{
  "index": "2005-A-4",
  "type": "ALG",
  "tag": [
    "ALG",
    "COMB",
    "NT"
  ],
  "difficulty": "",
  "question": "Let $H$ be an $n \\times n$ matrix all of whose entries are $\\pm 1$ and\nwhose rows are mutually orthogonal. Suppose $H$ has an $a \\times b$ submatrix\nwhose entries are all $1$. Show that $ab \\leq n$.",
  "solution": "\\textbf{First solution:}\nChoose a set of $a$ rows $r_1, \\dots, r_a$\ncontaining an $a \\times b$ submatrix whose\nentries are all 1. Then for $i,j \\in\\{1, \\dots, a\\}$, we have\n$r_i \\cdot r_j = n$ if $i=j$ and 0 otherwise. Hence\n\\[\n\\sum_{i,j=1}^a r_i \\cdot r_j = an.\n\\]\nOn the other hand, the term on the left is the dot product of\n$r_1 + \\cdots + r_a$ with itself, i.e., its squared length. Since\nthis vector has $a$ in each of its first $b$ coordinates, the dot product\nis at least $a^2 b$. Hence $an \\geq a^2 b$,\nwhence $n \\geq ab$ as desired.\n\n\\textbf{Second solution:}\n(by Richard Stanley)\nSuppose without loss of generality that the $a \\times b$ submatrix\noccupies the first $a$ rows and the first $b$ columns.\nLet $M$ be the submatrix occupying the first $a$ rows and the last\n$n-b$ columns. Then the hypothesis implies that the matrix\n$MM^T$ has $n-b$'s on the main diagonal and $-b$'s elsewhere.\nHence the column vector $v$ of length $a$ consisting of all 1's\nsatisfies $MM^T v = (n-ab)v$, so $n-ab$ is an eigenvalue of $MM^T$.\nBut $MM^T$ is semidefinite, so its eigenvalues are all nonnegative\nreal numbers. Hence $n-ab \\geq 0$.\n\n\\textbf{Remarks:}\nA matrix as in the problem is called a \\emph{Hadamard matrix}, because\nit meets the equality condition of Hadamard's inequality:\nany $n \\times n$ matrix with $\\pm 1$ entries has absolute determinant\nat most $n^{n/2}$, with equality if and only if the rows are mutually\northogonal\n(from the interpretation of the determinant as the volume of a paralellepiped\nwhose edges are parallel to the row vectors).\nNote that this implies  that the columns are also mutually orthogonal.\nA generalization of this problem, with a similar proof, is known\nas \\emph{Lindsey's lemma}: the sum of the entries in any\n$a \\times b$ submatrix of a Hadamard matrix is at most $\\sqrt{abn}$.\nStanley notes that Ryser (1981) asked for the smallest size of a Hadamard\nmatrix containing an $r \\times s$ submatrix of all 1's, and refers to\nthe URL \\texttt{www3.interscience.wiley.com/cgi-bin/ abstract/110550861/ABSTRACT} for more information.",
  "vars": [
    "H",
    "r_1",
    "r_i",
    "r_j",
    "r_a",
    "M",
    "v",
    "i",
    "j",
    "s"
  ],
  "params": [
    "n",
    "a",
    "b"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "H": "hadamardmatrix",
        "r_1": "firstrow",
        "r_i": "rowindexi",
        "r_j": "rowindexj",
        "r_a": "rowindexa",
        "M": "remainderblock",
        "v": "onesvector",
        "i": "indexi",
        "j": "indexj",
        "s": "subsize",
        "n": "dimension",
        "a": "numrows",
        "b": "numcols"
      },
      "question": "Let $hadamardmatrix$ be a $dimension \\times dimension$ matrix all of whose entries are $\\pm 1$ and whose rows are mutually orthogonal. Suppose $hadamardmatrix$ has a $numrows \\times numcols$ submatrix whose entries are all $1$. Show that $numrows numcols \\leq dimension$.",
      "solution": "\\textbf{First solution:}\\nChoose a set of $numrows$ rows $firstrow, \\dots, rowindexa$\\ncontaining a $numrows \\times numcols$ submatrix whose\\nentries are all 1. Then for $indexi,indexj \\in\\{1, \\dots, numrows\\}$, we have\\n$rowindexi \\cdot rowindexj = dimension$ if $indexi = indexj$ and $0$ otherwise. Hence\\n\\[\\n\\sum_{indexi,indexj=1}^{numrows} rowindexi \\cdot rowindexj = numrows dimension.\\n\\]\\nOn the other hand, the term on the left is the dot product of\\n$firstrow + \\cdots + rowindexa$ with itself, i.e., its squared length. Since\\nthis vector has $numrows$ in each of its first $numcols$ coordinates, the dot product\\nis at least $numrows^{2} numcols$. Hence $numrows dimension \\geq numrows^{2} numcols$,\\nwhence $dimension \\geq numrows numcols$ as desired.\\n\\n\\textbf{Second solution:}\\n(by Richard Stanley)\\nSuppose without loss of generality that the $numrows \\times numcols$ submatrix\\noccupies the first $numrows$ rows and the first $numcols$ columns.\\nLet $remainderblock$ be the submatrix occupying the first $numrows$ rows and the last\\n$dimension - numcols$ columns. Then the hypothesis implies that the matrix\\n$remainderblock\\,remainderblock^{T}$ has $(dimension - numcols)$'s on the main diagonal and $-numcols$'s elsewhere.\\nHence the column vector $onesvector$ of length $numrows$ consisting of all $1$'s\\nsatisfies $remainderblock\\,remainderblock^{T} onesvector = (dimension - numrows numcols) onesvector$, so $dimension - numrows numcols$ is an eigenvalue of $remainderblock\\,remainderblock^{T}$.\\nBut $remainderblock\\,remainderblock^{T}$ is semidefinite, so its eigenvalues are all nonnegative\\nreal numbers. Hence $dimension - numrows numcols \\ge 0$.\\n\\n\\textbf{Remarks:}\\nA matrix as in the problem is called a \\emph{Hadamard matrix}, because\\nit meets the equality condition of Hadamard's inequality:\\nany $dimension \\times dimension$ matrix with $\\pm 1$ entries has absolute determinant\\nat most $dimension^{dimension/2}$, with equality if and only if the rows are mutually\\northogonal (from the interpretation of the determinant as the volume of a parallelepiped\\nwhose edges are parallel to the row vectors).\\nNote that this implies that the columns are also mutually orthogonal.\\nA generalization of this problem, with a similar proof, is known\\nas \\emph{Lindsey's lemma}: the sum of the entries in any\\n$numrows \\times numcols$ submatrix of a Hadamard matrix is at most $\\sqrt{numrows numcols dimension}$.\\nStanley notes that Ryser (1981) asked for the smallest size of a Hadamard\\nmatrix containing an $r \\times subsize$ submatrix of all $1$'s, and refers to\\nthe URL \\texttt{www3.interscience.wiley.com/cgi-bin/ abstract/110550861/ABSTRACT} for more information."
    },
    "descriptive_long_confusing": {
      "map": {
        "H": "butterfly",
        "r_1": "tangerine",
        "r_i": "marathon",
        "r_j": "alligators",
        "r_a": "scarecrow",
        "M": "chocolate",
        "v": "lighthouse",
        "i": "pinecones",
        "j": "saxophone",
        "s": "backpacks",
        "n": "pineapple",
        "a": "waterfall",
        "b": "spaceship"
      },
      "question": "Let $butterfly$ be an $pineapple \\times pineapple$ matrix all of whose entries are $\\pm 1$ and\nwhose rows are mutually orthogonal. Suppose $butterfly$ has an $waterfall \\times spaceship$ submatrix\nwhose entries are all $1$. Show that $waterfall\\;spaceship \\leq pineapple$.",
      "solution": "\\textbf{First solution:}\nChoose a set of $waterfall$ rows $tangerine, \\dots, scarecrow$\ncontaining an $waterfall \\times spaceship$ submatrix whose\nentries are all 1. Then for $pinecones,saxophone \\in\\{1, \\dots, waterfall\\}$, we have\n$marathon \\cdot alligators = pineapple$ if $pinecones=saxophone$ and 0 otherwise. Hence\n\\[\n\\sum_{pinecones,saxophone=1}^{waterfall} marathon \\cdot alligators = waterfall\\,pineapple.\n\\]\nOn the other hand, the term on the left is the dot product of\n$tangerine + \\cdots + scarecrow$ with itself, i.e., its squared length. Since\nthis vector has $waterfall$ in each of its first $spaceship$ coordinates, the dot product\nis at least $waterfall^2\\,spaceship$. Hence $waterfall\\,pineapple \\geq waterfall^2\\,spaceship$,\nwhence $pineapple \\geq waterfall\\;spaceship$ as desired.\n\n\\textbf{Second solution:}\n(by Richard Stanley)\nSuppose without loss of generality that the $waterfall \\times spaceship$ submatrix\noccupies the first $waterfall$ rows and the first $spaceship$ columns.\nLet $chocolate$ be the submatrix occupying the first $waterfall$ rows and the last\n$pineapple-spaceship$ columns. Then the hypothesis implies that the matrix\n$chocolate choco late^T$ has $pineapple-spaceship$'s on the main diagonal and $-spaceship$'s elsewhere.\nHence the column vector $lighthouse$ of length $waterfall$ consisting of all 1's\nsatisfies $chocolate chocolate^T lighthouse = (pineapple-waterfall\\;spaceship)lighthouse$, so $pineapple-waterfall\\;spaceship$ is an eigenvalue of $chocolate chocolate^T$.\nBut $chocolate chocolate^T$ is semidefinite, so its eigenvalues are all nonnegative\nreal numbers. Hence $pineapple-waterfall\\;spaceship \\geq 0$.\n\n\\textbf{Remarks:}\nA matrix as in the problem is called a \\emph{Hadamard matrix}, because\nit meets the equality condition of Hadamard's inequality:\nany $pineapple \\times pineapple$ matrix with $\\pm 1$ entries has absolute determinant\nat most $pineapple^{pineapple/2}$, with equality if and only if the rows are mutually\northogonal\n(from the interpretation of the determinant as the volume of a paralellepiped\nwhose edges are parallel to the row vectors).\nNote that this implies  that the columns are also mutually orthogonal.\nA generalization of this problem, with a similar proof, is known\nas \\emph{Lindsey's lemma}: the sum of the entries in any\n$waterfall \\times spaceship$ submatrix of a Hadamard matrix is at most $\\sqrt{waterfall\\;spaceship\\;pineapple}$.\nStanley notes that Ryser (1981) asked for the smallest size of a Hadamard\nmatrix containing an $r \\times backpacks$ submatrix of all 1's, and refers to\nthe URL \\texttt{www3.interscience.wiley.com/cgi-bin/ abstract/110550861/ABSTRACT} for more information."
    },
    "descriptive_long_misleading": {
      "map": {
        "H": "nonorthmatrix",
        "r_1": "columnprime",
        "r_i": "columnith",
        "r_j": "columnjay",
        "r_a": "columnalpha",
        "M": "supermatrix",
        "v": "nullvector",
        "i": "endpoint",
        "j": "startpoint",
        "s": "antiscale",
        "n": "minilength",
        "a": "columncount",
        "b": "rowcount"
      },
      "question": "Let $nonorthmatrix$ be an $minilength \\times minilength$ matrix all of whose entries are $\\pm 1$ and\nwhose rows are mutually orthogonal. Suppose $nonorthmatrix$ has a $columncount \\times rowcount$ submatrix\nwhose entries are all $1$. Show that $columncount rowcount \\leq minilength$.",
      "solution": "\\textbf{First solution:}\nChoose a set of $columncount$ rows $columnprime, \\dots, columnalpha$\ncontaining a $columncount \\times rowcount$ submatrix whose\nentries are all 1. Then for $endpoint,startpoint \\in\\{1, \\dots, columncount\\}$, we have\n$columnith \\cdot columnjay = minilength$ if $endpoint=startpoint$ and 0 otherwise. Hence\n\\[\n\\sum_{endpoint,startpoint=1}^{columncount} columnith \\cdot columnjay = columncount minilength.\n\\]\nOn the other hand, the term on the left is the dot product of\n$columnprime + \\cdots + columnalpha$ with itself, i.e., its squared length. Since\nthis vector has $columncount$ in each of its first $rowcount$ coordinates, the dot product\nis at least $columncount^2 rowcount$. Hence $columncount minilength \\geq columncount^2 rowcount$,\nwhence $minilength \\geq columncount rowcount$ as desired.\n\n\\textbf{Second solution:}\n(by Richard Stanley)\nSuppose without loss of generality that the $columncount \\times rowcount$ submatrix\noccupies the first $columncount$ rows and the first $rowcount$ columns.\nLet $supermatrix$ be the submatrix occupying the first $columncount$ rows and the last\n$minilength-rowcount$ columns. Then the hypothesis implies that the matrix\n$supermatrix supermatrix^T$ has $minilength-rowcount$'s on the main diagonal and $-rowcount$'s elsewhere.\nHence the column vector $nullvector$ of length $columncount$ consisting of all 1's\nsatisfies $supermatrix supermatrix^T nullvector = (minilength-columncount rowcount)nullvector$, so $minilength-columncount rowcount$ is an eigenvalue of $supermatrix supermatrix^T$.\nBut $supermatrix supermatrix^T$ is semidefinite, so its eigenvalues are all nonnegative\nreal numbers. Hence $minilength-columncount rowcount \\geq 0$.\n\n\\textbf{Remarks:}\nA matrix as in the problem is called a \\emph{Hadamard matrix}, because\nit meets the equality condition of Hadamard's inequality:\nany $minilength \\times minilength$ matrix with $\\pm 1$ entries has absolute determinant\nat most $minilength^{minilength/2}$, with equality if and only if the rows are mutually\northogonal\n(from the interpretation of the determinant as the volume of a paralellepiped\nwhose edges are parallel to the row vectors).\nNote that this implies  that the columns are also mutually orthogonal.\nA generalization of this problem, with a similar proof, is known\nas \\emph{Lindsey's lemma}: the sum of the entries in any\n$columncount \\times rowcount$ submatrix of a Hadamard matrix is at most $\\sqrt{columncount rowcount minilength}$.\nStanley notes that Ryser (1981) asked for the smallest size of a Hadamard\nmatrix containing an $r \\times antiscale$ submatrix of all 1's, and refers to\nthe URL \\texttt{www3.interscience.wiley.com/cgi-bin/ abstract/110550861/ABSTRACT} for more information."
    },
    "garbled_string": {
      "map": {
        "H": "qarpsinl",
        "r_1": "vjtwqrsa",
        "r_i": "kxpqvntu",
        "r_j": "dhzmltfe",
        "r_a": "ygnobrca",
        "M": "zfurikpa",
        "v": "oxmdqsei",
        "i": "jqyrsemb",
        "j": "bxtlauhq",
        "s": "cmrqtnvo",
        "n": "heskoudr",
        "a": "ptfewyvn",
        "b": "ulmragci"
      },
      "question": "Let $qarpsinl$ be an $heskoudr \\times heskoudr$ matrix all of whose entries are $\\pm 1$ and\nwhose rows are mutually orthogonal. Suppose $qarpsinl$ has a $ptfewyvn \\times ulmragci$ submatrix\nwhose entries are all $1$. Show that $ptfewyvn ulmragci \\leq heskoudr$.",
      "solution": "\\textbf{First solution:}\nChoose a set of $ptfewyvn$ rows $vjtwqrsa, \\dots, ygnobrca$\ncontaining a $ptfewyvn \\times ulmragci$ submatrix whose\nentries are all 1. Then for $jqyrsemb,bxtlauhq \\in\\{1, \\dots, ptfewyvn\\}$, we have\n$kxpqvntu \\cdot dhzmltfe = heskoudr$ if $jqyrsemb=bxtlauhq$ and 0 otherwise. Hence\n\\[\n\\sum_{jqyrsemb,bxtlauhq=1}^{ptfewyvn} kxpqvntu \\cdot dhzmltfe = ptfewyvn heskoudr.\n\\]\nOn the other hand, the term on the left is the dot product of\n$vjtwqrsa + \\cdots + ygnobrca$ with itself, i.e., its squared length. Since\nthis vector has $ptfewyvn$ in each of its first $ulmragci$ coordinates, the dot product\nis at least $ptfewyvn^2 \\, ulmragci$. Hence $ptfewyvn heskoudr \\geq ptfewyvn^2 \\, ulmragci$,\nwhence $heskoudr \\geq ptfewyvn ulmragci$ as desired.\n\n\\textbf{Second solution:}\n(by Richard Stanley)\nSuppose without loss of generality that the $ptfewyvn \\times ulmragci$ submatrix\noccupies the first $ptfewyvn$ rows and the first $ulmragci$ columns.\nLet $zfurikpa$ be the submatrix occupying the first $ptfewyvn$ rows and the last\n$heskoudr-ulmragci$ columns. Then the hypothesis implies that the matrix\n$zfurikpa zfurikpa^{T}$ has $heskoudr-ulmragci$'s on the main diagonal and $-ulmragci$'s elsewhere.\nHence the column vector $oxmdqsei$ of length $ptfewyvn$ consisting of all 1's\nsatisfies $zfurikpa zfurikpa^{T} \\, oxmdqsei = (heskoudr-ptfewyvn ulmragci)\\, oxmdqsei$, so $heskoudr-ptfewyvn ulmragci$ is an eigenvalue of $zfurikpa zfurikpa^{T}$.\nBut $zfurikpa zfurikpa^{T}$ is semidefinite, so its eigenvalues are all nonnegative\nreal numbers. Hence $heskoudr-ptfewyvn ulmragci \\geq 0$.\n\n\\textbf{Remarks:}\nA matrix as in the problem is called a \\emph{Hadamard matrix}, because\nit meets the equality condition of Hadamard's inequality:\nany $heskoudr \\times heskoudr$ matrix with $\\pm 1$ entries has absolute determinant\nat most $heskoudr^{heskoudr/2}$, with equality if and only if the rows are mutually\northogonal\n(from the interpretation of the determinant as the volume of a paralellepiped\nwhose edges are parallel to the row vectors).\nNote that this implies  that the columns are also mutually orthogonal.\nA generalization of this problem, with a similar proof, is known\nas \\emph{Lindsey's lemma}: the sum of the entries in any\n$ptfewyvn \\times ulmragci$ submatrix of a Hadamard matrix is at most $\\sqrt{ptfewyvn ulmragci heskoudr}$.\nStanley notes that Ryser (1981) asked for the smallest size of a Hadamard\nmatrix containing an $r \\times cmrqtnvo$ submatrix of all 1's, and refers to\nthe URL \\texttt{www3.interscience.wiley.com/cgi-bin/ abstract/110550861/ABSTRACT} for more information."
    },
    "kernel_variant": {
      "question": "──────────────────────────────────  \nFix a non–zero real number $c$.  Let  \n\n\\[\nH=(h_{ij})_{1\\le i\\le k,\\;1\\le j\\le n}, \\qquad k\\le n,\\qquad h_{ij}\\in\\{+c,\\,-c\\},\n\\]\n\nbe a $k\\times n$ matrix whose rows are pairwise orthogonal with respect to the usual Euclidean inner product on $\\mathbb R^{\\,n}$, so that  \n\n\\[\nHH^{\\mathsf T}=n\\,c^{2}I_k. \\tag{1}\n\\]\n\nFor index–sets $R\\subseteq\\{1,\\dots ,k\\}$ and $C\\subseteq\\{1,\\dots ,n\\}$ write $a:=|R|$, $b:=|C|$, and denote by  \n\n\\[\nA:=H[R,C]\n\\]\n\nthe $a\\times b$ sub-matrix formed by the rows in $R$ and the columns in $C$.\n\n1.  (Spectral bound)  \n    Prove that the spectral norm (largest singular value) of $A$ satisfies  \n\n    \\[\n        \\lVert A\\rVert_2\\le |c|\\sqrt n. \\tag{$\\star$}\n    \\]\n\n2.  (Low–rank sub-matrices)  \n    Suppose that the rank of $A$ equals $r$ ($1\\le r\\le\\min\\{a,b\\}$).  Show that  \n\n    \\[\n        ab\\le r\\,n. \\tag{$\\star\\!\\star$}\n    \\]\n\n3.  (Lindsey’s inequality recovered)  \n    Deduce from $(\\star\\!\\star)$ that \\emph{if every entry of $A$ equals the same sign-choice of $c$} (so $A$ is a constant matrix, hence of rank $1$) then necessarily  \n\n    \\[\n        ab\\le n.\n    \\]\n\n──────────────────────────────────",
      "solution": "──────────────────────────────────  \nPreliminaries.  \nBecause each entry of $H$ has absolute value $|c|$, every row of $H$ has Euclidean length $|c|\\sqrt n$.  From (1) we also know that every non-zero singular value of $H$ equals $|c|\\sqrt n$; hence the operator (spectral) norm of $H$ is  \n\n\\[\n\\lVert H\\rVert_2=|c|\\sqrt n. \\tag{2}\n\\]\n\nLet $S_R$ denote the $a\\times k$ row-selector matrix that extracts rows in $R$, and let $P_C$ be the $n\\times b$ column-selector matrix that extracts columns in $C$.  Then  \n\n\\[\nA=S_RHP_C. \\tag{3}\n\\]\n\n————————————————————————  \n1.  Proof of $(\\star)$.  \nFor any $x\\in\\mathbb R^{\\,b}$ extend it to $\\bar x\\in\\mathbb R^{\\,n}$ by inserting zeros outside $C$, so $P_Cx=\\bar x$.  Then  \n\n\\[\nAx=S_RH\\bar x. \\tag{4}\n\\]\n\nTaking Euclidean norms and using (2),\n\n\\[\n\\lVert Ax\\rVert_2\\le\\lVert S_R\\rVert_2\\,\\lVert H\\rVert_2\\,\\lVert\\bar x\\rVert_2\n           \\le 1\\cdot |c|\\sqrt n\\cdot\\lVert x\\rVert_2,\n\\]\n\nbecause $S_R$ merely deletes rows and therefore has operator norm $1$.  Hence  \n\n\\[\n\\lVert A\\rVert_2=\\sup_{x\\neq 0}\\frac{\\lVert Ax\\rVert_2}{\\lVert x\\rVert_2}\\le |c|\\sqrt n.\n\\quad\\blacksquare\n\\]\n\n————————————————————————  \n2.  Proof of $(\\star\\!\\star)$.  \nLet the non-zero singular values of $A$ be $\\sigma_1,\\dots ,\\sigma_r$.  By $(\\star)$,  \n\n\\[\n\\sigma_i\\le |c|\\sqrt n\\quad\\text{for every }i. \\tag{5}\n\\]\n\nBecause the Frobenius norm equals the $\\ell^2$–sum of the singular values,  \n\n\\[\n\\lVert A\\rVert_F^{\\,2}=\\sigma_1^{2}+\\dots +\\sigma_r^{2}. \\tag{6}\n\\]\n\nSince every entry of $A$ equals $\\pm c$, there are $ab$ such entries, each of magnitude $|c|$, so  \n\n\\[\n\\lVert A\\rVert_F^{\\,2}=ab\\,c^{2}. \\tag{7}\n\\]\n\nCombining (5)–(7) gives  \n\n\\[\nab\\,c^{2}=\\sum_{i=1}^{r}\\sigma_i^{2}\\le r\\,(|c|\\sqrt n)^{2}=r\\,|c|^{2}n,\n\\]\n\nand division by $|c|^{2}>0$ yields  \n\n\\[\nab\\le r\\,n.\n\\quad\\blacksquare\n\\]\n\n————————————————————————  \n3.  Lindsey’s bound as a corollary.  \nSuppose every entry of $A$ equals the same sign-choice of $c$.  Then $A$ is a constant matrix and therefore has rank $1$.  Taking $r=1$ in $(\\star\\!\\star)$ gives  \n\n\\[\nab\\le n,\n\\]\n\nwhich is precisely Lindsey’s inequality.  Note that we used only the implication “constant $\\Longrightarrow$ rank $1$”; the converse is \\emph{not} asserted.  \n\\hfill$\\blacksquare$\n\n──────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.791710",
        "was_fixed": false,
        "difficulty_analysis": "──────────────────────────────────  \n•  The original problem dealt only with rank-one (constant) sub-matrices and could be settled by elementary vector arithmetic.  \n•  The current kernel variant already required recognising Lindsey’s inequality but still stayed within basic Cauchy–Schwarz arguments.  \n\nThe enhanced variant is markedly harder because  \n1.  It asks for a bound on the spectral norm of an arbitrary sub-matrix A; this forces the competitor to introduce operator norms, singular values, and their interaction with orthogonality outside the chosen rows and columns.  \n2.  Part 2 replaces the “all-ones’’ hypothesis by the far more general low-rank condition.  The solver must understand how the entire singular-value spectrum of A is constrained by HHᵀ = n c² I_k, and must relate the Frobenius and spectral norms to derive ab ≤ r n.  \n3.  Lindsey’s inequality now appears only as the easiest special case (r = 1); proving it is no longer the goal but a by-product of a broader, more sophisticated statement.  \n\nThus the enhanced variant introduces higher-order linear-algebraic notions (spectral norm, singular values, Frobenius norm), demands chaining several non-trivial inequalities, and generalises the original statement from rank 1 to arbitrary rank r, all of which substantially increase both conceptual depth and technical complexity."
      }
    },
    "original_kernel_variant": {
      "question": "──────────────────────────────────  \nFix a non–zero real number $c$.  Let  \n\n\\[\nH=(h_{ij})_{1\\le i\\le k,\\;1\\le j\\le n}, \\qquad k\\le n,\\qquad h_{ij}\\in\\{+c,\\,-c\\},\n\\]\n\nbe a $k\\times n$ matrix whose rows are pairwise orthogonal with respect to the usual Euclidean inner product on $\\mathbb R^{\\,n}$, so that  \n\n\\[\nHH^{\\mathsf T}=n\\,c^{2}I_k. \\tag{1}\n\\]\n\nFor index–sets $R\\subseteq\\{1,\\dots ,k\\}$ and $C\\subseteq\\{1,\\dots ,n\\}$ write $a:=|R|$, $b:=|C|$, and denote by  \n\n\\[\nA:=H[R,C]\n\\]\n\nthe $a\\times b$ sub-matrix formed by the rows in $R$ and the columns in $C$.\n\n1.  (Spectral bound)  \n    Prove that the spectral norm (largest singular value) of $A$ satisfies  \n\n    \\[\n        \\lVert A\\rVert_2\\le |c|\\sqrt n. \\tag{$\\star$}\n    \\]\n\n2.  (Low–rank sub-matrices)  \n    Suppose that the rank of $A$ equals $r$ ($1\\le r\\le\\min\\{a,b\\}$).  Show that  \n\n    \\[\n        ab\\le r\\,n. \\tag{$\\star\\!\\star$}\n    \\]\n\n3.  (Lindsey’s inequality recovered)  \n    Deduce from $(\\star\\!\\star)$ that \\emph{if every entry of $A$ equals the same sign-choice of $c$} (so $A$ is a constant matrix, hence of rank $1$) then necessarily  \n\n    \\[\n        ab\\le n.\n    \\]\n\n──────────────────────────────────",
      "solution": "──────────────────────────────────  \nPreliminaries.  \nBecause each entry of $H$ has absolute value $|c|$, every row of $H$ has Euclidean length $|c|\\sqrt n$.  From (1) we also know that every non-zero singular value of $H$ equals $|c|\\sqrt n$; hence the operator (spectral) norm of $H$ is  \n\n\\[\n\\lVert H\\rVert_2=|c|\\sqrt n. \\tag{2}\n\\]\n\nLet $S_R$ denote the $a\\times k$ row-selector matrix that extracts rows in $R$, and let $P_C$ be the $n\\times b$ column-selector matrix that extracts columns in $C$.  Then  \n\n\\[\nA=S_RHP_C. \\tag{3}\n\\]\n\n————————————————————————  \n1.  Proof of $(\\star)$.  \nFor any $x\\in\\mathbb R^{\\,b}$ extend it to $\\bar x\\in\\mathbb R^{\\,n}$ by inserting zeros outside $C$, so $P_Cx=\\bar x$.  Then  \n\n\\[\nAx=S_RH\\bar x. \\tag{4}\n\\]\n\nTaking Euclidean norms and using (2),\n\n\\[\n\\lVert Ax\\rVert_2\\le\\lVert S_R\\rVert_2\\,\\lVert H\\rVert_2\\,\\lVert\\bar x\\rVert_2\n           \\le 1\\cdot |c|\\sqrt n\\cdot\\lVert x\\rVert_2,\n\\]\n\nbecause $S_R$ merely deletes rows and therefore has operator norm $1$.  Hence  \n\n\\[\n\\lVert A\\rVert_2=\\sup_{x\\neq 0}\\frac{\\lVert Ax\\rVert_2}{\\lVert x\\rVert_2}\\le |c|\\sqrt n.\n\\quad\\blacksquare\n\\]\n\n————————————————————————  \n2.  Proof of $(\\star\\!\\star)$.  \nLet the non-zero singular values of $A$ be $\\sigma_1,\\dots ,\\sigma_r$.  By $(\\star)$,  \n\n\\[\n\\sigma_i\\le |c|\\sqrt n\\quad\\text{for every }i. \\tag{5}\n\\]\n\nBecause the Frobenius norm equals the $\\ell^2$–sum of the singular values,  \n\n\\[\n\\lVert A\\rVert_F^{\\,2}=\\sigma_1^{2}+\\dots +\\sigma_r^{2}. \\tag{6}\n\\]\n\nSince every entry of $A$ equals $\\pm c$, there are $ab$ such entries, each of magnitude $|c|$, so  \n\n\\[\n\\lVert A\\rVert_F^{\\,2}=ab\\,c^{2}. \\tag{7}\n\\]\n\nCombining (5)–(7) gives  \n\n\\[\nab\\,c^{2}=\\sum_{i=1}^{r}\\sigma_i^{2}\\le r\\,(|c|\\sqrt n)^{2}=r\\,|c|^{2}n,\n\\]\n\nand division by $|c|^{2}>0$ yields  \n\n\\[\nab\\le r\\,n.\n\\quad\\blacksquare\n\\]\n\n————————————————————————  \n3.  Lindsey’s bound as a corollary.  \nSuppose every entry of $A$ equals the same sign-choice of $c$.  Then $A$ is a constant matrix and therefore has rank $1$.  Taking $r=1$ in $(\\star\\!\\star)$ gives  \n\n\\[\nab\\le n,\n\\]\n\nwhich is precisely Lindsey’s inequality.  Note that we used only the implication “constant $\\Longrightarrow$ rank $1$”; the converse is \\emph{not} asserted.  \n\\hfill$\\blacksquare$\n\n──────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.604750",
        "was_fixed": false,
        "difficulty_analysis": "──────────────────────────────────  \n•  The original problem dealt only with rank-one (constant) sub-matrices and could be settled by elementary vector arithmetic.  \n•  The current kernel variant already required recognising Lindsey’s inequality but still stayed within basic Cauchy–Schwarz arguments.  \n\nThe enhanced variant is markedly harder because  \n1.  It asks for a bound on the spectral norm of an arbitrary sub-matrix A; this forces the competitor to introduce operator norms, singular values, and their interaction with orthogonality outside the chosen rows and columns.  \n2.  Part 2 replaces the “all-ones’’ hypothesis by the far more general low-rank condition.  The solver must understand how the entire singular-value spectrum of A is constrained by HHᵀ = n c² I_k, and must relate the Frobenius and spectral norms to derive ab ≤ r n.  \n3.  Lindsey’s inequality now appears only as the easiest special case (r = 1); proving it is no longer the goal but a by-product of a broader, more sophisticated statement.  \n\nThus the enhanced variant introduces higher-order linear-algebraic notions (spectral norm, singular values, Frobenius norm), demands chaining several non-trivial inequalities, and generalises the original statement from rank 1 to arbitrary rank r, all of which substantially increase both conceptual depth and technical complexity."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}