{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-linear-map-kernel-01",
  "mko_id": "mko-linear-map-kernel",
  "title": {
    "zh-Hant": "完整求解二階矩陣的核",
    "en": "The full kernel of a two-by-two map"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-linear-map-kernel-01",
    "source": {
      "mko_id": "mko-linear-map-kernel",
      "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f",
      "sha256": "01a060bc8fa8fe1e302b8dcb559e8b1a9d5f6607cde6d5973ab75c1e4d0d3abc",
      "english_sha256": "52a1b6b1a8e08eee92c49d6980e6b664033cf0b4e9173c76496b428aa1db8a5c"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "calc"
    ],
    "goal": {
      "zh-Hant": "T(x,y)=(x+2y,2x+4y)的核為span{(−2,1)}。",
      "en": "The kernel of T(x,y)=(x+2y,2x+4y) is span{(-2,1)}."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "A=[[1,2],[2,4]]，T:R²→R²，T(v)=Av，kernel目標為(0,0)。",
          "en": "A=[[1,2],[2,4]], T:R2 to R2, T(v)=Av, with kernel target (0,0)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1"
        ],
        "before_tex": "T\\binom{x}{y}=A\\binom{x}{y}",
        "after_tex": "T(x,y)=(x+2y,2x+4y)",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐row乘法寫出完整輸出，固定來源與目標均R²。",
          "en": "Row multiplication gives the full output, with source and target both R2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1"
          ],
          "explanation": {
            "zh-Hant": "不得把columns當成只測兩個輸入的證書。",
            "en": "The columns do not reduce the proof to testing two inputs."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "S01"
        ],
        "before_tex": "v\\in\\ker T",
        "after_tex": "x+2y=0,\\quad2x+4y=0",
        "relation": "equivalent",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "kernel定義要求輸出零向量，兩座標方程都必須為零。",
          "en": "Kernel membership requires zero output, so both coordinate equations must vanish."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "G1",
            "S01"
          ],
          "explanation": {
            "zh-Hant": "右側改成非零b是另一fiber，不是kernel。",
            "en": "A nonzero right side b defines another fiber, not the kernel."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S03",
        "uses": [
          "G1",
          "S02"
        ],
        "before_tex": "\\left[\\begin{array}{cc|c}1&2&0\\\\2&4&0\\end{array}\\right]",
        "after_tex": "R_2\\leftarrow R_2-2R_1:\\quad\\left[\\begin{array}{cc|c}1&2&0\\\\0&0&0\\end{array}\\right]",
        "relation": "equivalent",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "可逆row操作顯示第二方程完全冗餘，保留全部實解。",
          "en": "An invertible operation shows the second equation is redundant while preserving all real solutions."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "G1",
            "S02"
          ],
          "explanation": {
            "zh-Hant": "零row表示0=0，不表示y=0。",
            "en": "The zero row says 0=0, not y=0."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S04",
        "uses": [
          "G1",
          "S03"
        ],
        "before_tex": "x+2y=0",
        "after_tex": "y=t,\\quad x=-2t,\\quad t\\in\\mathbb R",
        "relation": "equivalent",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "y為任意實自由變數，解出x而不限制t。",
          "en": "y is an arbitrary real free variable; solving for x imposes no restriction on t."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "G1",
            "S03"
          ],
          "explanation": {
            "zh-Hant": "只取整數t會漏實向量，例如t=1/2。",
            "en": "Integer-only t would miss real vectors such as t=1/2."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S05",
        "uses": [
          "G1",
          "S01",
          "S04"
        ],
        "before_tex": "v=(-2t,t)",
        "after_tex": "v=t(-2,1),\\quad A(-2,1)^T=(0,0)^T",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "參數化給全部解的span，直接乘生成向量驗其在核中。",
          "en": "The parameterization spans all solutions, and direct multiplication verifies the generator lies in the kernel."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01",
            "S04"
          ],
          "explanation": {
            "zh-Hant": "需兼有所有解的表示與生成向量反代，不能只找到一個核元素。",
            "en": "Both full solution coverage and generator substitution are needed, not merely one kernel element."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S06",
        "uses": [
          "G1",
          "S04",
          "S05"
        ],
        "before_tex": "\\ker T=\\{t(-2,1):t\\in\\mathbb R\\}",
        "after_tex": "\\ker T=\\operatorname{span}\\{(-2,1)\\},\\quad\\dim\\ker T=1",
        "relation": "implies",
        "rule": "mko_definition",
        "conditions": [],
        "explanation": {
          "zh-Hant": "非零生成向量獨立，因此是一項完整kernel basis。",
          "en": "The nonzero generator is independent, so it forms a complete one-vector kernel basis."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S04",
            "S05"
          ],
          "explanation": {
            "zh-Hant": "無限多個t不表示nullity無限；維度是基底大小。",
            "en": "Infinitely many t values do not mean infinite nullity; dimension counts basis vectors."
          }
        },
        "leaf_basis": "mko_definition"
      },
      {
        "id": "S07",
        "uses": [
          "G1",
          "S04"
        ],
        "before_tex": "T(-1,1/2)",
        "after_tex": "T(-1,1/2)=(0,0),\\quad(-1,1/2)\\notin\\mathbb Z^2",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "半整數例子明示有限整數格搜尋不能覆蓋全實核。",
          "en": "A half-integer example shows that a finite integer grid cannot cover the real kernel."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S04"
          ],
          "explanation": {
            "zh-Hant": "此例僅說明符號證書，不以單例取代全稱參數化。",
            "en": "This illustrates rather than replaces the universal symbolic parameterization."
          }
        },
        "leaf_basis": "calc"
      },
      {
        "id": "S08",
        "uses": [
          "G1",
          "S01",
          "S04"
        ],
        "before_tex": "T(x,y)=(1,2)",
        "after_tex": "(x,y)=(1,0)+t(-2,1),\\quad t\\in\\mathbb R",
        "relation": "implies",
        "rule": "calc",
        "conditions": [],
        "explanation": {
          "zh-Hant": "非零fiber是核的仿射平移，不含零，不能稱子空間核。",
          "en": "The nonzero fiber is an affine translate of the kernel and omits zero, so it is not the kernel subspace."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "S01",
            "S04"
          ],
          "explanation": {
            "zh-Hant": "反代得固定輸出(1,2)而非0，區分兩種集合。",
            "en": "Substitution gives fixed output (1,2), not zero, distinguishing the sets."
          }
        },
        "leaf_basis": "calc"
      }
    ],
    "result": {
      "zh-Hant": "T(x,y)=(x+2y,2x+4y)的核為span{(−2,1)}。",
      "en": "The kernel of T(x,y)=(x+2y,2x+4y) is span{(-2,1)}."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "零row迫使y=0。",
          "en": "The zero row forces y=0."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "零row只是0=0，y為自由實變數。",
          "en": "A zero row is only 0=0; y remains free."
        }
      },
      {
        "claim": {
          "zh-Hant": "所有核元素只有整數倍(−2,1)。",
          "en": "All kernel elements are integer multiples of (-2,1)."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "t=1/2給合法實核元素。",
          "en": "t=1/2 gives a valid real kernel element."
        }
      },
      {
        "claim": {
          "zh-Hant": "T^{-1}((1,2))也是kernel。",
          "en": "The fiber T^-1((1,2)) is also the kernel."
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "它不含零且輸出非零。",
          "en": "It omits zero and has a nonzero output."
        }
      }
    ],
    "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。",
          "en": "Assume the real field, usual coordinate-vector arithmetic, matrix multiplication, and definitions of independence and span; elimination uses invertible elementary row operations."
        }
      },
      {
        "id": "mko_definition",
        "kind": "cited",
        "statement": {
          "zh-Hant": "本輪以精確2×2整數矩陣綁定T:R²→R²，使用符號null equations或精確row reduction求kernel。bounded coordinate search只回答該盒內，不可取代全實數kernel。非零fiber是affine solution set，不是kernel。",
          "en": "This slice binds an exact 2-by-2 integer matrix to T:R²→R². Symbolic null equations or exact row reduction describe the full kernel. A bounded coordinate search covers only its box and cannot replace the whole real kernel. A nonzero target fiber is an affine solution set, not the kernel."
        },
        "source": {
          "mko_id": "mko-linear-map-kernel",
          "sha256": "01a060bc8fa8fe1e302b8dcb559e8b1a9d5f6607cde6d5973ab75c1e4d0d3abc",
          "commit": "b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f"
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "kernel_first",
        "lhs": "-2+2*1",
        "rhs": "0",
        "step": "S05"
      },
      {
        "id": "kernel_second",
        "lhs": "2*(-2)+4",
        "rhs": "0",
        "step": "S05"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "精確2×2實線性映射及全稱null equation；有限盒搜尋不替代核。",
      "en": "An exact 2-by-2 real linear map and universal null equations; no bounded box replaces the kernel."
    },
    "md_symbols": [
      [
        "A",
        "完整2×2矩陣",
        "complete 2-by-2 matrix"
      ],
      [
        "t",
        "任意實自由變數",
        "arbitrary real free variable"
      ]
    ],
    "md_error_controls": [
      {
        "title": "零目標 / Zero target",
        "zh": "kernel固定Av=0；Av=b另列fiber。",
        "en": "The kernel fixes Av=0; Av=b is a separate fiber."
      },
      {
        "title": "自由變數域 / Free-variable domain",
        "zh": "t∈R而非有限盒或整數列。",
        "en": "t ranges over R, not a bounded box or integers only."
      }
    ],
    "md_open_items": []
  },
  "revision": {
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    "collection": "coverage-20261021",
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    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "reader_review": "142_of_142_source_bound",
    "wording_review": "original_Lead_decisions_resolved",
    "glossary_reader_paraphrases": [],
    "independent_AI_concurrence": "SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEW",
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  }
}