & breadcrumb earthworm & \\cdots & breadcrumb^{2}\n\\end{array}\\right)\n\\]\n\\( shoelace \\) has rank at most one, since any two rows (or columns) are clearly dependent. So there are \\( (floodgate-1) \\) zeros among the eigenvalues of \\( shoelace \\). Therefore the characteristic polynomial of \\( shoelace \\) is divisible by \\( latitude^{floodgate-1} \\). Hence\n\\[\n\\begin{aligned}\n\\operatorname{det}(latitude \\cdot raincloud-shoelace) & =latitude^{floodgate}-(\\operatorname{trace} shoelace) latitude^{floodgate-1} \\\\\n& =latitude^{floodgate-1}\\left(latitude-driftwood{ }^{2}-moonshine{ }^{2}-\\cdots-breadcrumb{ }^{2}\\right)\n\\end{aligned}\n\\]\nso\n\\[\n\\begin{aligned}\n\\operatorname{det}(shoelace+peppermint raincloud) & =(-1)^{floodgate} \\operatorname{det}(-peppermint raincloud-shoelace) \\\\\n& =peppermint^{floodgate-1}\\left(peppermint+driftwood{ }^{2}+moonshine{ }^{2}+\\cdots+breadcrumb{ }^{2}\\right)\n\\end{aligned}\n\\]\nand the other factor is \\( \\left(peppermint+driftwood{ }^{2}+moonshine{ }^{2}+\\cdots+breadcrumb{ }^{2}\\right) \\).\n\nSecond Solution. Assume for a moment that none of the \\( caravanser \\)'s are zero, and let\n\\[\ncrosswalk=\\left(\\begin{array}{lllll}\ndriftwood{ }^{2}+peppermint & driftwood moonshine & driftwood earthworm & \\cdots & driftwood breadcrumb \\\\\nmoonshine driftwood & moonshine{ }^{2}+peppermint & moonshine earthworm & \\cdots & moonshine breadcrumb \\\\\n\\cdots & \\cdots & \\cdots & \\cdots & \\cdots \\\\\nbreadcrumb driftwood & breadcrumb moonshine & breadcrumb earthworm & \\cdots & breadcrumb{ }^{2}+peppermint\n\\end{array}\\right)\n\\]\n\nSince the determinant is linear in the last row, we find\n\nNow in the first of these new determinants, subtract multiples of the last rows from the others to get\n\\[\n\\operatorname{det}\\left(\\begin{array}{lll:l}\npeppermint & & 0 & 0 \\\\\n0 & peppermint & peppermint & 0 \\\\\n\\hdashline breadcrumb driftwood & breadcrumb sailcloth & breadcrumb^{2}\n\\end{array}\\right) .\n\\]\n\nThen we have\n\\[\n\\operatorname{det} crosswalk=peppermint^{floodgate-1} breadcrumb^{2}+peppermint \\operatorname{det} overglaze\n\\]\n\nSince \\( \\operatorname{det} B_{1}=peppermint+driftwood{ }^{2} \\), the relation\n\\[\n\\operatorname{det} crosswalk=peppermint^{floodgate-1}\\left(peppermint+driftwood^{2}+\\cdots+breadcrumb^{2}\\right)\n\\]\nfollows easily by induction.\nAlthough this derivation depends on the assumption that the \\( caravanser \\)'s are not zero, the result remains valid for the case where some of the \\( caravanser \\)'s are zero, since \\( thumbtack \\) is evidently some polynomial in \\( peppermint \\) and the \\( caravanser \\)'s which agrees with \\( peppermint^{floodgate-1}\\left(peppermint+driftwood{ }^{2}+\\cdots+breadcrumb{ }^{2}\\right) \\) as long as none of the \\( caravanser \\)'s are zero. Therefore (1) must be a polynomial identity.\n\nAlternatively, we can regard the computation as taking place in the field \\( marshland\\left(peppermint, driftwood, moonshine, \\ldots, breadcrumb\\right) \\) where \\( peppermint \\) and the \\( soapstone \\) are independent indeterminates. In this field the condition \\( soapstone \\neq 0 \\) is satisfied.\n\nFor a discussion of such fields, see I. N. Herstein, Topics in Algebra, Blaisdell, Waltham, Mass., 1964."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "knownvalue",
        "a": "fixedscalar",
        "a_1": "lastcoeff",
        "a_2": "penultimate",
        "a_3": "antepenult",
        "a_n-1": "primaryone",
        "a_n": "initialcoef",
        "a_i": "variableidx",
        "k": "variablestar",
        "n": "tinysize",
        "B": "scalaronly",
        "I": "zeromatrix",
        "B_n": "vectorsolo",
        "B_n-1": "vectorprior",
        "D_n": "integralset",
        "Q": "irrational"
      },
      "question": "14. Prove that\n\\[\n\\left(\\begin{array}{lllll}\nlastcoeff^{2}+variablestar & lastcoeff penultimate & lastcoeff antepenult & \\ldots & lastcoeff initialcoef \\\\\npenultimate lastcoeff & penultimate^{2}+variablestar & penultimate antepenult & \\ldots & penultimate initialcoef \\\\\n\\ldots & \\ldots & \\ldots & \\ldots & \\ldots \\\\\ninitialcoef lastcoeff & initialcoef penultimate & initialcoef antepenult & \\ldots & initialcoef^{2}+variablestar\n\\end{array}\\right)\n\\]\nis divisible by \\( \\boldsymbol{variablestar}^{tinysize-1} \\) and find its other factor.",
      "solution": "First Solution. Let \\( scalaronly \\) be the matrix\n\\[\n\\left(\\begin{array}{lllll}\nlastcoeff^{2} & lastcoeff penultimate & lastcoeff antepenult & \\cdots & lastcoeff initialcoef \\\\\npenultimate lastcoeff & penultimate^{2} & penultimate antepenult & \\cdots & penultimate initialcoef \\\\\n\\cdots & \\cdots & \\cdots & \\cdots & \\cdots \\\\\ninitialcoef lastcoeff & initialcoef penultimate & initialcoef antepenult & \\cdots & initialcoef^{2}\n\\end{array}\\right)\n\\]\n\\( scalaronly \\) has rank at most one, since any two rows (or columns) are clearly dependent. So there are \\( (tinysize-1) \\) zeros among the eigenvalues of \\( scalaronly \\). Therefore the characteristic polynomial of \\( scalaronly \\) is divisible by \\( knownvalue^{tinysize-1} \\). Hence\n\\[\n\\begin{aligned}\n\\operatorname{det}(knownvalue \\cdot zeromatrix-scalaronly) & =knownvalue^{tinysize}-(\\operatorname{trace} scalaronly) knownvalue^{tinysize-1} \\\\\n& =knownvalue^{tinysize-1}\\left(knownvalue-lastcoeff{ }^{2}-penultimate{ }^{2}-\\cdots-initialcoef{ }^{2}\\right)\n\\end{aligned}\n\\]\nso\n\\[\n\\begin{aligned}\n\\operatorname{det}(scalaronly+variablestar\\, zeromatrix) & =(-1)^{tinysize} \\operatorname{det}(-variablestar\\, zeromatrix-scalaronly) \\\\\n& =variablestar^{tinysize-1}\\left(variablestar+lastcoeff{ }^{2}+penultimate{ }^{2}+\\cdots+initialcoef{ }^{2}\\right)\n\\end{aligned}\n\\]\nand the other factor is \\( \\left(variablestar+lastcoeff{ }^{2}+penultimate{ }^{2}+\\cdots+initialcoef{ }^{2}\\right) \\).\n\nSecond Solution. Assume for a moment that none of the fixedscalar's are zero, and let\n\\[\nvectorsolo=\\left(\\begin{array}{lllll}\nlastcoeff{ }^{2}+variablestar & lastcoeff penultimate & lastcoeff antepenult & \\cdots & lastcoeff initialcoef \\\\\npenultimate lastcoeff & penultimate{ }^{2}+variablestar & penultimate antepenult & \\cdots & penultimate initialcoef \\\\\n\\cdots & \\cdots & \\cdots & \\cdots & \\cdots \\\\\ninitialcoef lastcoeff & initialcoef penultimate & initialcoef antepenult & \\cdots & initialcoef{ }^{2}+variablestar\n\\end{array}\\right)\n\\]\n\nSince the determinant is linear in the last row, we find\n\nNow in the first of these new determinants, subtract multiples of the last rows from the others to get\n\\[\n\\operatorname{det}\\left(\\begin{array}{lll:l}\nvariablestar & & 0 & 0 \\\\\n0 & variablestar & variablestar & 0 \\\\\n\\hdashline initialcoef lastcoeff & initialcoef primaryone & initialcoef^{2}\n\\end{array}\\right) .\n\\]\n\nThen we have\n\\[\n\\operatorname{det} vectorsolo=variablestar^{tinysize-1} initialcoef^{2}+variablestar \\operatorname{det} vectorprior\n\\]\n\nSince \\( \\operatorname{det} B_{1}=variablestar+lastcoeff{ }^{2} \\), the relation\n\\[\n\\operatorname{det} vectorsolo=variablestar^{tinysize-1}\\left(variablestar+lastcoeff^{2}+\\cdots+initialcoef^{2}\\right)\n\\]\nfollows easily by induction.\nAlthough this derivation depends on the assumption that the fixedscalar's are not zero, the result remains valid for the case where some of the fixedscalar's are zero, since \\( integralset \\) is evidently some polynomial in variablestar and the fixedscalar's which agrees with \\( variablestar^{tinysize-1}\\left(variablestar+lastcoeff{ }^{2}+\\cdots+initialcoef{ }^{2}\\right) \\) as long as none of the fixedscalar's are zero. Therefore (1) must be a polynomial identity.\n\nAlternatively, we can regard the computation as taking place in the field \\( irrational\\left(variablestar, lastcoeff, penultimate, \\ldots, initialcoef\\right) \\) where \\( variablestar \\) and the \\( variableidx \\) are independent indeterminates. In this field the condition \\( variableidx \\neq 0 \\) is satisfied.\n\nFor a discussion of such fields, see I. N. Herstein, Topics in Algebra, Blaisdell, Waltham, Mass., 1964."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "a": "hjgrksla",
        "a_1": "mczbtequ",
        "a_2": "rdlwopsa",
        "a_3": "vynkfgzh",
        "a_n-1": "ksivpurm",
        "a_n": "fqhxaljo",
        "a_i": "ztemwqub",
        "k": "lzvmuyhc",
        "n": "epqsdktj",
        "B": "njcwroae",
        "I": "pavhxlge",
        "B_n": "xutdrclm",
        "B_n-1": "sybqomtn",
        "D_n": "bzvawxye",
        "Q": "wlsrcfuv"
      },
      "question": "14. Prove that\n\\[\n\\left(\\begin{array}{lllll}\nmczbtequ^{2}+lzvmuyhc & mczbtequ rdlwopsa & mczbtequ vynkfgzh & \\ldots & mczbtequ fqhxaljo \\\\\nrdlwopsa mczbtequ & rdlwopsa^{2}+lzvmuyhc & rdlwopsa vynkfgzh & \\ldots & rdlwopsa fqhxaljo \\\\\n\\ldots & \\ldots & \\ldots & \\ldots & \\ldots \\\\\nfqhxaljo mczbtequ & fqhxaljo rdlwopsa & fqhxaljo vynkfgzh & \\ldots & fqhxaljo^{2}+lzvmuyhc\n\\end{array}\\right)\n\\]\nis divisible by \\( \\boldsymbol{lzvmuyhc}^{epqsdktj-1} \\) and find its other factor.",
      "solution": "First Solution. Let \\( njcwroae \\) be the matrix\n\\[\n\\left(\\begin{array}{lllll}\nmczbtequ^{2} & mczbtequ rdlwopsa & mczbtequ vynkfgzh & \\cdots & mczbtequ fqhxaljo \\\\\nrdlwopsa mczbtequ & rdlwopsa^{2} & rdlwopsa vynkfgzh & \\cdots & rdlwopsa fqhxaljo \\\\\n\\cdots & \\cdots & \\cdots & \\cdots & \\cdots \\\\\nfqhxaljo mczbtequ & fqhxaljo rdlwopsa & fqhxaljo vynkfgzh & \\cdots & fqhxaljo^{2}\n\\end{array}\\right)\n\\]\n\\( njcwroae \\) has rank at most one, since any two rows (or columns) are clearly dependent. So there are \\( (epqsdktj-1) \\) zeros among the eigenvalues of \\( njcwroae \\). Therefore the characteristic polynomial of \\( njcwroae \\) is divisible by \\( qzxwvtnp^{epqsdktj-1} \\). Hence\n\\[\n\\begin{aligned}\n\\operatorname{det}(qzxwvtnp \\cdot pavhxlge-njcwroae) & =qzxwvtnp^{epqsdktj}-(\\operatorname{trace} njcwroae) qzxwvtnp^{epqsdktj-1} \\\\\n& =qzxwvtnp^{epqsdktj-1}\\left(qzxwvtnp-mczbtequ^{2}-rdlwopsa^{2}-\\cdots-fqhxaljo^{2}\\right)\n\\end{aligned}\n\\]\nso\n\\[\n\\begin{aligned}\n\\operatorname{det}(njcwroae+lzvmuyhc pavhxlge) & =(-1)^{epqsdktj} \\operatorname{det}(-lzvmuyhc pavhxlge-njcwroae) \\\\\n& =lzvmuyhc^{epqsdktj-1}\\left(lzvmuyhc+mczbtequ^{2}+rdlwopsa^{2}+vynkfgzh^{2}+\\cdots+fqhxaljo^{2}\\right)\n\\end{aligned}\n\\]\nand the other factor is \\( \\left(lzvmuyhc+mczbtequ^{2}+rdlwopsa^{2}+\\cdots+fqhxaljo^{2}\\right) \\).\n\nSecond Solution. Assume for a moment that none of the \\( hjgrksla \\) 's are zero, and let\n\\[\nxutdrclm=\\left(\\begin{array}{lllll}\nmczbtequ^{2}+lzvmuyhc & mczbtequ rdlwopsa & mczbtequ vynkfgzh & \\cdots & mczbtequ fqhxaljo \\\\\nrdlwopsa mczbtequ & rdlwopsa^{2}+lzvmuyhc & rdlwopsa vynkfgzh & \\cdots & rdlwopsa fqhxaljo \\\\\n\\cdots & \\cdots & \\cdots & \\cdots & \\cdots \\\\\nfqhxaljo mczbtequ & fqhxaljo rdlwopsa & fqhxaljo vynkfgzh & \\cdots & fqhxaljo^{2}+lzvmuyhc\n\\end{array}\\right)\n\\]\n\nSince the determinant is linear in the last row, we find\n\nNow in the first of these new determinants, subtract multiples of the last rows from the others to get\n\\[\n\\operatorname{det}\\left(\\begin{array}{lll:l}\nlzvmuyhc & & 0 & 0 \\\\\n0 & lzvmuyhc & lzvmuyhc & 0 \\\\\n\\hdashline fqhxaljo mczbtequ & fqhxaljo ksivpurm & fqhxaljo^{2}\n\\end{array}\\right) .\n\\]\n\nThen we have\n\\[\n\\operatorname{det} xutdrclm=lzvmuyhc^{epqsdktj-1} fqhxaljo^{2}+lzvmuyhc \\operatorname{det} sybqomtn\n\\]\n\nSince \\( \\operatorname{det} xutdrclm_{1}=lzvmuyhc+mczbtequ^{2} \\), the relation\n\\[\n\\operatorname{det} xutdrclm=lzvmuyhc^{epqsdktj-1}\\left(lzvmuyhc+mczbtequ^{2}+\\cdots+fqhxaljo^{2}\\right)\n\\]\nfollows easily by induction.\nAlthough this derivation depends on the assumption that the \\( hjgrksla \\) 's are not zero, the result remains valid for the case where some of the \\( hjgrksla \\) 's are zero, since \\( bzvawxye \\) is evidently some polynomial in \\( lzvmuyhc \\) and the \\( hjgrksla \\) 's which agrees with \\( lzvmuyhc^{epqsdktj-1}\\left(lzvmuyhc+mczbtequ^{2}+\\cdots+fqhxaljo^{2}\\right) \\) as long as none of the \\( hjgrksla \\) 's are zero. Therefore (1) must be a polynomial identity.\n\nAlternatively, we can regard the computation as taking place in the field \\( wlsrcfuv\\left(lzvmuyhc, mczbtequ, rdlwopsa, \\ldots, fqhxaljo\\right) \\) where \\( lzvmuyhc \\) and the \\( ztemwqub \\) are independent indeterminates.\n\nFor a discussion of such fields, see I. N. Herstein, Topics in Algebra, Blaisdell, Waltham, Mass., 1964."
    },
    "kernel_variant": {
      "question": "Let m and r be integers with 1 ≤ r < m.  \nFor k = 1,…,r fix (not necessarily distinct) real column–vectors  \n\n  u⁽ᵏ⁾ = (u₁⁽ᵏ⁾,…,u_m⁽ᵏ⁾)ᵗ,  v⁽ᵏ⁾ = (v₁⁽ᵏ⁾,…,v_m⁽ᵏ⁾)ᵗ\n\nand assemble the two m × r matrices  \n\n  U := [u⁽¹⁾ u⁽²⁾ … u⁽ʳ⁾],  V := [v⁽¹⁾ v⁽²⁾ … v⁽ʳ⁾].\n\nFor indeterminates t₀,t₁,…,t_r put  \n\n  T := diag(t₁,…,t_r),  A(t₀,…,t_r) := t₀I_m + U T Vᵗ.     (1)\n\nAll algebraic statements below are meant in the polynomial ring  \nℝ[t₀,…,t_r, u_i^{(k)}, v_j^{(k)}].\n\n1. (Factorisation of the determinant)  \n Prove that det A is divisible by t₀^{m−r} and that  \n\n  det A(t₀,…,t_r) = t₀^{\\,m-r} det(t₀ I_r + T VᵗU).        (2)\n\n2. (Characteristic polynomial)  \n Let χ_A(λ) := det(λ I_m − A). Show that  \n\n  χ_A(λ) = (λ−t₀)^{\\,m-r} · det[(λ−t₀) I_r − T VᵗU].      (3)\n\n3. (Spectrum in the general case)  \n Using (3) describe all eigenvalues of A(t₀,…,t_r) for arbitrary real\nchoices of the parameters, and give their algebraic multiplicities.\n\n4. From now on impose the additional non-degeneracy conditions  \n\n  det T ≠ 0  (i.e. t_k ≠ 0 for k = 1,…,r)  and  det VᵗU ≠ 0,\n\nand fix a real number t₀ ≠ 0 that is not the negative of any eigenvalue of T VᵗU  \n(equivalently det(t₀ I_r + T VᵗU) ≠ 0).\n\n (a)  Prove that A(t₀,…,t_r) is invertible and establish the\nWoodbury–type formula  \n\n  A^{-1} = t₀^{−1}\\Bigl[I_m − U\\bigl(t₀ T^{−1}+VᵗU\\bigr)^{−1}Vᵗ\\Bigr]. (4)\n\n (b)  Compute the adjugate of A:\n\n  adj A  \n  = t₀^{\\,m-r-1} det T · det\\bigl(t₀ T^{−1}+VᵗU\\bigr)  \n    × \\Bigl[I_m − U\\bigl(t₀ T^{−1}+VᵗU\\bigr)^{−1}Vᵗ\\Bigr].     (5)\n\n5. (Symmetric orthonormal specialisation)  \n Assume in addition that U = V and that the r columns of U are\northonormal, i.e. UᵗU = I_r.\n\n (a)  Prove that A is symmetric and construct an orthonormal basis\nof ℝ^m consisting of eigenvectors of A.\n\n (b)  Determine the complete spectrum of A and, for every real\nspecialisation of the parameters, give the algebraic and geometric\nmultiplicities of the eigenvalues. In particular show that\n\n  spec A = {t₀ (mult m−r)} ∪ {t₀+t_k  (k = 1,…,r)},     (6)\n\nand discuss in detail how both multiplicities change when some of the\nparameters t_k coincide.\n\n\n\n────────────────────────────────────────────────────────────────────────",
      "solution": "Throughout we repeatedly use Sylvester’s determinant identity  \n\n det(I_m + U C Vᵗ) = det(I_r + C VᵗU)                (S)\n\nwhich holds for all conformable matrices over any commutative ring.\n\n1.  Put C := t₀^{−1}T.  Then (1) rewrites as  \n\n A = t₀(I_m + U C Vᵗ).                    (7)\n\nTaking determinants and applying (S) gives  \n\n det A = t₀^{m} det(I_m + U C Vᵗ)  \n     = t₀^{m} det(I_r + C VᵗU)  \n     = t₀^{m} det(I_r + t₀^{−1}T VᵗU)  \n     = t₀^{m-r} det(t₀I_r + T VᵗU),  \n\nestablishing (2) and the divisibility by t₀^{m−r}.\n\n2.  Put μ := λ−t₀.  Then  \n\n λI_m − A = μI_m − U T Vᵗ  \n      = μ[I_m − U(T/μ)Vᵗ].  \n\nThus  \n\n χ_A(λ) = μ^{\\,m} det(I_m − U(T/μ)Vᵗ).  \n\nApply (S) with C := −T/μ to obtain  \n\n χ_A(λ) = μ^{\\,m} det(I_r − (T/μ) VᵗU)  \n     = μ^{\\,m-r} det(μ I_r − T VᵗU)  \n     = (λ−t₀)^{\\,m-r} det[(λ−t₀)I_r − T VᵗU],  \n\nwhich is (3).\n\n3.  Denote by μ₁,…,μ_r the (possibly repeated) eigenvalues of the\nr × r matrix T VᵗU.  Formula (3) implies\n\n •  λ = t₀ is an eigenvalue of A with algebraic multiplicity m−r.  \n •  For each j = 1,…,r the number λ = t₀+μ_j is an eigenvalue of A,\n  having the same multiplicity as μ_j for T VᵗU.\n\nWhen several μ_j coincide their multiplicities add, and if some μ_j = 0\nthe corresponding eigenvalue merges with λ = t₀; the combined\nmultiplicity is the sum of the individual ones.\n\n4(a).  Because det(t₀I_r+T VᵗU) ≠ 0 by hypothesis, the factor in (2) is\nnon–zero, hence det A ≠ 0 and A is invertible.\n\nWrite again A = t₀(I_m + U C Vᵗ) with C = t₀^{−1}T.  Our additional\nassumption det T ≠ 0 guarantees that C is invertible, so the Woodbury\nidentity applies:\n\n (I_m + U C Vᵗ)^{−1} = I_m − U(C^{−1}+VᵗU)^{−1}Vᵗ.\n\nMultiplication by t₀^{−1} yields (4).\n\n4(b).  Combine det A from (2) with A^{-1} from (4):\n\n adj A = (det A)·A^{-1}  \n     = t₀^{m-r} det(t₀I_r+T VᵗU)·t₀^{−1}  \n     × [I_m − U(t₀T^{−1}+VᵗU)^{−1}Vᵗ].  \n\nUsing det(t₀I_r+T VᵗU)=det T·det(t₀T^{−1}+VᵗU) (valid because det T ≠ 0)\ngives (5).\n\n5.  Additional assumptions: U = V, UᵗU = I_r.\n\nA is a sum of symmetric rank-1 matrices, hence symmetric.  Let 𝒮 := im U\n(dim 𝒮 = r) and 𝒮^{⊥} its orthogonal complement.\n\n •  For x ∈ 𝒮^{⊥} one has Uᵗx = 0, so A x = t₀x; every vector of\n  𝒮^{⊥} is an eigenvector with eigenvalue t₀.\n\n •  The columns u⁽ᵏ⁾ (k = 1,…,r) are an orthonormal basis of 𝒮.  For such\n  a vector\n\n   A u⁽ᵏ⁾ = t₀u⁽ᵏ⁾ + t_k u⁽ᵏ⁾(u⁽ᵏ⁾)ᵗu⁽ᵏ⁾ = (t₀+t_k)u⁽ᵏ⁾,\n\n  because (u⁽ʲ⁾)ᵗu⁽ᵏ⁾ = 0 for j ≠ k.  Thus u⁽ᵏ⁾ is an eigenvector with\n  eigenvalue t₀+t_k.\n\nAn orthonormal eigenbasis of ℝ^m is obtained by adjoining any\northonormal basis of 𝒮^{⊥} to the columns of U, and (6) holds:\n\n •  λ = t₀ with algebraic = geometric multiplicity m−r;  \n •  For each k,  λ = t₀+t_k  has algebraic = geometric multiplicity 1.\n\nCoincidences among the parameters t_k act as follows:  \nAssume t_{k₁}=⋯=t_{k_s}=τ for some τ and distinct indices k₁,…,k_s.  \nThe vectors u^{(k₁)},…,u^{(k_s)} are still orthonormal, hence linearly\nindependent.  Consequently\n\n  λ = t₀+τ has algebraic multiplicity s,  \n  λ = t₀+τ has geometric multiplicity s,\n\nwhile the multiplicities for the remaining eigenvalues are unchanged.\nIn the extreme case t₁ = … = t_r, the spectrum consists of exactly two\neigenvalues,\n\n t₀ (mult m−r) and t₀+t₁ (mult r),\n\nboth with equal algebraic and geometric multiplicities.\n\n\n\n────────────────────────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.378283",
        "was_fixed": false,
        "difficulty_analysis": "•  Several new parameters (t₀,…,t_r) and up to r rank-one perturbations are introduced, replacing the single parameter and single outer product of the original problem.  \n•  The solver must invoke Sylvester’s determinant theorem, factor the characteristic polynomial, trace multiplicities, and handle adjugate matrices—techniques absent from the original statement.  \n•  Items (3)–(5) demand spectral analysis, explicit inversion via the Sherman–Morrison–Woodbury formula, and decomposition with respect to an orthogonal direct sum, blending linear algebra, formal algebraic identities, and polynomial factorisation.  \n•  The appearance of the full parameter matrix T as well as the assumption VᵗU invertible forces the contestant to juggle non-commutative matrix products, rather than the mere scalar dot-product in the current kernel variant.  \n•  Altogether the enhanced variant requires several intertwined ideas—rank considerations, determinant lemmas, eigen-structure, adjugate properties—far beyond the single-line calculation that sufficed previously, and therefore constitutes a substantially harder problem."
      }
    },
    "original_kernel_variant": {
      "question": "Let m and r be integers with 1 ≤ r < m.  \nFor k = 1,…,r fix (not necessarily distinct) real column–vectors  \n\n  u⁽ᵏ⁾ = (u₁⁽ᵏ⁾,…,u_m⁽ᵏ⁾)ᵗ,  v⁽ᵏ⁾ = (v₁⁽ᵏ⁾,…,v_m⁽ᵏ⁾)ᵗ\n\nand assemble the two m × r matrices  \n\n  U := [u⁽¹⁾ u⁽²⁾ … u⁽ʳ⁾],  V := [v⁽¹⁾ v⁽²⁾ … v⁽ʳ⁾].\n\nFor indeterminates t₀,t₁,…,t_r put  \n\n  T := diag(t₁,…,t_r),  A(t₀,…,t_r) := t₀I_m + U T Vᵗ.     (1)\n\nAll algebraic statements below are meant in the polynomial ring  \nℝ[t₀,…,t_r, u_i^{(k)}, v_j^{(k)}].\n\n1. (Factorisation of the determinant)  \n Prove that det A is divisible by t₀^{m−r} and that  \n\n  det A(t₀,…,t_r) = t₀^{\\,m-r} det(t₀ I_r + T VᵗU).        (2)\n\n2. (Characteristic polynomial)  \n Let χ_A(λ) := det(λ I_m − A). Show that  \n\n  χ_A(λ) = (λ−t₀)^{\\,m-r} · det[(λ−t₀) I_r − T VᵗU].      (3)\n\n3. (Spectrum in the general case)  \n Using (3) describe all eigenvalues of A(t₀,…,t_r) for arbitrary real\nchoices of the parameters, and give their algebraic multiplicities.\n\n4. From now on impose the additional non-degeneracy conditions  \n\n  det T ≠ 0  (i.e. t_k ≠ 0 for k = 1,…,r)  and  det VᵗU ≠ 0,\n\nand fix a real number t₀ ≠ 0 that is not the negative of any eigenvalue of T VᵗU  \n(equivalently det(t₀ I_r + T VᵗU) ≠ 0).\n\n (a)  Prove that A(t₀,…,t_r) is invertible and establish the\nWoodbury–type formula  \n\n  A^{-1} = t₀^{−1}\\Bigl[I_m − U\\bigl(t₀ T^{−1}+VᵗU\\bigr)^{−1}Vᵗ\\Bigr]. (4)\n\n (b)  Compute the adjugate of A:\n\n  adj A  \n  = t₀^{\\,m-r-1} det T · det\\bigl(t₀ T^{−1}+VᵗU\\bigr)  \n    × \\Bigl[I_m − U\\bigl(t₀ T^{−1}+VᵗU\\bigr)^{−1}Vᵗ\\Bigr].     (5)\n\n5. (Symmetric orthonormal specialisation)  \n Assume in addition that U = V and that the r columns of U are\northonormal, i.e. UᵗU = I_r.\n\n (a)  Prove that A is symmetric and construct an orthonormal basis\nof ℝ^m consisting of eigenvectors of A.\n\n (b)  Determine the complete spectrum of A and, for every real\nspecialisation of the parameters, give the algebraic and geometric\nmultiplicities of the eigenvalues. In particular show that\n\n  spec A = {t₀ (mult m−r)} ∪ {t₀+t_k  (k = 1,…,r)},     (6)\n\nand discuss in detail how both multiplicities change when some of the\nparameters t_k coincide.\n\n\n\n────────────────────────────────────────────────────────────────────────",
      "solution": "Throughout we repeatedly use Sylvester’s determinant identity  \n\n det(I_m + U C Vᵗ) = det(I_r + C VᵗU)                (S)\n\nwhich holds for all conformable matrices over any commutative ring.\n\n1.  Put C := t₀^{−1}T.  Then (1) rewrites as  \n\n A = t₀(I_m + U C Vᵗ).                    (7)\n\nTaking determinants and applying (S) gives  \n\n det A = t₀^{m} det(I_m + U C Vᵗ)  \n     = t₀^{m} det(I_r + C VᵗU)  \n     = t₀^{m} det(I_r + t₀^{−1}T VᵗU)  \n     = t₀^{m-r} det(t₀I_r + T VᵗU),  \n\nestablishing (2) and the divisibility by t₀^{m−r}.\n\n2.  Put μ := λ−t₀.  Then  \n\n λI_m − A = μI_m − U T Vᵗ  \n      = μ[I_m − U(T/μ)Vᵗ].  \n\nThus  \n\n χ_A(λ) = μ^{\\,m} det(I_m − U(T/μ)Vᵗ).  \n\nApply (S) with C := −T/μ to obtain  \n\n χ_A(λ) = μ^{\\,m} det(I_r − (T/μ) VᵗU)  \n     = μ^{\\,m-r} det(μ I_r − T VᵗU)  \n     = (λ−t₀)^{\\,m-r} det[(λ−t₀)I_r − T VᵗU],  \n\nwhich is (3).\n\n3.  Denote by μ₁,…,μ_r the (possibly repeated) eigenvalues of the\nr × r matrix T VᵗU.  Formula (3) implies\n\n •  λ = t₀ is an eigenvalue of A with algebraic multiplicity m−r.  \n •  For each j = 1,…,r the number λ = t₀+μ_j is an eigenvalue of A,\n  having the same multiplicity as μ_j for T VᵗU.\n\nWhen several μ_j coincide their multiplicities add, and if some μ_j = 0\nthe corresponding eigenvalue merges with λ = t₀; the combined\nmultiplicity is the sum of the individual ones.\n\n4(a).  Because det(t₀I_r+T VᵗU) ≠ 0 by hypothesis, the factor in (2) is\nnon–zero, hence det A ≠ 0 and A is invertible.\n\nWrite again A = t₀(I_m + U C Vᵗ) with C = t₀^{−1}T.  Our additional\nassumption det T ≠ 0 guarantees that C is invertible, so the Woodbury\nidentity applies:\n\n (I_m + U C Vᵗ)^{−1} = I_m − U(C^{−1}+VᵗU)^{−1}Vᵗ.\n\nMultiplication by t₀^{−1} yields (4).\n\n4(b).  Combine det A from (2) with A^{-1} from (4):\n\n adj A = (det A)·A^{-1}  \n     = t₀^{m-r} det(t₀I_r+T VᵗU)·t₀^{−1}  \n     × [I_m − U(t₀T^{−1}+VᵗU)^{−1}Vᵗ].  \n\nUsing det(t₀I_r+T VᵗU)=det T·det(t₀T^{−1}+VᵗU) (valid because det T ≠ 0)\ngives (5).\n\n5.  Additional assumptions: U = V, UᵗU = I_r.\n\nA is a sum of symmetric rank-1 matrices, hence symmetric.  Let 𝒮 := im U\n(dim 𝒮 = r) and 𝒮^{⊥} its orthogonal complement.\n\n •  For x ∈ 𝒮^{⊥} one has Uᵗx = 0, so A x = t₀x; every vector of\n  𝒮^{⊥} is an eigenvector with eigenvalue t₀.\n\n •  The columns u⁽ᵏ⁾ (k = 1,…,r) are an orthonormal basis of 𝒮.  For such\n  a vector\n\n   A u⁽ᵏ⁾ = t₀u⁽ᵏ⁾ + t_k u⁽ᵏ⁾(u⁽ᵏ⁾)ᵗu⁽ᵏ⁾ = (t₀+t_k)u⁽ᵏ⁾,\n\n  because (u⁽ʲ⁾)ᵗu⁽ᵏ⁾ = 0 for j ≠ k.  Thus u⁽ᵏ⁾ is an eigenvector with\n  eigenvalue t₀+t_k.\n\nAn orthonormal eigenbasis of ℝ^m is obtained by adjoining any\northonormal basis of 𝒮^{⊥} to the columns of U, and (6) holds:\n\n •  λ = t₀ with algebraic = geometric multiplicity m−r;  \n •  For each k,  λ = t₀+t_k  has algebraic = geometric multiplicity 1.\n\nCoincidences among the parameters t_k act as follows:  \nAssume t_{k₁}=⋯=t_{k_s}=τ for some τ and distinct indices k₁,…,k_s.  \nThe vectors u^{(k₁)},…,u^{(k_s)} are still orthonormal, hence linearly\nindependent.  Consequently\n\n  λ = t₀+τ has algebraic multiplicity s,  \n  λ = t₀+τ has geometric multiplicity s,\n\nwhile the multiplicities for the remaining eigenvalues are unchanged.\nIn the extreme case t₁ = … = t_r, the spectrum consists of exactly two\neigenvalues,\n\n t₀ (mult m−r) and t₀+t₁ (mult r),\n\nboth with equal algebraic and geometric multiplicities.\n\n\n\n────────────────────────────────────────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.326069",
        "was_fixed": false,
        "difficulty_analysis": "•  Several new parameters (t₀,…,t_r) and up to r rank-one perturbations are introduced, replacing the single parameter and single outer product of the original problem.  \n•  The solver must invoke Sylvester’s determinant theorem, factor the characteristic polynomial, trace multiplicities, and handle adjugate matrices—techniques absent from the original statement.  \n•  Items (3)–(5) demand spectral analysis, explicit inversion via the Sherman–Morrison–Woodbury formula, and decomposition with respect to an orthogonal direct sum, blending linear algebra, formal algebraic identities, and polynomial factorisation.  \n•  The appearance of the full parameter matrix T as well as the assumption VᵗU invertible forces the contestant to juggle non-commutative matrix products, rather than the mere scalar dot-product in the current kernel variant.  \n•  Altogether the enhanced variant requires several intertwined ideas—rank considerations, determinant lemmas, eigen-structure, adjugate properties—far beyond the single-line calculation that sufficed previously, and therefore constitutes a substantially harder problem."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}