{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-vector-space-dimension-01",
  "mko_id": "mko-vector-space-dimension",
  "title": {
    "zh-Hant": "先證基底再讀取維度",
    "en": "Prove a basis before reading dimension"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-vector-space-dimension-01",
    "source": {
      "mko_id": "mko-vector-space-dimension",
      "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f",
      "sha256": "0d547787130d8aa91b60c2231053891922efe324656000b1e67b8cd7f2658dd8",
      "english_sha256": "4160447f99df20950bd70ea71b06a3a8a3bcf0dfe817b1224bc3e955bfad0ef0"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "calc"
    ],
    "goal": {
      "zh-Hant": "B=((1,1),(1,−1))是R²有序基底，故dim R²=2。",
      "en": "B=((1,1),(1,-1)) is an ordered basis of R2, so dim R2=2."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "V=R²；b₁=(1,1)，b₂=(1,−1)；P的columns依序為b₁、b₂。",
          "en": "V=R2; b1=(1,1), b2=(1,-1); P has b1,b2 as its ordered columns."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1"
        ],
        "before_tex": "b_1=(1,1),\\quad b_2=(1,-1)",
        "after_tex": "P=\\begin{pmatrix}1&1\\\\1&-1\\end{pmatrix}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "完整兩個columns均列出，基底判定必同時看獨立與生成。",
          "en": "Both columns are complete; basis qualification requires independence and spanning."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1"
          ],
          "explanation": {
            "zh-Hant": "矩陣有兩row、兩column，母空間固定R²。",
            "en": "There are two rows and two columns in the fixed ambient R2."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "S01"
        ],
        "before_tex": "P=\\begin{pmatrix}1&1\\\\1&-1\\end{pmatrix}",
        "after_tex": "R_2\\leftarrow R_2-R_1:\\quad\\begin{pmatrix}1&1\\\\0&-2\\end{pmatrix}",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "第二row減第一row是可逆操作，逆操作是加回第一row。",
          "en": "Subtracting row one from row two is invertible, with addition as its inverse."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01"
          ],
          "explanation": {
            "zh-Hant": "−2為精確值，不是接近零的浮點pivot。",
            "en": "The pivot -2 is exact, not a floating-point near-zero value."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S03",
        "uses": [
          "G1",
          "S02"
        ],
        "before_tex": "\\begin{pmatrix}1&1\\\\0&-2\\end{pmatrix}",
        "after_tex": "R_2\\leftarrow-\\tfrac12R_2:\\quad\\begin{pmatrix}1&1\\\\0&1\\end{pmatrix}\\ ;\\ R_1\\leftarrow R_1-R_2:\\quad I_2",
        "relation": "implies",
        "rule": "calc",
        "conditions": [
          "the pivot -2 is nonzero"
        ],
        "explanation": {
          "zh-Hant": "乘−1/2合法因−2非零，接著清掉上方1，得完整RREF。",
          "en": "Scaling by -1/2 is valid because -2 is nonzero; clearing the upper entry gives the full RREF."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S02"
          ],
          "explanation": {
            "zh-Hant": "兩操作均可逆，所以齊次解與相容性均保留。",
            "en": "Both operations are invertible, preserving homogeneous solutions and consistency."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S04",
        "uses": [
          "G1",
          "S01",
          "S03"
        ],
        "before_tex": "\\alpha b_1+\\beta b_2=0",
        "after_tex": "P\\binom{\\alpha}{\\beta}=0\\iff\\alpha=\\beta=0",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "RREF有兩pivot且無自由變數，唯一齊次解是零，證線性獨立。",
          "en": "Two pivots and no free variables leave only the zero homogeneous solution, proving independence."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "獨立只給維度至少2；生成仍須另證。",
            "en": "Independence gives only a lower bound of 2; spanning still needs proof."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S05",
        "uses": [
          "G1",
          "S01"
        ],
        "before_tex": "(x,y)\\in\\mathbb R^2",
        "after_tex": "(x,y)=\\frac{x+y}{2}b_1+\\frac{x-y}{2}b_2",
        "relation": "implies",
        "rule": "calc",
        "conditions": [
          "2 is nonzero in R"
        ],
        "explanation": {
          "zh-Hant": "對任意實座標給明示係數，兩座標反代為x、y，證完整生成。",
          "en": "Explicit coefficients for arbitrary real coordinates substitute back to x,y, proving full spanning."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01"
          ],
          "explanation": {
            "zh-Hant": "除2合法；x、y未受有限網格限制。",
            "en": "Division by 2 is valid and x,y are not restricted to a finite grid."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S06",
        "uses": [
          "G1",
          "S04",
          "S05"
        ],
        "before_tex": "B\\text{ independent and spans }V",
        "after_tex": "B\\text{ basis};\\quad\\dim V=|B|=2",
        "relation": "implies",
        "rule": "mko_definition",
        "conditions": [],
        "explanation": {
          "zh-Hant": "先完成基底資格，再以假設的有限維不變定理讀取維度。",
          "en": "Only after basis qualification do we read dimension using explicitly assumed finite-dimensional invariance."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S04",
            "S05"
          ],
          "explanation": {
            "zh-Hant": "維度2不是因清單有兩項，而是兩項已證為基底。",
            "en": "The dimension is 2 because the vectors form a basis, not merely because the list has length two."
          }
        },
        "leaf_basis": "mko_definition"
      },
      {
        "id": "S07",
        "uses": [
          "G1",
          "S06"
        ],
        "before_tex": "C=((1,0))",
        "after_tex": "C\\text{ independent},\\quad\\operatorname{span}C=\\{(x,0):x\\in\\mathbb R\\}\\ne V",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "單一非零向量獨立，但不能生成(0,1)，故大小1不等於dim V。",
          "en": "A single nonzero vector is independent but misses (0,1), so size 1 is not dim V."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S06"
          ],
          "explanation": {
            "zh-Hant": "此例區分獨立的下界與確切維度。",
            "en": "This separates an independence lower bound from exact dimension."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S08",
        "uses": [
          "G1",
          "S06"
        ],
        "before_tex": "D=((1,0),(0,1),(1,1))",
        "after_tex": "D\\text{ spans }V,\\quad(1,0)+(0,1)-(1,1)=0",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "前兩項生成V，但三項有非平凡關係，故生成清單大小3只是上界。",
          "en": "The first two span V, but the three have a nontrivial relation, so list size 3 is only an upper bound."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S06"
          ],
          "explanation": {
            "zh-Hant": "三係數1、1、−1不全為零，不能稱三項為基底。",
            "en": "Coefficients 1,1,-1 are not all zero, so the list is not a basis."
          }
        },
        "leaf_basis": "calc"
      }
    ],
    "result": {
      "zh-Hant": "B=((1,1),(1,−1))是R²有序基底，故dim R²=2。",
      "en": "B=((1,1),(1,-1)) is an ordered basis of R2, so dim R2=2."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "一個獨立向量表示dim R²=1。",
          "en": "One independent vector means dim R2=1."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "尚未生成(0,1)，只給下界1。",
          "en": "It need not span (0,1); it supplies only lower bound 1."
        }
      },
      {
        "claim": {
          "zh-Hant": "三項生成清單表示dim R²=3。",
          "en": "A spanning list of three means dim R2=3."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "上述D有非平凡線性關係。",
          "en": "The displayed D is dependent."
        }
      },
      {
        "claim": {
          "zh-Hant": "det P=0因此本例維度0。",
          "en": "det P=0, hence dimension zero here."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "det P=−2不是0，完整RREF為I₂。",
          "en": "det P=-2, not zero, and the full RREF is I2."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "calc",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "假設通常實數域與座標向量運算、矩陣乘法、線性獨立及生成的定義；所用消去均為可逆初等列運算。另明示假設有限維基底大小不變定理。",
          "en": "Assume the real field, usual coordinate-vector arithmetic, matrix multiplication, and definitions of independence and span; elimination uses invertible elementary row operations. Also explicitly assume finite-dimensional basis-cardinality invariance."
        }
      },
      {
        "id": "mko_definition",
        "kind": "cited",
        "statement": {
          "zh-Hant": "本輪只處理R²/R³完整座標族。先精確核線性獨立與生成，成立後才以基底大小給dimension。獨立族只給下界，生成族只給上界；一般基底基數不變依有限維定理。",
          "en": "This slice handles complete coordinate families in R² or R³. Check exact linear independence and spanning before using basis size as dimension. An independent family supplies only a lower bound and a spanning family only an upper bound; basis-cardinality invariance invokes the finite-dimensional theorem."
        },
        "source": {
          "mko_id": "mko-vector-space-dimension",
          "sha256": "0d547787130d8aa91b60c2231053891922efe324656000b1e67b8cd7f2658dd8",
          "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f"
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "det",
        "lhs": "1*(-1)-1*1",
        "rhs": "-2",
        "step": "S02"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "完整R²座標族與精確有理消去；不由例子證一般基底不變定理。",
      "en": "Complete R2 coordinate families and exact rational elimination; no general invariance theorem inferred from examples."
    },
    "md_symbols": [
      [
        "P",
        "基底column矩陣",
        "basis-column matrix"
      ],
      [
        "B",
        "有序候選基底",
        "ordered candidate basis"
      ]
    ],
    "md_error_controls": [
      {
        "title": "兩項資格 / Two qualifications",
        "zh": "獨立與生成均通過才稱基底。",
        "en": "A basis needs both independence and spanning."
      },
      {
        "title": "定理層 / Theorem layer",
        "zh": "基底大小不變是明示前提，不由例子推全稱。",
        "en": "Basis-cardinality invariance is an explicit prerequisite, not inferred universally from examples."
      }
    ],
    "md_open_items": []
  },
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}