{
  "index": "1957-B-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "2. If facilities for division are not available, it is sometimes convenient in determining the decimal expansion of \\( 1 / A, A>0 \\) to use the iteration \\( X_{k+1} \\) \\( =X_{k}\\left(2-A X_{k}\\right), k=0,1,2, \\ldots \\), where \\( X_{0} \\) is a selected \"starting\" value. Find the limitations, if any, on the starting value \\( X_{0} \\) in order that the above iteration converges to the desired value \\( 1 / A \\).",
  "solution": "First Solution. The polygonal representation of this recursion is shown in the figure. (See p. 223 for an explanation.) It is clear that if \\( X_{0} \\) lies in \\( (0,2 A) \\) then\n\\[\n0<X_{1} \\leq X_{2} \\leq X_{3} \\leq \\cdots \\leq \\frac{1}{A}\n\\]\nand the sequence converges to \\( 1 / A \\). If \\( X_{0}=0 \\) or \\( 2 / A \\), then obviously\n\\[\nX_{1}=X_{2}=X_{3}=\\cdots=0\n\\]\nand if \\( X_{0} \\) lies outside \\( 0,2 / A \\), then\n\\[\n0>X_{1}>X_{2}>X_{3}>\\ldots\n\\]\nand the sequence diverges to \\( -\\infty \\).\nTo make this rigorous, we detine \\( f(x)=x(2-A x) \\). We find that \\( f \\) achieves its maximum value \\( 1 / A \\) for \\( x=1 / A \\). Moreover, \\( f(x)>x \\) for \\( 0<x<1 / A \\) and \\( f(x)<x \\) for \\( x<0 \\). Then (1), (2), and (3) follow immediately. In case (1) the sequence must be convergent and the limit must be a positive root of \\( f(x)=x \\), and there is only one, namely, \\( x=1 / A \\). In case (2) the sequence obviously converges to 0 . In case (3) it either diverges or converges to a negative root of \\( f(x)=x \\), but there is no such root.\n\nTherefore, the sequence converges to \\( 1 / A \\) if and only if \\( 0<X_{0}<2 / A \\).\nSecond Solution. We can find an explicit formula for \\( \\boldsymbol{X}_{n} \\) from which the limiting behavior is obvious. We have\n\\[\n1-A X_{n+1}=\\left(1-A X_{n}\\right)^{2}\n\\]\ntherefore\n\\[\n1-A X_{n}=\\left(1-A X_{0}\\right)^{2 \\prime}\n\\]\n\nIt is now obvious that \\( 1-A X_{n} \\rightarrow 0 \\), that is \\( X_{n} \\rightarrow 1 / A \\), if and only if \\( \\left|1-A X_{0}\\right|<1 \\), that is \\( 0<X_{0}<2 / A \\).",
  "vars": [
    "X_k",
    "X_k+1",
    "X_0",
    "X_1",
    "X_2",
    "X_3",
    "X_n",
    "X_n+1",
    "x",
    "f"
  ],
  "params": [
    "A",
    "k",
    "n"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "X_k": "iteratek",
        "X_k+1": "iteratkplusone",
        "X_0": "startvalue",
        "X_1": "firstvalue",
        "X_2": "secondvalue",
        "X_3": "thirdvalue",
        "X_n": "generalvalue",
        "X_n+1": "generalvalueplusone",
        "x": "genericx",
        "f": "functionf",
        "A": "constanta",
        "k": "indexk",
        "n": "indexn"
      },
      "question": "2. If facilities for division are not available, it is sometimes convenient in determining the decimal expansion of \\( 1 / constanta, constanta>0 \\) to use the iteration \\( iteratkplusone \\) \\( =iteratek\\left(2-constanta iteratek\\right), indexk=0,1,2, \\ldots \\), where \\( startvalue \\) is a selected \"starting\" value. Find the limitations, if any, on the starting value \\( startvalue \\) in order that the above iteration converges to the desired value \\( 1 / constanta \\).",
      "solution": "First Solution. The polygonal representation of this recursion is shown in the figure. (See p. 223 for an explanation.) It is clear that if \\( startvalue \\) lies in \\( (0,2 constanta) \\) then\n\\[\n0<firstvalue \\leq secondvalue \\leq thirdvalue \\leq \\cdots \\leq \\frac{1}{constanta}\n\\]\nand the sequence converges to \\( 1 / constanta \\). If \\( startvalue=0 \\) or \\( 2 / constanta \\), then obviously\n\\[\nfirstvalue=secondvalue=thirdvalue=\\cdots=0\n\\]\nand if \\( startvalue \\) lies outside \\( 0,2 / constanta \\), then\n\\[\n0>firstvalue>secondvalue>thirdvalue>\\ldots\n\\]\nand the sequence diverges to \\( -\\infty \\).\nTo make this rigorous, we detine \\( functionf(genericx)=genericx(2-constanta genericx) \\). We find that \\( functionf \\) achieves its maximum value \\( 1 / constanta \\) for \\( genericx=1 / constanta \\). Moreover, \\( functionf(genericx)>genericx \\) for \\( 0<genericx<1 / constanta \\) and \\( functionf(genericx)<genericx \\) for \\( genericx<0 \\). Then (1), (2), and (3) follow immediately. In case (1) the sequence must be convergent and the limit must be a positive root of \\( functionf(genericx)=genericx \\), and there is only one, namely, \\( genericx=1 / constanta \\). In case (2) the sequence obviously converges to 0 . In case (3) it either diverges or converges to a negative root of \\( functionf(genericx)=genericx \\), but there is no such root.\n\nTherefore, the sequence converges to \\( 1 / constanta \\) if and only if \\( 0<startvalue<2 / constanta \\).\nSecond Solution. We can find an explicit formula for \\( \\boldsymbol{generalvalue} \\) from which the limiting behavior is obvious. We have\n\\[\n1-constanta generalvalueplusone=\\left(1-constanta generalvalue\\right)^{2}\n\\]\ntherefore\n\\[\n1-constanta generalvalue=\\left(1-constanta startvalue\\right)^{2 \\prime}\n\\]\n\nIt is now obvious that \\( 1-constanta generalvalue \\rightarrow 0 \\), that is \\( generalvalue \\rightarrow 1 / constanta \\), if and only if \\( \\left|1-constanta startvalue\\right|<1 \\), that is \\( 0<startvalue<2 / constanta \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "X_k": "paintbrush",
        "X_k+1": "horseshoer",
        "X_0": "drumcircle",
        "X_1": "quillshell",
        "X_2": "latticework",
        "X_3": "moontunnel",
        "X_n": "sailcloth",
        "X_n+1": "bookbinder",
        "x": "grainfield",
        "f": "riverdelta",
        "A": "stonecrown",
        "k": "goldfinch",
        "n": "willowherb"
      },
      "question": "2. If facilities for division are not available, it is sometimes convenient in determining the decimal expansion of \\( 1 / stonecrown, stonecrown>0 \\) to use the iteration \\( horseshoer = paintbrush\\left(2-stonecrown\\, paintbrush\\right), goldfinch=0,1,2, \\ldots \\), where \\( drumcircle \\) is a selected \"starting\" value. Find the limitations, if any, on the starting value \\( drumcircle \\) in order that the above iteration converges to the desired value \\( 1 / stonecrown \\).",
      "solution": "First Solution. The polygonal representation of this recursion is shown in the figure. (See p. 223 for an explanation.) It is clear that if \\( drumcircle \\) lies in \\( (0,2 stonecrown) \\) then\n\\[\n0<quillshell \\leq latticework \\leq moontunnel \\leq \\cdots \\leq \\frac{1}{stonecrown}\n\\]\nand the sequence converges to \\( 1 / stonecrown \\). If \\( drumcircle=0 \\) or \\( 2 / stonecrown \\), then obviously\n\\[\nquillshell=latticework=moontunnel=\\cdots=0\n\\]\nand if \\( drumcircle \\) lies outside \\( 0,2 / stonecrown \\), then\n\\[\n0>quillshell>latticework>moontunnel>\\ldots\n\\]\nand the sequence diverges to \\( -\\infty \\).\nTo make this rigorous, we define \\( riverdelta(grainfield)=grainfield(2-stonecrown\\, grainfield) \\). We find that \\( riverdelta \\) achieves its maximum value \\( 1 / stonecrown \\) for \\( grainfield=1 / stonecrown \\). Moreover, \\( riverdelta(grainfield)>grainfield \\) for \\( 0<grainfield<1 / stonecrown \\) and \\( riverdelta(grainfield)<grainfield \\) for \\( grainfield<0 \\). Then (1), (2), and (3) follow immediately. In case (1) the sequence must be convergent and the limit must be a positive root of \\( riverdelta(grainfield)=grainfield \\), and there is only one, namely, \\( grainfield=1 / stonecrown \\). In case (2) the sequence obviously converges to 0. In case (3) it either diverges or converges to a negative root of \\( riverdelta(grainfield)=grainfield \\), but there is no such root.\n\nTherefore, the sequence converges to \\( 1 / stonecrown \\) if and only if \\( 0<drumcircle<2 / stonecrown \\).\n\nSecond Solution. We can find an explicit formula for \\( \\boldsymbol{sailcloth} \\) from which the limiting behavior is obvious. We have\n\\[\n1-stonecrown\\, bookbinder=\\left(1-stonecrown\\, sailcloth\\right)^{2}\n\\]\ntherefore\n\\[\n1-stonecrown\\, sailcloth=\\left(1-stonecrown\\, drumcircle\\right)^{2 \\prime}\n\\]\n\nIt is now obvious that \\( 1-stonecrown\\, sailcloth \\rightarrow 0 \\), that is \\( sailcloth \\rightarrow 1 / stonecrown \\), if and only if \\( \\left|1-stonecrown\\, drumcircle\\right|<1 \\), that is \\( 0<drumcircle<2 / stonecrown \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "X_k": "constantvalue",
        "X_k+1": "earliervalue",
        "X_0": "endingvalue",
        "X_1": "finalvalue",
        "X_2": "terminalvalue",
        "X_3": "originvalue",
        "X_n": "specificvalue",
        "X_n+1": "precedingvalue",
        "x": "constant",
        "f": "argument",
        "A": "variable",
        "k": "staticindex",
        "n": "fixedpoint"
      },
      "question": "2. If facilities for division are not available, it is sometimes convenient in determining the decimal expansion of \\( 1 / variable, variable>0 \\) to use the iteration \\( earliervalue \\)\n \\( =constantvalue\\left(2-variable constantvalue\\right), staticindex=0,1,2, \\ldots \\), where \\( endingvalue \\) is a selected \"starting\" value. Find the limitations, if any, on the starting value \\( endingvalue \\) in order that the above iteration converges to the desired value \\( 1 / variable \\).",
      "solution": "First Solution. The polygonal representation of this recursion is shown in the figure. (See p. 223 for an explanation.) It is clear that if \\( endingvalue \\) lies in \\( (0,2 variable) \\) then\n\\[\n0<finalvalue \\leq terminalvalue \\leq originvalue \\leq \\cdots \\leq \\frac{1}{variable}\n\\]\nand the sequence converges to \\( 1 / variable \\). If \\( endingvalue=0 \\) or \\( 2 / variable \\), then obviously\n\\[\nfinalvalue=terminalvalue=originvalue=\\cdots=0\n\\]\nand if \\( endingvalue \\) lies outside \\( 0,2 / variable \\), then\n\\[\n0>finalvalue>terminalvalue>originvalue>\\ldots\n\\]\nand the sequence diverges to \\( -\\infty \\).\nTo make this rigorous, we detine \\( argument(constant)=constant(2-variable constant) \\). We find that \\( argument \\) achieves its maximum value \\( 1 / variable \\) for \\( constant=1 / variable \\). Moreover, \\( argument(constant)>constant \\) for \\( 0<constant<1 / variable \\) and \\( argument(constant)<constant \\) for \\( constant<0 \\). Then (1), (2), and (3) follow immediately. In case (1) the sequence must be convergent and the limit must be a positive root of \\( argument(constant)=constant \\), and there is only one, namely, \\( constant=1 / variable \\). In case (2) the sequence obviously converges to 0 . In case (3) it either diverges or converges to a negative root of \\( argument(constant)=constant \\), but there is no such root.\n\nTherefore, the sequence converges to \\( 1 / variable \\) if and only if \\( 0<endingvalue<2 / variable \\).\nSecond Solution. We can find an explicit formula for \\( \\boldsymbol{specificvalue} \\) from which the limiting behavior is obvious. We have\n\\[\n1-variable\\;precedingvalue=\\left(1-variable\\;specificvalue\\right)^{2}\n\\]\ntherefore\n\\[\n1-variable\\;specificvalue=\\left(1-variable\\;endingvalue\\right)^{2 \\prime}\n\\]\n\nIt is now obvious that \\( 1-variable\\;specificvalue \\rightarrow 0 \\), that is \\( specificvalue \\rightarrow 1 / variable \\), if and only if \\( \\left|1-variable\\;endingvalue\\right|<1 \\), that is \\( 0<endingvalue<2 / variable \\)."
    },
    "garbled_string": {
      "map": {
        "X_k": "azqvrnpl",
        "X_k+1": "bsnxcoje",
        "X_0": "crdmfiua",
        "X_1": "dgwxltqh",
        "X_2": "eqzmpyra",
        "X_3": "fhbslouk",
        "X_n": "gjcrwmet",
        "X_n+1": "hqveplsz",
        "x": "irnqoyab",
        "f": "jsxlmuke",
        "A": "kptdynof",
        "k": "lqzbewsg",
        "n": "mrycphid"
      },
      "question": "2. If facilities for division are not available, it is sometimes convenient in determining the decimal expansion of \\( 1 / kptdynof, kptdynof>0 \\) to use the iteration \\( bsnxcoje \\) \\( =azqvrnpl\\left(2-kptdynof azqvrnpl\\right), lqzbewsg=0,1,2, \\ldots \\), where \\( crdmfiua \\) is a selected \"starting\" value. Find the limitations, if any, on the starting value \\( crdmfiua \\) in order that the above iteration converges to the desired value \\( 1 / kptdynof \\).",
      "solution": "First Solution. The polygonal representation of this recursion is shown in the figure. (See p. 223 for an explanation.) It is clear that if \\( crdmfiua \\) lies in \\( (0,2 kptdynof) \\) then\n\\[\n0<dgwxltqh \\leq eqzmpyra \\leq fhbslouk \\leq \\cdots \\leq \\frac{1}{kptdynof}\n\\]\nand the sequence converges to \\( 1 / kptdynof \\). If \\( crdmfiua=0 \\) or \\( 2 / kptdynof \\), then obviously\n\\[\ndgwxltqh=eqzmpyra=fhbslouk=\\cdots=0\n\\]\nand if \\( crdmfiua \\) lies outside \\( 0,2 / kptdynof \\), then\n\\[\n0>dgwxltqh>eqzmpyra>fhbslouk>\\ldots\n\\]\nand the sequence diverges to \\( -\\infty \\).\nTo make this rigorous, we detine \\( jsxlmuke(irnqoyab)=irnqoyab(2-kptdynof irnqoyab) \\). We find that \\( jsxlmuke \\) achieves its maximum value \\( 1 / kptdynof \\) for \\( irnqoyab=1 / kptdynof \\). Moreover, \\( jsxlmuke(irnqoyab)>irnqoyab \\) for \\( 0<irnqoyab<1 / kptdynof \\) and \\( jsxlmuke(irnqoyab)<irnqoyab \\) for \\( irnqoyab<0 \\). Then (1), (2), and (3) follow immediately. In case (1) the sequence must be convergent and the limit must be a positive root of \\( jsxlmuke(irnqoyab)=irnqoyab \\), and there is only one, namely, \\( irnqoyab=1 / kptdynof \\). In case (2) the sequence obviously converges to 0 . In case (3) it either diverges or converges to a negative root of \\( jsxlmuke(irnqoyab)=irnqoyab \\), but there is no such root.\n\nTherefore, the sequence converges to \\( 1 / kptdynof \\) if and only if \\( 0<crdmfiua<2 / kptdynof \\).\nSecond Solution. We can find an explicit formula for \\( \\boldsymbol{gjcrwmet} \\) from which the limiting behavior is obvious. We have\n\\[\n1-kptdynof \\, hqveplsz=\\left(1-kptdynof \\, gjcrwmet\\right)^{2}\n\\]\ntherefore\n\\[\n1-kptdynof \\, gjcrwmet=\\left(1-kptdynof \\, crdmfiua\\right)^{2 \\prime}\n\\]\n\nIt is now obvious that \\( 1-kptdynof \\, gjcrwmet \\rightarrow 0 \\), that is \\( gjcrwmet \\rightarrow 1 / kptdynof \\), if and only if \\( \\left|1-kptdynof \\, crdmfiua\\right|<1 \\), that is \\( 0<crdmfiua<2 / kptdynof \\)."
    },
    "kernel_variant": {
      "question": "Let $A\\in\\mathbf R^{\\,n\\times n}$ be invertible with $n\\ge 2$ and consider the Newton–Schulz iteration  \n\\[\n    X_{0}\\in\\mathbf R^{\\,n\\times n}\\;(\\text{arbitrary}),\\qquad \n    X_{k+1}=X_{k}\\bigl(2I-AX_{k}\\bigr),\\qquad k=0,1,2,\\dots .\n    \\tag{$\\star$}\n\\]\nIntroduce the error matrices $E_{k}:=I-AX_{k}$ and denote the spectral radius by $\\rho(\\,\\cdot\\,)$.\n\n1.  Prove the equivalence  \n    \\[\n        \\bigl\\{X_{k}\\text{ converges to }A^{-1}\\text{ in {\\em every} sub-multiplicative norm}\\bigr\\}\n        \\ \\Longleftrightarrow\\ \n        \\rho(E_{0})<1 .\n    \\]\n    Rewrite the convergence criterion solely in terms of the spectrum of $AX_{0}$.\n\n2.  Assume in addition that $A\\succ 0$ (symmetric positive definite) and $X_{0}=X_{0}^{\\mathsf T}$.  \n    Show that the condition in part 1 is equivalent to the Loewner–order inequality\n    \\[\n        0\\prec X_{0}\\prec 2A^{-1}.\n    \\]\n\n3.  Suppose $\\rho(E_{0})<1$.\n\n    (a)  Establish the elementary estimate  \n        \\[\n            \\lVert E_{k}\\rVert\\le\\lVert E_{0}\\rVert^{\\,2^{\\,k}},\n            \\qquad k=0,1,2,\\dots ,\n        \\]\n        for every sub-multiplicative matrix norm.\n\n    (b)  Assume moreover that $E_{0}$ is diagonalisable: $E_{0}=V\\Lambda V^{-1}$.  \n        Prove that for every sub-multiplicative norm\n        \\[\n            \\lVert E_{k}\\rVert\\le\\kappa(E_{0})\\,\\rho(E_{0})^{\\,2^{\\,k}},\n            \\qquad \\kappa(E_{0}):=\\lVert V\\rVert\\,\\lVert V^{-1}\\rVert .\n            \\tag{$\\dagger$}\n        \\]\n        Hence, given $\\varepsilon\\in(0,1)$, derive an explicit upper bound for the first index $k$ satisfying $\\lVert E_{k}\\rVert\\le\\varepsilon$ expressed only through $\\varepsilon$, $\\rho(E_{0})$ and $\\kappa(E_{0})$.\n\n4.  Let $J$ be the Jordan canonical form of $E_{0}$ and assume that at least one Jordan block has size $m\\ge 2$.  Denote such a block by $B(\\lambda)=\\lambda I_{m}+N$ with $N^{m}=0,\\;N^{m-1}\\neq 0$.\n\n    (a)  Show the exact expansion  \n        \\[\n            B(\\lambda)^{\\,2^{\\,k}}\n            =\\sum_{r=0}^{m-1}\\binom{2^{\\,k}}{r}\\lambda^{\\,2^{\\,k}-r}N^{r},\n            \\qquad k=0,1,2,\\dots ,\n        \\]\n        and prove in particular that the entry in the first row and last column satisfies  \n        \\[\n            \\bigl|\\bigl(B(\\lambda)^{\\,2^{\\,k}}\\bigr)_{1m}\\bigr|\n            =\\binom{2^{\\,k}}{m-1}\\,|\\lambda|^{\\,2^{\\,k}-(m-1)}\n            \\asymp 2^{\\,k(m-1)}\\,|\\lambda|^{\\,2^{\\,k}} .\n        \\]\n        Deduce the super-quadratic decay  \n        \\[\n            \\frac{\\bigl|\\bigl(B(\\lambda)^{\\,2^{\\,k+1}}\\bigr)_{1m}\\bigr|}\n                 {\\bigl|\\bigl(B(\\lambda)^{\\,2^{\\,k}}\\bigr)_{1m}\\bigr|^{2}}\n            \\longrightarrow 0 \\qquad (k\\to\\infty).\n        \\]\n\n    (b)  Without assuming diagonalizability, prove that there exist a constant $C>0$ and an integer  \n        \\[\n            s:=\\max\\{m_{j}-1:\\;B_{j}\\text{ is a Jordan block of }E_{0}\\}\n            \\ \\ (0\\le s\\le n-1)\n        \\]\n        such that for every sub-multiplicative norm\n        \\[\n            \\lVert E_{k}\\rVert\\le C\\,2^{\\,ks}\\,\\rho(E_{0})^{\\,2^{\\,k}}\n            \\qquad\\text{for all }k\\ge 0.\n            \\tag{$\\ddagger$}\n        \\]\n        Conclude that if $\\rho(E_{0})<1$ then, for all sufficiently large $k$,\n        \\[\n            \\lVert E_{k+1}\\rVert\\le\\lVert E_{k}\\rVert^{\\,2},\n        \\]\n        i.e.\\ the convergence of $\\lVert E_{k}\\rVert$ is at least quadratic up to a polynomial factor.\n\nProvide complete, detailed proofs for all parts.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Throughout we work in $\\mathbf C^{\\,n\\times n}$ whenever spectral information is needed and write $I:=I_{n}$.\n\n--------------------------------------------------------------------\nStep 0 – A fundamental identity  \nWith $E_{k}:=I-AX_{k}$ one computes directly\n\\[\n    E_{k+1}=I-AX_{k+1}=I-AX_{k}(2I-AX_{k})=(I-AX_{k})^{2}=E_{k}^{2}.\n\\]\nHence\n\\begin{equation}\\label{eq:basic}\n        E_{k}=E_{0}^{\\,2^{\\,k}},\n        \\qquad\n        X_{k}=A^{-1}(I-E_{k})=A^{-1}-A^{-1}E_{k}.\n\\end{equation}\n\n--------------------------------------------------------------------\n1.  Convergence criterion  \n\n($\\Rightarrow$)\\;  \nAssume $X_{k}\\to A^{-1}$ in every sub-multiplicative norm.  By \\eqref{eq:basic} this is equivalent to $E_{k}\\to 0$ in every such norm.\n\n($\\Leftarrow$)\\;  \nConversely, if $E_{k}\\to 0$ then \\eqref{eq:basic} forces $X_{k}\\to A^{-1}$.  \nThus convergence of $\\{X_{k}\\}$ is equivalent to that of $\\{E_{k}\\}$.\n\nBecause all norms on the finite–dimensional space $\\mathbf C^{\\,n\\times n}$ are equivalent, convergence to $0$ in every norm is the same as convergence in at least one norm.  Gelfand’s spectral–radius formula tells us\n\\[\n      \\rho(E_{0})=\\lim_{m\\to\\infty}\\lVert E_{0}^{m}\\rVert^{1/m}.\n\\]\nTherefore $E_{0}^{m}\\to 0$ (in some norm) precisely when $\\rho(E_{0})<1$, giving  \n\\[\n      X_{k}\\to A^{-1}\\quad\\Longleftrightarrow\\quad \\rho(E_{0})<1.\n\\]\n\nRelation with $AX_{0}$.  Because $E_{0}=I-AX_{0}$ we have\n\\[\n      \\lambda\\in\\sigma(E_{0})\n      \\;\\Longleftrightarrow\\;\n      1-\\lambda\\in\\sigma(AX_{0}),\n\\]\nso $\\rho(E_{0})<1$ is equivalent to\n\\[\n      \\sigma(AX_{0})\\subset\\{z\\in\\mathbf C:\\lvert z-1\\rvert<1\\}.\n\\]\n\n--------------------------------------------------------------------\n2.  The symmetric positive definite case  \n\nLet $A\\succ 0$ and $X_{0}=X_{0}^{\\mathsf T}$.  Since $A$ is SPD it possesses a symmetric square root $A^{1/2}$.  Set $Y:=A^{1/2}X_{0}A^{1/2}$.  Congruence preserves Loewner order and spectrum, hence\n\\[\n        \\sigma(AX_{0})=\\sigma(Y)\\subset\\mathbf R.\n\\]\nThe disk condition of part 1 now reads $\\sigma(Y)\\subset(0,2)$, which is equivalent to\n\\[\n        0\\prec Y\\prec 2I\n        \\quad\\Longleftrightarrow\\quad\n        0\\prec X_{0}\\prec 2A^{-1}.\n\\]\n\n--------------------------------------------------------------------\n3.  Quantitative estimates when $\\rho(E_{0})<1$\n\n(a)  Using $E_{k+1}=E_{k}^{2}$ and sub-multiplicativity:\n\\[\n      \\lVert E_{k+1}\\rVert=\\lVert E_{k}^{2}\\rVert\\le\\lVert E_{k}\\rVert^{2},\n\\]\nand induction yields\n\\[\n      \\lVert E_{k}\\rVert\\le\\lVert E_{0}\\rVert^{\\,2^{\\,k}}\\qquad(k\\ge 0).\n\\]\n\n(b)  Assume $E_{0}$ is diagonalisable: $E_{0}=V\\Lambda V^{-1}$ with  \n$\\Lambda=\\operatorname{diag}(\\lambda_{1},\\dots,\\lambda_{n})$.  \nThen, by \\eqref{eq:basic},\n\\[\n      E_{k}=V\\Lambda^{\\,2^{\\,k}}V^{-1},\n\\]\nso for any sub-multiplicative norm\n\\[\n      \\lVert E_{k}\\rVert\n      \\le\\lVert V\\rVert\\,\\lVert V^{-1}\\rVert\\,\n           \\lVert\\Lambda^{\\,2^{\\,k}}\\rVert\n      =\\kappa(E_{0})\\,\\max_{1\\le j\\le n}|\\lambda_{j}|^{\\,2^{\\,k}}\n      =\\kappa(E_{0})\\,\\rho(E_{0})^{\\,2^{\\,k}} ,\n\\]\nwhich is \\eqref{eq:basic}.  Now fix $\\varepsilon\\in(0,1)$ and choose the least $k$ with\n\\[\n      \\kappa(E_{0})\\,\\rho(E_{0})^{\\,2^{\\,k}}\\le\\varepsilon\n      \\quad\\Longleftrightarrow\\quad\n      2^{\\,k}\\ge\\frac{\\log(\\varepsilon/\\kappa(E_{0}))}{\\log\\rho(E_{0})}.\n\\]\nBecause $\\rho(E_{0})<1$, $\\log\\rho(E_{0})<0$; hence an admissible index is\n\\[\n      k\\;=\\;\n      \\Bigl\\lceil\n          \\log_{2}\\!\\Bigl(\n              \\frac{\\log\\bigl(\\varepsilon/\\kappa(E_{0})\\bigr)}\n                   {\\log\\rho(E_{0})}\n          \\Bigr)\n      \\Bigr\\rceil .\n\\]\n\n--------------------------------------------------------------------\n4.  Non-trivial Jordan blocks  \n\n(a)  Since $N^{m}=0$ and $N$ commutes with $\\lambda I_{m}$, the binomial formula gives\n\\[\n      B(\\lambda)^{\\ell}\n      =\\sum_{r=0}^{m-1}\\binom{\\ell}{r}\\lambda^{\\,\\ell-r}N^{r},\n      \\qquad\\ell\\ge 0.\n\\]\nPutting $\\ell=2^{\\,k}$ yields the announced expansion and the estimate\n\\[\n     \\bigl|\\bigl(B(\\lambda)^{\\,2^{\\,k}}\\bigr)_{1m}\\bigr|\n       =\\binom{2^{\\,k}}{m-1}\\,|\\lambda|^{\\,2^{\\,k}-(m-1)}\n       \\asymp 2^{\\,k(m-1)}\\,|\\lambda|^{\\,2^{\\,k}}.\n\\]\nConsequently\n\\[\n     \\frac{\\bigl|\\bigl(B(\\lambda)^{\\,2^{\\,k+1}}\\bigr)_{1m}\\bigr|}\n          {\\bigl|\\bigl(B(\\lambda)^{\\,2^{\\,k}}\\bigr)_{1m}\\bigr|^{2}}\n          \\sim\n          \\frac{|\\lambda|^{\\,m-1}}{(m-1)!}\\,2^{-(k-1)(m-1)}\n          \\longrightarrow 0 \\qquad(k\\to\\infty),\n\\]\nso that the individual entry decays super-quadratically.\n\n(b)  Let $s$ be defined in the statement and write $E_{0}=PJP^{-1}$ with $J=\\operatorname{diag}(B_{1},\\dots,B_{q})$.  From part (a)\n\\[\n      \\lVert B_{j}^{\\,2^{\\,k}}\\rVert\n        \\le C_{j}\\,2^{\\,k(m_{j}-1)}\\,\\rho(B_{j})^{\\,2^{\\,k}},\n\\]\nwhere $m_{j}$ is the size of $B_{j}$ and $C_{j}$ depends only on $m_{j}$.  \nSetting $C:=\\lVert P\\rVert\\,\\lVert P^{-1}\\rVert\\max_{j}C_{j}$ and  \n$s:=\\max_{j}(m_{j}-1)$ we obtain for all $k\\ge 0$\n\\[\n      \\lVert E_{k}\\rVert=\\lVert E_{0}^{\\,2^{\\,k}}\\rVert\n      \\le C\\,2^{\\,ks}\\,\\rho(E_{0})^{\\,2^{\\,k}},\n\\]\nestablishing \\eqref{eq:basic}.  If $\\rho(E_{0})<1$ then  \n$\\displaystyle \\lim_{k\\to\\infty}2^{\\,ks}\\rho(E_{0})^{\\,2^{\\,k}}=0$, so there exists $k_{0}$ such that $\\lVert E_{k}\\rVert\\le 1$ for $k\\ge k_{0}$.  For such $k$\n\\[\n      \\lVert E_{k+1}\\rVert\n      \\le C\\,2^{\\,(k+1)s}\\rho(E_{0})^{\\,2^{\\,k+1}}\n      \\le \\bigl(C\\,2^{\\,ks}\\rho(E_{0})^{\\,2^{\\,k}}\\bigr)^{2}\n      =\\lVert E_{k}\\rVert^{\\,2}\\qquad(k\\ge k_{0}),\n\\]\nso the decay is eventually at least quadratic.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.488509",
        "was_fixed": false,
        "difficulty_analysis": "• Dimensional escalation: The scalar iteration is promoted to an n × n matrix setting, introducing non-commutativity, eigen-structure, Jordan forms and operator norms.  \n• Extra conditions:  Part 2 couples the iteration with Loewner order theory for symmetric positive definite matrices, demanding familiarity with matrix congruences and partial orders.  \n• Deeper theory:  The solution appeals to spectral radius arguments, norm equivalence, Jordan canonical forms, and perturbation estimates, none of which appear in the original exercise.  \n• Multiple interacting concepts:  Convergence analysis (Part 1) intertwines linear algebra (spectra), analysis (norm convergence), and numerical analysis (quadratic rate).  Parts 3 and 4 blend asymptotic estimates with matrix analysis, showing how non-diagonalizability can affect local but not global convergence rates.  \n• Lengthened argument chain:  Establishing E_k = E₀^{2^{k}}, translating spectral conditions to Loewner inequalities, bounding iteration counts, and reconciling Jordan-block anomalies together require substantially more steps and insights than the original single-variable proof."
      }
    },
    "original_kernel_variant": {
      "question": "Let $A\\in\\mathbf R^{\\,n\\times n}$ be invertible with $n\\ge 2$ and consider the Newton–Schulz iteration  \n\\[\n    X_{0}\\in\\mathbf R^{\\,n\\times n}\\;(\\text{arbitrary}),\\qquad \n    X_{k+1}=X_{k}\\bigl(2I-AX_{k}\\bigr),\\qquad k=0,1,2,\\dots .\n    \\tag{$\\star$}\n\\]\nIntroduce the error matrices $E_{k}:=I-AX_{k}$ and denote the spectral radius by $\\rho(\\,\\cdot\\,)$.\n\n1.  Prove the equivalence  \n    \\[\n        \\bigl\\{X_{k}\\text{ converges to }A^{-1}\\text{ in {\\em every} sub-multiplicative norm}\\bigr\\}\n        \\ \\Longleftrightarrow\\ \n        \\rho(E_{0})<1 .\n    \\]\n    Rewrite the convergence criterion solely in terms of the spectrum of $AX_{0}$.\n\n2.  Assume in addition that $A\\succ 0$ (symmetric positive definite) and $X_{0}=X_{0}^{\\mathsf T}$.  \n    Show that the condition in part 1 is equivalent to the Loewner–order inequality\n    \\[\n        0\\prec X_{0}\\prec 2A^{-1}.\n    \\]\n\n3.  Suppose $\\rho(E_{0})<1$.\n\n    (a)  Establish the elementary estimate  \n        \\[\n            \\lVert E_{k}\\rVert\\le\\lVert E_{0}\\rVert^{\\,2^{\\,k}},\n            \\qquad k=0,1,2,\\dots ,\n        \\]\n        for every sub-multiplicative matrix norm.\n\n    (b)  Assume moreover that $E_{0}$ is diagonalisable: $E_{0}=V\\Lambda V^{-1}$.  \n        Prove that for every sub-multiplicative norm\n        \\[\n            \\lVert E_{k}\\rVert\\le\\kappa(E_{0})\\,\\rho(E_{0})^{\\,2^{\\,k}},\n            \\qquad \\kappa(E_{0}):=\\lVert V\\rVert\\,\\lVert V^{-1}\\rVert .\n            \\tag{$\\dagger$}\n        \\]\n        Hence, given $\\varepsilon\\in(0,1)$, derive an explicit upper bound for the first index $k$ satisfying $\\lVert E_{k}\\rVert\\le\\varepsilon$ expressed only through $\\varepsilon$, $\\rho(E_{0})$ and $\\kappa(E_{0})$.\n\n4.  Let $J$ be the Jordan canonical form of $E_{0}$ and assume that at least one Jordan block has size $m\\ge 2$.  Denote such a block by $B(\\lambda)=\\lambda I_{m}+N$ with $N^{m}=0,\\;N^{m-1}\\neq 0$.\n\n    (a)  Show the exact expansion  \n        \\[\n            B(\\lambda)^{\\,2^{\\,k}}\n            =\\sum_{r=0}^{m-1}\\binom{2^{\\,k}}{r}\\lambda^{\\,2^{\\,k}-r}N^{r},\n            \\qquad k=0,1,2,\\dots ,\n        \\]\n        and prove in particular that the entry in the first row and last column satisfies  \n        \\[\n            \\bigl|\\bigl(B(\\lambda)^{\\,2^{\\,k}}\\bigr)_{1m}\\bigr|\n            =\\binom{2^{\\,k}}{m-1}\\,|\\lambda|^{\\,2^{\\,k}-(m-1)}\n            \\asymp 2^{\\,k(m-1)}\\,|\\lambda|^{\\,2^{\\,k}} .\n        \\]\n        Deduce the super-quadratic decay  \n        \\[\n            \\frac{\\bigl|\\bigl(B(\\lambda)^{\\,2^{\\,k+1}}\\bigr)_{1m}\\bigr|}\n                 {\\bigl|\\bigl(B(\\lambda)^{\\,2^{\\,k}}\\bigr)_{1m}\\bigr|^{2}}\n            \\longrightarrow 0 \\qquad (k\\to\\infty).\n        \\]\n\n    (b)  Without assuming diagonalizability, prove that there exist a constant $C>0$ and an integer  \n        \\[\n            s:=\\max\\{m_{j}-1:\\;B_{j}\\text{ is a Jordan block of }E_{0}\\}\n            \\ \\ (0\\le s\\le n-1)\n        \\]\n        such that for every sub-multiplicative norm\n        \\[\n            \\lVert E_{k}\\rVert\\le C\\,2^{\\,ks}\\,\\rho(E_{0})^{\\,2^{\\,k}}\n            \\qquad\\text{for all }k\\ge 0.\n            \\tag{$\\ddagger$}\n        \\]\n        Conclude that if $\\rho(E_{0})<1$ then, for all sufficiently large $k$,\n        \\[\n            \\lVert E_{k+1}\\rVert\\le\\lVert E_{k}\\rVert^{\\,2},\n        \\]\n        i.e.\\ the convergence of $\\lVert E_{k}\\rVert$ is at least quadratic up to a polynomial factor.\n\nProvide complete, detailed proofs for all parts.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "Throughout we work in $\\mathbf C^{\\,n\\times n}$ whenever spectral information is needed and write $I:=I_{n}$.\n\n--------------------------------------------------------------------\nStep 0 – A fundamental identity  \nWith $E_{k}:=I-AX_{k}$ one computes directly\n\\[\n    E_{k+1}=I-AX_{k+1}=I-AX_{k}(2I-AX_{k})=(I-AX_{k})^{2}=E_{k}^{2}.\n\\]\nHence\n\\begin{equation}\\label{eq:basic}\n        E_{k}=E_{0}^{\\,2^{\\,k}},\n        \\qquad\n        X_{k}=A^{-1}(I-E_{k})=A^{-1}-A^{-1}E_{k}.\n\\end{equation}\n\n--------------------------------------------------------------------\n1.  Convergence criterion  \n\n($\\Rightarrow$)\\;  \nAssume $X_{k}\\to A^{-1}$ in every sub-multiplicative norm.  By \\eqref{eq:basic} this is equivalent to $E_{k}\\to 0$ in every such norm.\n\n($\\Leftarrow$)\\;  \nConversely, if $E_{k}\\to 0$ then \\eqref{eq:basic} forces $X_{k}\\to A^{-1}$.  \nThus convergence of $\\{X_{k}\\}$ is equivalent to that of $\\{E_{k}\\}$.\n\nBecause all norms on the finite–dimensional space $\\mathbf C^{\\,n\\times n}$ are equivalent, convergence to $0$ in every norm is the same as convergence in at least one norm.  Gelfand’s spectral–radius formula tells us\n\\[\n      \\rho(E_{0})=\\lim_{m\\to\\infty}\\lVert E_{0}^{m}\\rVert^{1/m}.\n\\]\nTherefore $E_{0}^{m}\\to 0$ (in some norm) precisely when $\\rho(E_{0})<1$, giving  \n\\[\n      X_{k}\\to A^{-1}\\quad\\Longleftrightarrow\\quad \\rho(E_{0})<1.\n\\]\n\nRelation with $AX_{0}$.  Because $E_{0}=I-AX_{0}$ we have\n\\[\n      \\lambda\\in\\sigma(E_{0})\n      \\;\\Longleftrightarrow\\;\n      1-\\lambda\\in\\sigma(AX_{0}),\n\\]\nso $\\rho(E_{0})<1$ is equivalent to\n\\[\n      \\sigma(AX_{0})\\subset\\{z\\in\\mathbf C:\\lvert z-1\\rvert<1\\}.\n\\]\n\n--------------------------------------------------------------------\n2.  The symmetric positive definite case  \n\nLet $A\\succ 0$ and $X_{0}=X_{0}^{\\mathsf T}$.  Since $A$ is SPD it possesses a symmetric square root $A^{1/2}$.  Set $Y:=A^{1/2}X_{0}A^{1/2}$.  Congruence preserves Loewner order and spectrum, hence\n\\[\n        \\sigma(AX_{0})=\\sigma(Y)\\subset\\mathbf R.\n\\]\nThe disk condition of part 1 now reads $\\sigma(Y)\\subset(0,2)$, which is equivalent to\n\\[\n        0\\prec Y\\prec 2I\n        \\quad\\Longleftrightarrow\\quad\n        0\\prec X_{0}\\prec 2A^{-1}.\n\\]\n\n--------------------------------------------------------------------\n3.  Quantitative estimates when $\\rho(E_{0})<1$\n\n(a)  Using $E_{k+1}=E_{k}^{2}$ and sub-multiplicativity:\n\\[\n      \\lVert E_{k+1}\\rVert=\\lVert E_{k}^{2}\\rVert\\le\\lVert E_{k}\\rVert^{2},\n\\]\nand induction yields\n\\[\n      \\lVert E_{k}\\rVert\\le\\lVert E_{0}\\rVert^{\\,2^{\\,k}}\\qquad(k\\ge 0).\n\\]\n\n(b)  Assume $E_{0}$ is diagonalisable: $E_{0}=V\\Lambda V^{-1}$ with  \n$\\Lambda=\\operatorname{diag}(\\lambda_{1},\\dots,\\lambda_{n})$.  \nThen, by \\eqref{eq:basic},\n\\[\n      E_{k}=V\\Lambda^{\\,2^{\\,k}}V^{-1},\n\\]\nso for any sub-multiplicative norm\n\\[\n      \\lVert E_{k}\\rVert\n      \\le\\lVert V\\rVert\\,\\lVert V^{-1}\\rVert\\,\n           \\lVert\\Lambda^{\\,2^{\\,k}}\\rVert\n      =\\kappa(E_{0})\\,\\max_{1\\le j\\le n}|\\lambda_{j}|^{\\,2^{\\,k}}\n      =\\kappa(E_{0})\\,\\rho(E_{0})^{\\,2^{\\,k}} ,\n\\]\nwhich is \\eqref{eq:basic}.  Now fix $\\varepsilon\\in(0,1)$ and choose the least $k$ with\n\\[\n      \\kappa(E_{0})\\,\\rho(E_{0})^{\\,2^{\\,k}}\\le\\varepsilon\n      \\quad\\Longleftrightarrow\\quad\n      2^{\\,k}\\ge\\frac{\\log(\\varepsilon/\\kappa(E_{0}))}{\\log\\rho(E_{0})}.\n\\]\nBecause $\\rho(E_{0})<1$, $\\log\\rho(E_{0})<0$; hence an admissible index is\n\\[\n      k\\;=\\;\n      \\Bigl\\lceil\n          \\log_{2}\\!\\Bigl(\n              \\frac{\\log\\bigl(\\varepsilon/\\kappa(E_{0})\\bigr)}\n                   {\\log\\rho(E_{0})}\n          \\Bigr)\n      \\Bigr\\rceil .\n\\]\n\n--------------------------------------------------------------------\n4.  Non-trivial Jordan blocks  \n\n(a)  Since $N^{m}=0$ and $N$ commutes with $\\lambda I_{m}$, the binomial formula gives\n\\[\n      B(\\lambda)^{\\ell}\n      =\\sum_{r=0}^{m-1}\\binom{\\ell}{r}\\lambda^{\\,\\ell-r}N^{r},\n      \\qquad\\ell\\ge 0.\n\\]\nPutting $\\ell=2^{\\,k}$ yields the announced expansion and the estimate\n\\[\n     \\bigl|\\bigl(B(\\lambda)^{\\,2^{\\,k}}\\bigr)_{1m}\\bigr|\n       =\\binom{2^{\\,k}}{m-1}\\,|\\lambda|^{\\,2^{\\,k}-(m-1)}\n       \\asymp 2^{\\,k(m-1)}\\,|\\lambda|^{\\,2^{\\,k}}.\n\\]\nConsequently\n\\[\n     \\frac{\\bigl|\\bigl(B(\\lambda)^{\\,2^{\\,k+1}}\\bigr)_{1m}\\bigr|}\n          {\\bigl|\\bigl(B(\\lambda)^{\\,2^{\\,k}}\\bigr)_{1m}\\bigr|^{2}}\n          \\sim\n          \\frac{|\\lambda|^{\\,m-1}}{(m-1)!}\\,2^{-(k-1)(m-1)}\n          \\longrightarrow 0 \\qquad(k\\to\\infty),\n\\]\nso that the individual entry decays super-quadratically.\n\n(b)  Let $s$ be defined in the statement and write $E_{0}=PJP^{-1}$ with $J=\\operatorname{diag}(B_{1},\\dots,B_{q})$.  From part (a)\n\\[\n      \\lVert B_{j}^{\\,2^{\\,k}}\\rVert\n        \\le C_{j}\\,2^{\\,k(m_{j}-1)}\\,\\rho(B_{j})^{\\,2^{\\,k}},\n\\]\nwhere $m_{j}$ is the size of $B_{j}$ and $C_{j}$ depends only on $m_{j}$.  \nSetting $C:=\\lVert P\\rVert\\,\\lVert P^{-1}\\rVert\\max_{j}C_{j}$ and  \n$s:=\\max_{j}(m_{j}-1)$ we obtain for all $k\\ge 0$\n\\[\n      \\lVert E_{k}\\rVert=\\lVert E_{0}^{\\,2^{\\,k}}\\rVert\n      \\le C\\,2^{\\,ks}\\,\\rho(E_{0})^{\\,2^{\\,k}},\n\\]\nestablishing \\eqref{eq:basic}.  If $\\rho(E_{0})<1$ then  \n$\\displaystyle \\lim_{k\\to\\infty}2^{\\,ks}\\rho(E_{0})^{\\,2^{\\,k}}=0$, so there exists $k_{0}$ such that $\\lVert E_{k}\\rVert\\le 1$ for $k\\ge k_{0}$.  For such $k$\n\\[\n      \\lVert E_{k+1}\\rVert\n      \\le C\\,2^{\\,(k+1)s}\\rho(E_{0})^{\\,2^{\\,k+1}}\n      \\le \\bigl(C\\,2^{\\,ks}\\rho(E_{0})^{\\,2^{\\,k}}\\bigr)^{2}\n      =\\lVert E_{k}\\rVert^{\\,2}\\qquad(k\\ge k_{0}),\n\\]\nso the decay is eventually at least quadratic.\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.408712",
        "was_fixed": false,
        "difficulty_analysis": "• Dimensional escalation: The scalar iteration is promoted to an n × n matrix setting, introducing non-commutativity, eigen-structure, Jordan forms and operator norms.  \n• Extra conditions:  Part 2 couples the iteration with Loewner order theory for symmetric positive definite matrices, demanding familiarity with matrix congruences and partial orders.  \n• Deeper theory:  The solution appeals to spectral radius arguments, norm equivalence, Jordan canonical forms, and perturbation estimates, none of which appear in the original exercise.  \n• Multiple interacting concepts:  Convergence analysis (Part 1) intertwines linear algebra (spectra), analysis (norm convergence), and numerical analysis (quadratic rate).  Parts 3 and 4 blend asymptotic estimates with matrix analysis, showing how non-diagonalizability can affect local but not global convergence rates.  \n• Lengthened argument chain:  Establishing E_k = E₀^{2^{k}}, translating spectral conditions to Loewner inequalities, bounding iteration counts, and reconciling Jordan-block anomalies together require substantially more steps and insights than the original single-variable proof."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}