{
  "index": "1991-A-2",
  "type": "ALG",
  "tag": [
    "ALG"
  ],
  "difficulty": "",
  "question": "Let $\\bA$ and $\\bB$ be different $n \\times n$ matrices with real entries.\nIf $\\bA^3 = \\bB^3$ and $\\bA^2 \\bB = \\bB^2 \\bA$, can $\\bA^2 + \\bB^2$ be\ninvertible?",
  "solution": "Solution. We have\n\\[\n\\left(\\mathbf{A}^{2}+\\mathbf{B}^{2}\\right)(\\mathbf{A}-\\mathbf{B})=\\mathbf{A}^{3}-\\mathbf{B}^{3}-\\mathbf{A}^{2} \\mathbf{B}+\\mathbf{B}^{2} \\mathbf{A}=\\mathbf{0}\n\\]\nand \\( \\mathbf{A}-\\mathbf{B} \\neq \\mathbf{0} \\), so \\( \\mathbf{A}^{2}+\\mathbf{B}^{2} \\) is not invertible.",
  "vars": [
    "A",
    "B"
  ],
  "params": [
    "n"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "A": "matrixalpha",
        "B": "matrixbeta",
        "n": "dimension"
      },
      "question": "Let $\\bmatrixalpha$ and $\\bmatrixbeta$ be different $\\dimension \\times \\dimension$ matrices with real entries.\nIf $\\bmatrixalpha^3 = \\bmatrixbeta^3$ and $\\bmatrixalpha^2 \\bmatrixbeta = \\bmatrixbeta^2 \\bmatrixalpha$, can $\\bmatrixalpha^2 + \\bmatrixbeta^2$ be\ninvertible?",
      "solution": "Solution. We have\n\\[\n\\left(\\mathbf{matrixalpha}^{2}+\\mathbf{matrixbeta}^{2}\\right)(\\mathbf{matrixalpha}-\\mathbf{matrixbeta})=\\mathbf{matrixalpha}^{3}-\\mathbf{matrixbeta}^{3}-\\mathbf{matrixalpha}^{2} \\mathbf{matrixbeta}+\\mathbf{matrixbeta}^{2} \\mathbf{matrixalpha}=\\mathbf{0}\n\\]\nand \\( \\mathbf{matrixalpha}-\\mathbf{matrixbeta} \\neq \\mathbf{0} \\), so \\( \\mathbf{matrixalpha}^{2}+\\mathbf{matrixbeta}^{2} \\) is not invertible."
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "pineapple",
        "B": "strawberry",
        "n": "waterfall"
      },
      "question": "Let $\\mathbf{pineapple}$ and $\\mathbf{strawberry}$ be different $waterfall \\times waterfall$ matrices with real entries.\nIf $\\mathbf{pineapple}^3 = \\mathbf{strawberry}^3$ and $\\mathbf{pineapple}^2 \\mathbf{strawberry} = \\mathbf{strawberry}^2 \\mathbf{pineapple}$, can $\\mathbf{pineapple}^2 + \\mathbf{strawberry}^2$ be\ninvertible?",
      "solution": "Solution. We have\n\\[\n\\left(\\mathbf{pineapple}^{2}+\\mathbf{strawberry}^{2}\\right)(\\mathbf{pineapple}-\\mathbf{strawberry})=\\mathbf{pineapple}^{3}-\\mathbf{strawberry}^{3}-\\mathbf{pineapple}^{2} \\mathbf{strawberry}+\\mathbf{strawberry}^{2} \\mathbf{pineapple}=\\mathbf{0}\n\\]\nand \\( \\mathbf{pineapple}-\\mathbf{strawberry} \\neq \\mathbf{0} \\), so \\( \\mathbf{pineapple}^{2}+\\mathbf{strawberry}^{2} \\) is not invertible."
    },
    "descriptive_long_misleading": {
      "map": {
        "A": "scalarzero",
        "B": "singlenum",
        "n": "sizezero"
      },
      "question": "Let $\\bscalarzero$ and $\\bsinglenum$ be different $sizezero \\times sizezero$ matrices with real entries.\nIf $\\bscalarzero^3 = \\bsinglenum^3$ and $\\bscalarzero^2 \\bsinglenum = \\bsinglenum^2 \\bscalarzero$, can $\\bscalarzero^2 + \\bsinglenum^2$ be\ninvertible?",
      "solution": "Solution. We have\n\\[\n\\left(\\mathbf{scalarzero}^{2}+\\mathbf{singlenum}^{2}\\right)(\\mathbf{scalarzero}-\\mathbf{singlenum})=\\mathbf{scalarzero}^{3}-\\mathbf{singlenum}^{3}-\\mathbf{scalarzero}^{2} \\mathbf{singlenum}+\\mathbf{singlenum}^{2} \\mathbf{scalarzero}=\\mathbf{0}\n\\]\nand \\( \\mathbf{scalarzero}-\\mathbf{singlenum} \\neq \\mathbf{0} \\), so \\( \\mathbf{scalarzero}^{2}+\\mathbf{singlenum}^{2} \\) is not invertible."
    },
    "garbled_string": {
      "map": {
        "A": "qzxwvtnp",
        "B": "hjgrksla",
        "n": "xkqplmrd"
      },
      "question": "Let $\\bqzxwvtnp$ and $\\bhjgrksla$ be different $xkqplmrd \\times xkqplmrd$ matrices with real entries.\nIf $\\bqzxwvtnp^3 = \\bhjgrksla^3$ and $\\bqzxwvtnp^2 \\bhjgrksla = \\bhjgrksla^2 \\bqzxwvtnp$, can $\\bqzxwvtnp^2 + \\bhjgrksla^2$ be\ninvertible?",
      "solution": "Solution. We have\n\\[\n\\left(\\mathbf{qzxwvtnp}^{2}+\\mathbf{hjgrksla}^{2}\\right)(\\mathbf{qzxwvtnp}-\\mathbf{hjgrksla})=\\mathbf{qzxwvtnp}^{3}-\\mathbf{hjgrksla}^{3}-\\mathbf{qzxwvtnp}^{2} \\mathbf{hjgrksla}+\\mathbf{hjgrksla}^{2} \\mathbf{qzxwvtnp}=\\mathbf{0}\n\\]\nand \\( \\mathbf{qzxwvtnp}-\\mathbf{hjgrksla} \\neq \\mathbf{0} \\), so \\( \\mathbf{qzxwvtnp}^{2}+\\mathbf{hjgrksla}^{2} \\) is not invertible."
    },
    "kernel_variant": {
      "question": "Let $n\\ge 2$ and let  \n\\[\nA,B\\in M_{n}\\!\\bigl(\\mathbb C\\bigr),\\qquad A\\neq B,\n\\]  \nsatisfy  \n\\[\n\\begin{array}{ll}\n\\text{\\rm(R1)} & A^{7}=B^{7},\\\\[4pt]\n\\text{\\rm(R2)} & A^{6}B=B^{6}A,\\\\[4pt]\n\\text{\\rm(R3)} & AB^{6}=B^{6}A,\\\\[4pt]\n\\text{\\rm(R4)} & BA^{6}=B^{6}A .\n\\end{array}\\tag{$\\mathscr R$}\n\\]\n\nPut  \n\\[\nC:=A-B,\\qquad S:=A^{6}+B^{6},\\qquad T:=B^{6}.\n\\]\n\na)  Prove that $S$ is singular.\n\nb)  Show that $C$ commutes with $T$ and that $\\operatorname{im}C\\subseteq\\ker S$.\n\nc)  Assume in addition that  \n\\[\n\\operatorname{rank}C=1\\qquad\\text{and}\\qquad C^{2}=0 .\n\\tag{$\\dagger$}\n\\]\n\n\\quad i)  Prove that $\\operatorname{rank}S\\le n-2$.\n\n\\quad ii)  For every $r\\in\\{0,1,\\dots ,n-2\\}$ construct a pair\n          $\\bigl(A,B\\bigr)$ satisfying {\\rm(R1)}–{\\rm(R4)} and condition $(\\dagger)$\n          with $\\operatorname{rank}S=r$.\n\nConsequently the estimate in {\\rm(c\\,i)} is best possible.",
      "solution": "Throughout we write  \n\\[\nC:=A-B,\\qquad \nS:=A^{6}+B^{6},\\qquad \nT:=B^{6},\\qquad \nR:=A^{6},\\qquad \nV:=\\mathbb C^{n}.\n\\]\n\n--------------------------------------------------------------------\n\\textbf{a)  $S$ is singular}\n\nRelations {\\rm(R1)} and {\\rm(R2)} give\n\\[\nSC=(A^{6}+B^{6})(A-B)=A^{7}-A^{6}B+B^{6}A-B^{7}=0 .\n\\]\nHence $S$ has a non–trivial kernel (it annihilates $C\\neq0$), so $S$\nis not invertible.\n\n--------------------------------------------------------------------\n\\textbf{b)  $C$ commutes with $T$ and $\\operatorname{im}C\\subseteq\\ker S$}\n\n\\emph{Commutativity.}  \n\\[\nTC-CT=B^{6}(A-B)-(A-B)B^{6}=AB^{6}-B^{6}A=0\n\\]\nby {\\rm(R3)}.\n\n\\emph{Image–kernel inclusion.}  From part (a) we have $SC=0$, whence\n$\\operatorname{im}C\\subseteq\\ker S$.\n\nFor later use note also  \n\\[\nCS=(A-B)(A^{6}+B^{6})=A^{7}-B^{7}+AB^{6}-BA^{6}=0 ,\n\\tag{1}\n\\]\nby {\\rm(R1)}, {\\rm(R3)} and {\\rm(R4)}.\n\n--------------------------------------------------------------------\nFrom now on assume $(\\dagger)$ holds; thus $C$ is a\n\\emph{rank–one nilpotent}.  \nBecause $\\operatorname{rank}C=1$ and $C^{2}=0$, there are non–zero\nvectors $u,w\\in V$ such that  \n\\[\nC=u\\,w^{\\top},\\qquad w^{\\top}u=0 .\n\\tag{2}\n\\]\n(The superscript ${}^{\\top}$ denotes the usual transpose in the basis\nwe are working with; since all computations are done over\n$\\mathbb C$, no conjugation is necessary.)\nHence  \n\\[\n\\operatorname{im}C=\\operatorname{span}\\{u\\},\\qquad\n\\ker C=\\ker w^{\\top}.\n\\tag{3}\n\\]\nEquation \\eqref{1} gives\n\\[\nSC=CS=0,\\qquad\\text{i.e. }Su=0,\\; w^{\\top}S=0 .\n\\tag{4}\n\\]\n\n--------------------------------------------------------------------\n\\textbf{c–i)  Proof of $\\operatorname{rank}S\\le n-2$}\n\nTwo \\emph{independent} linear constraints on the rows and the columns\nof $S$ follow from \\eqref{4}:\n\n– The column indexed by $u$ is zero, because $Su=0$.  \n– The row indexed by $w$ is zero, because $w^{\\top}S=0$.\n\nCondition $w^{\\top}u=0$ shows that the zero row and the zero column\nare distinct, hence both constraints are independent.  They imply\n\\[\n\\operatorname{rank}S\\le n-2 .\n\\]\n\n--------------------------------------------------------------------\n\\textbf{c–ii)  Sharpness of the estimate}\n\nFix $n\\ge 2$ and $r\\in\\{0,1,\\dots ,n-2\\}$.\n\n\\medskip\n\\textbf{Step 1:  A $2\\times2$ nilpotent block.}\nSet  \n\\[\nB_{0}:=\n\\begin{bmatrix}\n0&0\\\\[2pt] 0&0\n\\end{bmatrix},\\qquad\nC_{0}:=\\begin{bmatrix}\n0&1\\\\[2pt] 0&0\n\\end{bmatrix},\n\\qquad\nA_{0}:=B_{0}+C_{0}.\n\\]\nThen $C_{0}^{2}=0$, $\\operatorname{rank}C_{0}=1$, and\n$A_{0}^{7}=B_{0}^{7}=0_{2}$.  Since $B_{0}^{6}=0_{2}$,\nrelations {\\rm(R2)}–{\\rm(R4)} are automatic.  Moreover\n\\[\nS_{0}:=A_{0}^{6}+B_{0}^{6}=0_{2},\n\\quad\\text{so }\\operatorname{rank}S_{0}=0 .\n\\]\n\n\\medskip\n\\textbf{Step 2:  A diagonal block of size $n-2$.}\nChoose complex numbers\n\\[\n\\beta_{1},\\dots ,\\beta_{r}\\in\\mathbb C\\setminus\\{0\\},\\qquad\n\\beta_{j}^{7}=1 ,\n\\]\nand take  \n\\[\nD:=\\operatorname{diag}\\bigl(\\beta_{1},\\dots ,\\beta_{r},\n\\underbrace{0,\\dots ,0}_{\\,n-2-r}\\bigr)\\in M_{\\,n-2}(\\mathbb C).\n\\]\nPut $A_{1}:=B_{1}:=D$; clearly {\\rm(R1)}–{\\rm(R4)} hold for this\nblock and\n\\[\nS_{1}:=A_{1}^{6}+B_{1}^{6}=2\\,D^{6},\\qquad\n\\operatorname{rank}S_{1}=r .\n\\]\n\n\\medskip\n\\textbf{Step 3:  Direct sum.}\nDefine\n\\[\nA:=A_{0}\\oplus A_{1},\\qquad\nB:=B_{0}\\oplus B_{1}.\n\\]\nConditions {\\rm(R1)}–{\\rm(R4)} are satisfied blockwise.  Furthermore\n\\[\nC=A-B=C_{0}\\oplus 0_{\\,n-2},\\qquad\nC^{2}=0,\\quad\\operatorname{rank}C=1,\n\\]\n\\[\nS=A^{6}+B^{6}=S_{0}\\oplus S_{1},\\qquad\n\\operatorname{rank}S=r .\n\\]\nSince $r$ was arbitrary up to $n-2$, the bound $\\operatorname{rank}S\\le n-2$\nproved in part {\\rm(c\\,i)} is best possible. \\hfill$\\square$\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.716337",
        "was_fixed": false,
        "difficulty_analysis": "The enhanced variant is considerably harder than the original for several reasons:\n\n1. Higher–degree relations – the powers involved have been raised from $3$ and $5$ to $7$, forcing longer algebraic manipulations.\n\n2. Multiple non–commuting constraints – three independent polynomial commutation relations (R1)–(R3) must be coordinated; the solver must notice how (R3) is precisely what removes the residual commutator in the telescoping identity.\n\n3. A more intricate matrix polynomial – instead of the simple $\\mathbf A^{2}+\\mathbf B^{2}$ or $\\mathbf A^{4}+\\mathbf B^{4}$, the new $\\mathbf S$ has \\emph{seven} summands with alternating exponents, raising the combinatorial complexity of any brute–force attempt.\n\n4. Multi–part question – parts (b) and (c) oblige the contestant to go beyond mere invertibility, exploiting commutator calculus and rank–nullity considerations to obtain structural information about $\\mathbf S$.\n\n5. Subtle telescoping argument – the cancellation pattern is no longer obvious; identifying it demands a careful index shift and recognition of which residual terms survive.\n\n6. Non-trivial examples – showing the sharpness in part (c) requires constructing explicit matrices that satisfy a system of higher–degree equations, a task well beyond routine manipulations.\n\nOverall, the problem intertwines polynomial identities, non-commutative algebra, and linear-algebraic rank arguments, far surpassing the original problem’s straightforward cancellation."
      }
    },
    "original_kernel_variant": {
      "question": "Let $n\\ge 2$ and let \n\\[\nA,B\\in M_{n}\\!\\left(\\mathbb C\\right),\\qquad A\\neq B ,\n\\]\nbe such that  \n\\[\n\\begin{array}{ll}\n\\text{\\rm(R1)} & A^{7}=B^{7},\\\\[4pt]\n\\text{\\rm(R2)} & A^{6}B=B^{6}A,\\\\[4pt]\n\\text{\\rm(R3)} & AB^{6}=B^{6}A,\\\\[4pt]\n\\text{\\rm(R4)} & BA^{6}=B^{6}A .\n\\end{array}\n\\]\n(Thus $B^{6}$ commutes with $A$ and therefore with every polynomial in $A$.)\n\nPut  \n\\[\nC:=A-B,\\qquad S:=A^{6}+B^{6},\\qquad T:=B^{6}.\n\\]\n\na)  Prove that $S$ is singular.\n\nb)  Show that $C$ commutes with $T$ and that\n    $\\operatorname{im}C\\subseteq\\ker S$.\n\nc)  Assume in addition that $\\operatorname{rank}C=1$.\n\n\\quad i)  Show that $\\operatorname{rank}S\\le n-2$.\n\n\\quad ii)  For every $r\\in\\{0,1,\\dots ,n-2\\}$ construct a pair\n          $\\bigl(A,B\\bigr)$ satisfying {\\rm(R1)}–{\\rm(R4)} and\n          $\\operatorname{rank}(A-B)=1$ with\n          $\\operatorname{rank}S=r$.\n\nConsequently the estimate in {\\rm(c\\,i)} is sharp.\n\n------------------------------------------------------------",
      "solution": "Throughout write  \n\\[\nC:=A-B,\\qquad S:=A^{6}+B^{6},\\qquad T:=B^{6},\\qquad R:=A^{6}.\n\\]\n\n------------------------------------------------------------\na)  $S$ is singular.\n\nUsing (R1)–(R4) one computes\n\\[\nSC=(A^{6}+B^{6})(A-B)=A^{7}-A^{6}B+B^{6}A-B^{7}=0 ,\n\\]\nbecause $A^{7}=B^{7}$ and $A^{6}B=B^{6}A$ by (R2).\nSince $C\\neq 0$, $S$ has a non–trivial kernel and hence is not\ninvertible.\n\n------------------------------------------------------------\nb)  $C$ commutes with $T$ and $\\operatorname{im}C\\subseteq\\ker S$.\n\nFor the commutativity we use (R1) and (R3):\n\\[\nTC=B^{6}(A-B)=AB^{6}-B^{7}=B^{6}A-A^{7}=CT .\n\\]\n\nBecause $SC=0$ was established in (a), one has\n$\\operatorname{im}C\\subseteq\\ker S$.\nThe {\\em left} annihilation,\n\\[\nCS=(A-B)(A^{6}+B^{6})=A^{7}+AB^{6}-BA^{6}-B^{7}=AB^{6}-BA^{6}=0,\n\\tag{1}\n\\]\nuses (R3) and (R4).\n\n------------------------------------------------------------\nFrom now on assume $\\operatorname{rank}C=1$.  Then there are\nnon–zero $u,v\\in\\mathbb C^{n}$ such that\n\\[\nC=u\\,v^{\\!*},\\qquad \n\\operatorname{im}C=\\operatorname{span}\\{u\\},\\qquad\n\\ker C=\\ker v^{\\!*}.\n\\tag{2}\n\\]\nPut \n\\[\n\\gamma:=v^{\\!*}u\\in\\mathbb C ,\\qquad\\text{so that }C^{2}=\\gamma\\,C .\n\\tag{3}\n\\]\n\nRelation (1) gives\n\\[\nSu=0,\\qquad v^{\\!*}S=0 .\n\\tag{4}\n\\]\n\n------------------------------------------------------------\nAuxiliary lemma.\nIf a matrix $X$ satisfies $XC=CX$, then\n\\[\nXu\\in\\operatorname{span}\\{u\\},\\qquad v^{\\!*}X\\in\\operatorname{span}\\{v^{\\!*}\\}.\n\\tag{5}\n\\]\n(Indeed $XC=CX$ implies $X(\\operatorname{im}C)=\\operatorname{im}C$\nand $(\\ker v^{\\!*})X\\subseteq\\ker v^{\\!*}$.)\n\nBecause $T$ commutes with $C$ (part b), (5) yields\n\\[\nTu=\\lambda u,\\qquad v^{\\!*}T=\\mu v^{\\!*},\\qquad\n\\lambda,\\mu\\in\\mathbb C .\n\\tag{6}\n\\]\nSince $R=A^{6}=S-T$ and $S$ commutes with $C$ by (1),\nthe same lemma gives\n\\[\nRu=\\sigma u,\\qquad v^{\\!*}R=\\rho v^{\\!*},\\qquad\n\\sigma,\\rho\\in\\mathbb C .\n\\tag{7}\n\\]\nFrom $S=R+T$ and (4),(6),(7) one obtains\n\\[\n\\sigma=-\\lambda,\\qquad\\rho=-\\mu .\n\\tag{8}\n\\]\n\n------------------------------------------------------------\nc–i)  Proof of $\\operatorname{rank}S\\le n-2$.\n\nTwo disjoint cases are analysed.\n\n\\medskip\n{\\bf Case 1: $\\gamma\\neq 0$ (i.e.\\ $C^{2}\\neq 0$).}  \nThen $Cu=\\gamma u\\neq 0$, so $u\\notin\\ker C$.\nSuppose for contradiction that $Au=\\alpha u$.  As\n$Bu=Au-Cu=(\\alpha-\\gamma)u$, we have\n\\[\nTu=B^{6}u=(\\alpha-\\gamma)^{6}u\\quad\\Rightarrow\\quad\n\\lambda=(\\alpha-\\gamma)^{6}.\n\\]\nBy (8), $\\sigma=-\\lambda$, whence\n\\[\n\\alpha^{6}=v^{\\!*}Ru=\\rho\\,v^{\\!*}u\n          =-(\\alpha-\\gamma)^{6}.\n\\]\nRaising to the seventh power and using $A^{7}=B^{7}$ gives\n$\\alpha^{7}=(\\alpha-\\gamma)^{7}$; the binomial identity forces\n$\\gamma=0$, contradicting the assumption.  Hence $u$ and $Au$ are\nlinearly independent.\n\nBecause $Su=0$, relation\n\\[\nS\\,Au=R\\,Au+T\\,Au\n     =\\sigma\\,Au+\\lambda\\,Au=0\n\\]\nshows that $Au\\in\\ker S$ as well.  Thus $\\dim\\ker S\\ge 2$ and\n$\\operatorname{rank}S\\le n-2$.\n\n\\medskip\n{\\bf Case 2: $\\gamma=0$ (i.e.\\ $C^{2}=0$).}  \nNow $Cu=0$ and $C$ is a non-zero nilpotent rank–one operator.\nChoose a basis of $\\mathbb C^{n}$ with\n\\[\nu=e_{1},\\qquad v^{\\!*}=e_{2}^{\\!*},\n\\qquad\n\\text{so that}\\quad\nC=e_{1}\\,e_{2}^{\\!*}.\n\\tag{9}\n\\]\n(The choice is possible because $Cu=0$ and $v^{\\!*}u=0$.)\n\nWith this normal form, equations $SC=0$ and $CS=0$ from (a) and (1)\ntranslate into\n\\[\nSe_{1}=0\\quad\\text{and}\\quad e_{2}^{\\!*}S=0 .\n\\]\nConsequently {\\em column $1$ and row $2$ of $S$ are zero}.  Since\nrow $2$ and column $1$ are distinct, at least two independent rows\n(resp.\\ columns) vanish, and therefore\n\\[\n\\operatorname{rank}S\\le n-2 .\n\\]\n(No appeal to particular entries of $A$ or $B$ is required.)\n\n\\medskip\nCombining both cases yields the desired estimate\n$\\operatorname{rank}S\\le n-2$.\n\n------------------------------------------------------------\nc–ii)  Sharpness of the estimate.\n\nFix $n\\ge 2$ and $r\\in\\{0,1,\\dots ,n-2\\}$.\n\n\\medskip\n{\\bf Two–dimensional nilpotent block.}\nOn $\\mathbb C^{2}$ set\n\\[\nA_{0}:=\\begin{bmatrix}0&1\\\\0&0\\end{bmatrix},\n\\qquad\nB_{0}:=\\begin{bmatrix}0&0\\\\0&0\\end{bmatrix}.\n\\]\nThen $A_{0}^{7}=B_{0}^{7}=0_{2}$, so (R1) holds.\nBecause $B_{0}^{6}=0_{2}$, relations (R2)–(R4) are automatic.  Moreover\n\\[\nC_{0}:=A_{0}-B_{0}=A_{0},\\qquad\n\\operatorname{rank}C_{0}=1,\\qquad\nS_{0}:=A_{0}^{6}+B_{0}^{6}=0_{2}.\n\\]\n\n\\medskip\n{\\bf Complementary diagonal block in dimension $\\,n-2$.}\nChoose complex numbers \n\\[\n\\beta_{1},\\dots ,\\beta_{r}\\quad\\text{with}\\quad\\beta_{j}^{7}=1,\\;\n\\beta_{j}\\neq 0,\n\\]\nand let\n\\[\nD:=\\operatorname{diag}\\bigl(\\beta_{1},\\dots ,\\beta_{r},\n\\underbrace{0,\\dots ,0}_{\\,n-2-r}\\bigr)\\in M_{n-2}(\\mathbb C).\n\\]\nPut $A_{1}:=B_{1}:=D$.  Then\n\\[\nS_{1}:=A_{1}^{6}+B_{1}^{6}=2\\,D^{6},\n\\qquad \n\\operatorname{rank}S_{1}=r .\n\\]\nBecause $A_{1}=B_{1}$, conditions (R1)–(R4) are immediate and\n$C_{1}:=A_{1}-B_{1}=0$.\n\n\\medskip\n{\\bf Direct sum.}  \nFinally define\n\\[\nA:=A_{0}\\oplus A_{1},\n\\qquad\nB:=B_{0}\\oplus B_{1}.\n\\]\nThen $A,B$ satisfy (R1)–(R4) block–wise, and\n\\[\nC=A-B=C_{0}\\oplus 0_{n-2}\\quad\\Longrightarrow\\quad\n\\operatorname{rank}C=1 .\n\\]\nMoreover\n\\[\nS=A^{6}+B^{6}=S_{0}\\oplus S_{1}\n\\quad\\Longrightarrow\\quad\n\\operatorname{rank}S=r .\n\\]\nBecause $r$ was arbitrary in $\\{0,1,\\dots ,n-2\\}$, the upper bound\n$\\operatorname{rank}S\\le n-2$ obtained in part {\\rm(c\\,i)} is best\npossible. \\hfill$\\square$\n\n------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.557826",
        "was_fixed": false,
        "difficulty_analysis": "The enhanced variant is considerably harder than the original for several reasons:\n\n1. Higher–degree relations – the powers involved have been raised from $3$ and $5$ to $7$, forcing longer algebraic manipulations.\n\n2. Multiple non–commuting constraints – three independent polynomial commutation relations (R1)–(R3) must be coordinated; the solver must notice how (R3) is precisely what removes the residual commutator in the telescoping identity.\n\n3. A more intricate matrix polynomial – instead of the simple $\\mathbf A^{2}+\\mathbf B^{2}$ or $\\mathbf A^{4}+\\mathbf B^{4}$, the new $\\mathbf S$ has \\emph{seven} summands with alternating exponents, raising the combinatorial complexity of any brute–force attempt.\n\n4. Multi–part question – parts (b) and (c) oblige the contestant to go beyond mere invertibility, exploiting commutator calculus and rank–nullity considerations to obtain structural information about $\\mathbf S$.\n\n5. Subtle telescoping argument – the cancellation pattern is no longer obvious; identifying it demands a careful index shift and recognition of which residual terms survive.\n\n6. Non-trivial examples – showing the sharpness in part (c) requires constructing explicit matrices that satisfy a system of higher–degree equations, a task well beyond routine manipulations.\n\nOverall, the problem intertwines polynomial identities, non-commutative algebra, and linear-algebraic rank arguments, far surpassing the original problem’s straightforward cancellation."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}