{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-real-bilinear-form-01", "mko_id": "mko-real-bilinear-form", "title": { "zh-Hant": "實雙線性形式：以矩陣公式 x^T A y 分槽證明雙線性並精確求值", "en": "Real bilinear form: proving bilinearity slot by slot from x^T A y, with exact evaluation" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-real-bilinear-form-01", "source": { "mko_id": "mko-real-bilinear-form", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "fa2ccad4a26e9cf94b590f02594c9e1c0db0ba86fc50c60ee410e2b934d4c7d5", "english_sha256": "f3af9343056ca2070bf7df334b6c362b6abb13d0b8c5cb69981beec9201b8e72" }, "mode": "theorem_proof", "audience_prerequisites": [ "matrix_product_def", "distributive", "finite_sum_rules", "vector_space_axioms", "linear_map_def" ], "goal": { "zh-Hant": "設 A=[[1,2],[−1,3]]，B(x,y)=x^T A y（x、y∈ℝ²）。分別證明 B 對第一槽與第二槽都是線性的（雙線性）；對 x=(2,−1)、y=(1,4) 精確求 B(x,y)=7 與 B(y,x)=−20，說明雙線性不要求對稱；並證明黑箱函數 f(x,y)=x1y1+(x2³−x2)y2 雖通過若干加法樣本卻不是雙線性。", "en": "Let A=[[1,2],[−1,3]] and B(x,y)=x^T A y for x, y∈ℝ². Prove separately that B is linear in the first slot and in the second slot (bilinear); evaluate exactly B(x,y)=7 and B(y,x)=−20 for x=(2,−1), y=(1,4), showing that bilinearity does not require symmetry; and prove that the black-box function f(x,y)=x1y1+(x2³−x2)y2 passes some additivity samples but is not bilinear." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "A=[[1,2],[−1,3]]，即 a11=1、a12=2、a21=−1、a22=3。x、y 視為 ℝ² 的列向量（2×1），B(x,y)=x^T A y。", "en": "A=[[1,2],[−1,3]], that is a11=1, a12=2, a21=−1, a22=3. x and y are column vectors of ℝ² (2×1), and B(x,y)=x^T A y." } }, { "id": "G2", "statement": { "zh-Hant": "例子：x=(2,−1)、y=(1,4)、x′=(1,1)、α=3、β=−2。", "en": "Example: x=(2,−1), y=(1,4), x′=(1,1), α=3, β=−2." } }, { "id": "G3", "statement": { "zh-Hant": "對照黑箱：f(x,y)=x1y1+(x2³−x2)y2。", "en": "Black box for contrast: f(x,y)=x1y1+(x2³−x2)y2." } } ], "steps": [ { "id": "S01", "uses": [ "G1", "linear_map_def" ], "before_tex": "B\\ \\text{bilinear}", "after_tex": "\\forall y:\\ x\\mapsto B(x,y)\\ \\text{linear}\\ \\land\\ \\forall x:\\ y\\mapsto B(x,y)\\ \\text{linear}", "relation": "definition", "rule": "mko_bilinear_def", "conditions": [], "explanation": { "zh-Hant": "依 MKO，雙線性有兩項分開的責任：固定第二輸入時第一槽線性，固定第一輸入時第二槽線性。兩項都要對所有向量與所有實數係數成立。", "en": "By the MKO, bilinearity has two separate obligations: with the second input fixed the first slot is linear, and with the first input fixed the second slot is linear. Both must hold for all vectors and all real coefficients." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "mko_bilinear_def", "linear_map_def" ], "explanation": { "zh-Hant": "此步只展開定義，兩槽之後分開核（S04 到 S06、S07）。", "en": "This step only unfolds the definition; the two slots are checked separately later (S04 to S06, S07)." } }, "leaf_basis": "mko_bilinear_def + def_unfold" }, { "id": "S02", "uses": [ "G1", "matrix_product_def" ], "before_tex": "x^{T}Ay\\ \\ (1\\times2)(2\\times2)(2\\times1)", "after_tex": "B(x,y)=\\sum_{i=1}^{2}\\sum_{j=1}^{2}x_i\\,a_{ij}\\,y_j", "relation": "equivalent", "rule": "matrix_product_def", "conditions": [ "\\text{dimensions compatible}" ], "explanation": { "zh-Hant": "維度相容：x^T 是 1×2、A 是 2×2、y 是 2×1，乘積是 1×1，即一個實數。用兩次矩陣乘法定義：(Ay)_i=Σ_j a_ij y_j，再 x^T(Ay)=Σ_i x_i (Ay)_i，合起來是雙重和。", "en": "Dimensions are compatible: x^T is 1×2, A is 2×2, y is 2×1, so the product is 1×1, a real number. Apply the definition of the matrix product twice: (Ay)_i=Σ_j a_ij y_j, then x^T(Ay)=Σ_i x_i (Ay)_i, which together give a double sum." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "matrix_product_def", "distributive" ], "explanation": { "zh-Hant": "兩式是同一個數的兩種寫法，雙向成立。", "en": "Both are expressions of the same number, so the step is two-way." } }, "leaf_basis": "matrix_product_def + distributive + finite_sum_rules" }, { "id": "S03", "uses": [ "S02", "G1" ], "before_tex": "\\sum_{i,j}x_i\\,a_{ij}\\,y_j", "after_tex": "x_1y_1+2x_1y_2-x_2y_1+3x_2y_2", "relation": "equivalent", "rule": "finite_sum_rules", "conditions": [], "explanation": { "zh-Hant": "把四項 (i,j)=(1,1)、(1,2)、(2,1)、(2,2) 寫出並代入 a11=1、a12=2、a21=−1、a22=3。N1 以格點核對這個多線性恆等式。", "en": "Write out the four terms (i,j)=(1,1), (1,2), (2,1), (2,2) and substitute a11=1, a12=2, a21=−1, a22=3. N1 checks this multilinear identity on a grid." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "finite_sum_rules" ], "explanation": { "zh-Hant": "有限和的逐項展開是等值改寫，雙向成立。", "en": "Writing out a finite sum term by term rewrites equal values, so the step is two-way." } }, "leaf_basis": "finite_sum_rules + numeric_evaluation" }, { "id": "S04", "uses": [ "S02", "vector_space_axioms" ], "before_tex": "B(\\alpha x+\\beta x',y)", "after_tex": "\\sum_{i,j}(\\alpha x_i+\\beta x'_i)\\,a_{ij}\\,y_j", "relation": "equivalent", "rule": "vector_space_axioms", "conditions": [], "explanation": { "zh-Hant": "第一槽：固定 y，任取 x、x′∈ℝ² 與 α、β∈ℝ。向量 αx+βx′ 的第 i 個座標是 αx_i+βx′_i（座標運算），代入 S02 的雙重和。", "en": "First slot: fix y and take any x, x′∈ℝ² and α, β∈ℝ. The i-th coordinate of αx+βx′ is αx_i+βx′_i (coordinate operations); substitute into the double sum of S02." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "vector_space_axioms", "matrix_product_def" ], "explanation": { "zh-Hant": "代入同一個向量的座標，兩邊相等，雙向成立。", "en": "Substituting the coordinates of the same vector keeps the two sides equal, two-way." } }, "leaf_basis": "vector_space_axioms + eq_replace" }, { "id": "S05", "uses": [ "S04", "distributive" ], "before_tex": "\\sum_{i,j}(\\alpha x_i+\\beta x'_i)\\,a_{ij}\\,y_j", "after_tex": "\\sum_{i,j}\\left(\\alpha\\,x_i a_{ij} y_j+\\beta\\,x'_i a_{ij} y_j\\right)", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "每一項用分配律展開 (αx_i+βx′_i)a_ij y_j=αx_i a_ij y_j+βx′_i a_ij y_j。", "en": "Expand each term by distributivity: (αx_i+βx′_i)a_ij y_j=αx_i a_ij y_j+βx′_i a_ij y_j." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive" ], "explanation": { "zh-Hant": "展開與因式分解互逆，雙向成立。", "en": "Expanding and factoring are inverse, so the step is two-way." } }, "leaf_basis": "distributive + mul_assoc + mul_comm" }, { "id": "S06", "uses": [ "S05", "finite_sum_rules" ], "before_tex": "\\sum_{i,j}\\left(\\alpha\\,x_i a_{ij} y_j+\\beta\\,x'_i a_{ij} y_j\\right)", "after_tex": "\\alpha B(x,y)+\\beta B(x',y)", "relation": "equivalent", "rule": "finite_sum_rules", "conditions": [], "explanation": { "zh-Hant": "有限和可拆成兩個和（Σ(a+b)=Σa+Σb），常數 α、β 可提出（Σca=cΣa），剩下的兩個和依 S02 分別是 B(x,y) 與 B(x′,y)。所以第一槽線性。N2 以格點核對這個恆等式。", "en": "A finite sum splits into two sums (Σ(a+b)=Σa+Σb), and the constants α, β come out (Σca=cΣa); by S02 the remaining sums are B(x,y) and B(x′,y). So the first slot is linear. N2 checks this identity on a grid." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "finite_sum_rules" ], "explanation": { "zh-Hant": "拆和與提出常數都可倒推，雙向成立。", "en": "Splitting sums and pulling out constants can be reversed, so the step is two-way." } }, "leaf_basis": "finite_sum_rules + eq_replace" }, { "id": "S07", "uses": [ "S02" ], "before_tex": "B(x,\\alpha y+\\beta y')", "after_tex": "\\alpha B(x,y)+\\beta B(x,y')", "relation": "theorem_application", "rule": "L_slot2", "conditions": [], "explanation": { "zh-Hant": "第二槽要另外證明（引理 L_slot2，S07.1 到 S07.3）：A 不對稱，所以不能把第一槽的結論「交換 x、y」搬過來。", "en": "The second slot must be proved separately (Lemma L_slot2, S07.1 to S07.3): A is not symmetric, so the first-slot result cannot be carried over by 'swapping x and y'." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S02", "L_slot2" ], "explanation": { "zh-Hant": "只需 S02 的雙重和表示與引理。", "en": "It needs only the double-sum form of S02 and the lemma." } }, "leaf_basis": "L_slot2" }, { "id": "S07.1", "uses": [ "S02", "vector_space_axioms" ], "before_tex": "B(x,\\alpha y+\\beta y')", "after_tex": "\\sum_{i,j}x_i\\,a_{ij}\\,(\\alpha y_j+\\beta y'_j)", "relation": "equivalent", "rule": "vector_space_axioms", "conditions": [], "explanation": { "zh-Hant": "固定 x，αy+βy′ 的第 j 個座標是 αy_j+βy′_j，代入雙重和。", "en": "Fix x; the j-th coordinate of αy+βy′ is αy_j+βy′_j, substituted into the double sum." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "vector_space_axioms", "matrix_product_def" ], "explanation": { "zh-Hant": "代入同一向量的座標，雙向成立；此子步不使用 L_slot2。", "en": "Substituting the coordinates of the same vector is two-way; this sub-step does not use L_slot2." } }, "leaf_basis": "vector_space_axioms + eq_replace" }, { "id": "S07.2", "uses": [ "S07.1", "distributive" ], "before_tex": "\\sum_{i,j}x_i\\,a_{ij}\\,(\\alpha y_j+\\beta y'_j)", "after_tex": "\\sum_{i,j}\\left(\\alpha\\,x_i a_{ij} y_j+\\beta\\,x_i a_{ij} y'_j\\right)", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "每一項用分配律展開，常數 α、β 移到前面（乘法交換與結合）。", "en": "Expand each term by distributivity and move α, β to the front (commutativity and associativity of multiplication)." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "mul_comm" ], "explanation": { "zh-Hant": "展開與因式分解互逆；此子步不使用 L_slot2。", "en": "Expanding and factoring are inverse; this sub-step does not use L_slot2." } }, "leaf_basis": "distributive + mul_comm + mul_assoc" }, { "id": "S07.3", "uses": [ "S07.2", "finite_sum_rules" ], "before_tex": "\\sum_{i,j}\\left(\\alpha\\,x_i a_{ij} y_j+\\beta\\,x_i a_{ij} y'_j\\right)", "after_tex": "\\alpha B(x,y)+\\beta B(x,y')", "relation": "equivalent", "rule": "finite_sum_rules", "conditions": [], "explanation": { "zh-Hant": "拆成兩個和並提出常數，依 S02 得 αB(x,y)+βB(x,y′)。S07.1 到 S07.3 合起來證得引理 L_slot2。N3 以格點核對。", "en": "Split into two sums and pull out the constants; by S02 this is αB(x,y)+βB(x,y′). S07.1 to S07.3 together prove Lemma L_slot2. N3 checks it on a grid." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "finite_sum_rules" ], "explanation": { "zh-Hant": "拆和與提出常數可倒推；此子步不使用 L_slot2。", "en": "Splitting sums and pulling out constants can be reversed; this sub-step does not use L_slot2." } }, "leaf_basis": "finite_sum_rules + eq_replace" }, { "id": "S08", "uses": [ "S01", "S06", "S07" ], "before_tex": "\\text{slot 1 linear (S06)},\\ \\text{slot 2 linear (S07)}", "after_tex": "B\\ \\text{is a bilinear form on}\\ \\mathbb{R}^2", "relation": "implies", "rule": "forall_intro", "conditions": [], "explanation": { "zh-Hant": "S04 到 S07 中 x、x′、y、y′、α、β 都是任意取的，沒有額外假設，依全稱推廣兩槽對所有輸入都線性，所以 B 雙線性。論證只用到 a_ij 是固定的數，所以固定 m×n 實矩陣在相容的 ℝᵐ×ℝⁿ 上都給雙線性映射；若稱為同一 ℝⁿ 上的雙線性形式，矩陣取 n×n。本題是 n=2 的方形情形。", "en": "In S04 to S07, x, x′, y, y′, α, β were arbitrary with no extra assumption, so by generalization both slots are linear for all inputs, and B is bilinear. The argument used only that the a_ij are fixed numbers, so a fixed m-by-n real matrix gives a bilinear map on the compatible spaces ℝᵐ×ℝⁿ. A bilinear form on one ℝⁿ uses an n-by-n matrix; this example is the square n=2 case." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S01", "S06", "S07", "forall_intro" ], "explanation": { "zh-Hant": "目標 G：B 雙線性。足夠條件 H：兩槽各自線性；H⇒G 由 S01。H 由 S04 到 S07 正向證成，是全稱代數證書，不是樣本。", "en": "Goal G: B is bilinear. Sufficient condition H: each slot is linear; H⇒G by S01. H is proved forwards in S04 to S07 as a universal algebraic certificate, not a sample." } }, "leaf_basis": "forall_intro + and_intro_elim + mko_bilinear_def" }, { "id": "S09", "uses": [ "S02", "G1", "G2" ], "before_tex": "B(x,y)=x^{T}(Ay),\\ x=(2,-1),\\ y=(1,4)", "after_tex": "B(x,y)=7", "relation": "implies", "rule": "matrix_product_def", "conditions": [], "explanation": { "zh-Hant": "先算 Ay（S09.1），再與 x 作內積式的乘法（S09.2）。", "en": "First compute Ay (S09.1), then multiply by x^T (S09.2)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S02", "matrix_product_def" ], "explanation": { "zh-Hant": "精確整數計算，可用 S03 的展開式交叉核對（N13）。", "en": "Exact integer computation, cross-checked by the expansion of S03 (N13)." } }, "leaf_basis": "matrix_product_def + numeric_evaluation" }, { "id": "S09.1", "uses": [ "G1", "G2" ], "before_tex": "Ay=\\begin{pmatrix}1&2\\\\-1&3\\end{pmatrix}\\begin{pmatrix}1\\\\4\\end{pmatrix}", "after_tex": "\\begin{pmatrix}9\\\\11\\end{pmatrix}", "relation": "equivalent", "rule": "matrix_product_def", "conditions": [], "explanation": { "zh-Hant": "(Ay)_1=1·1+2·4=9，(Ay)_2=(−1)·1+3·4=11。", "en": "(Ay)_1=1·1+2·4=9 and (Ay)_2=(−1)·1+3·4=11." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "matrix_product_def", "numeric_evaluation" ], "explanation": { "zh-Hant": "兩邊是同一個向量。", "en": "Both sides are the same vector." } }, "leaf_basis": "matrix_product_def + numeric_evaluation" }, { "id": "S09.2", "uses": [ "S09.1", "G2" ], "before_tex": "x^{T}(Ay)=(2,-1)\\begin{pmatrix}9\\\\11\\end{pmatrix}", "after_tex": "2\\cdot9+(-1)\\cdot11=7", "relation": "equivalent", "rule": "matrix_product_def", "conditions": [], "explanation": { "zh-Hant": "2·9+(−1)·11=18−11=7。", "en": "2·9+(−1)·11=18−11=7." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "matrix_product_def", "numeric_evaluation" ], "explanation": { "zh-Hant": "兩邊是同一個數。", "en": "Both sides are the same number." } }, "leaf_basis": "matrix_product_def + numeric_evaluation" }, { "id": "S10", "uses": [ "S02", "G1", "G2" ], "before_tex": "B(y,x)=y^{T}(Ax),\\ Ax=(1\\cdot2+2\\cdot(-1),\\ -1\\cdot2+3\\cdot(-1))", "after_tex": "Ax=(0,-5),\\ B(y,x)=1\\cdot0+4\\cdot(-5)=-20", "relation": "implies", "rule": "matrix_product_def", "conditions": [], "explanation": { "zh-Hant": "交換兩個輸入：Ax=(2−2,−2−3)=(0,−5)，y^T(Ax)=1·0+4·(−5)=−20。", "en": "Swap the two inputs: Ax=(2−2,−2−3)=(0,−5) and y^T(Ax)=1·0+4·(−5)=−20." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "matrix_product_def", "numeric_evaluation" ], "explanation": { "zh-Hant": "精確整數計算（N7）。", "en": "Exact integer computation (N7)." } }, "leaf_basis": "matrix_product_def + numeric_evaluation" }, { "id": "S11", "uses": [ "S08", "S09", "S10" ], "before_tex": "B(x,y)=7,\\ B(y,x)=-20", "after_tex": "B\\ \\text{bilinear but not symmetric}", "relation": "implies", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "7≠−20，所以 B 不對稱；但 S08 已證 B 雙線性。對稱是額外條件，不是雙線性的一部分；內積還要求對稱與正定，所以 B 不是內積。", "en": "7≠−20, so B is not symmetric; yet S08 proves B bilinear. Symmetry is an extra condition, not part of bilinearity; an inner product additionally needs symmetry and positive definiteness, so B is not an inner product." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S08", "S09", "S10" ], "explanation": { "zh-Hant": "一個不等的值對就否證對稱（全稱命題）；雙線性不受影響。", "en": "One unequal pair of values refutes symmetry (a universal statement); bilinearity is unaffected." } }, "leaf_basis": "numeric_evaluation + quantifier_negation" }, { "id": "S12", "uses": [ "S06", "S09", "G2" ], "before_tex": "\\alpha x+\\beta x'=(4,-5)", "after_tex": "B((4,-5),y)=-19=3\\cdot7+(-2)\\cdot20", "relation": "implies", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "αx+βx′=(3·2−2·1, 3·(−1)−2·1)=(4,−5)。用 Ay=(9,11)：B((4,−5),y)=36−55=−19；B(x′,y)=9+11=20，αB(x,y)+βB(x′,y)=21−40=−19。這個實例與 S06 一致，但只是例子，不是證明。", "en": "αx+βx′=(3·2−2·1, 3·(−1)−2·1)=(4,−5). With Ay=(9,11): B((4,−5),y)=36−55=−19; B(x′,y)=9+11=20, so αB(x,y)+βB(x′,y)=21−40=−19. This instance agrees with S06 but is an example, not a proof." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S06", "numeric_evaluation" ], "explanation": { "zh-Hant": "一般結論來自 S06；此步只核一個實例（N9）。", "en": "The general conclusion comes from S06; this step checks one instance (N9)." } }, "leaf_basis": "numeric_evaluation + matrix_product_def" }, { "id": "S13", "uses": [ "G3", "linear_map_def" ], "before_tex": "x=x'=(0,1),\\ y=(0,1)", "after_tex": "f(x+x',y)=6\\neq0=f(x,y)+f(x',y)", "relation": "implies", "rule": "contradiction", "conditions": [], "explanation": { "zh-Hant": "f((0,2),(0,1))=0+(8−2)·1=6，而 f((0,1),(0,1))=0+(1−1)·1=0，兩者相加為 0。取 α=β=1，第一槽的線性失敗，所以 f 不是雙線性。", "en": "f((0,2),(0,1))=0+(8−2)·1=6, while f((0,1),(0,1))=0+(1−1)·1=0, and the sum of two such values is 0. With α=β=1 linearity in the first slot fails, so f is not bilinear." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G3", "contradiction", "linear_map_def" ], "explanation": { "zh-Hant": "線性是全稱命題，一個反例就否證。", "en": "Linearity is a universal statement, and one counterexample refutes it." } }, "leaf_basis": "contradiction + quantifier_negation + numeric_evaluation" }, { "id": "S14", "uses": [ "G3", "S13" ], "before_tex": "x_2,\\ x'_2,\\ x_2+x'_2\\in\\{-1,0,1\\}", "after_tex": "f(x+x',y)=f(x,y)+f(x',y)\\ \\text{on these samples}", "relation": "implies", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "當 t∈{−1,0,1} 時 t³−t=0，所以只要 x2、x′2 與 x2+x′2 都在 {−1,0,1}，f 在這些輸入上等於 x1y1，可加性樣本全部通過（例如 x=(1,1)、x′=(2,−1)）。但 S13 已證 f 不是雙線性：通過有限樣本不能證明黑箱函數雙線性，必須有像 S04 到 S07 那樣的全稱證書。", "en": "When t∈{−1,0,1}, t³−t=0, so whenever x2, x′2 and x2+x′2 all lie in {−1,0,1}, f equals x1y1 on those inputs and every additivity sample passes (for example x=(1,1), x′=(2,−1)). Yet S13 shows f is not bilinear: passing finitely many samples cannot prove a black-box function bilinear; a universal certificate like S04 to S07 is needed." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G3", "S13" ], "explanation": { "zh-Hant": "此步只說明樣本通過；結論「不是雙線性」來自 S13。", "en": "This step only records that the samples pass; the conclusion 'not bilinear' comes from S13." } }, "leaf_basis": "numeric_evaluation + power_laws" } ], "result": { "zh-Hant": "B(x,y)=x^T A y=x1y1+2x1y2−x2y1+3x2y2 對第一槽與第二槽分別線性（全稱代數證書），所以是雙線性形式；固定實矩陣在相容的兩個座標空間上皆給雙線性映射，同一空間上的形式使用方形矩陣。B(x,y)=7、B(y,x)=−20，B 雙線性但不對稱，也不是內積。黑箱 f(x,y)=x1y1+(x2³−x2)y2 在 x2∈{−1,0,1} 的樣本上看似可加，但 f((0,2),(0,1))=6≠0，不是雙線性。", "en": "B(x,y)=x^T A y=x1y1+2x1y2−x2y1+3x2y2 is linear in the first slot and in the second slot (a universal algebraic certificate), so it is a bilinear form; a fixed real matrix gives such a bilinear map on compatible coordinate spaces, using a square matrix for a form on one space. B(x,y)=7 and B(y,x)=−20, so B is bilinear but not symmetric, and not an inner product. The black box f(x,y)=x1y1+(x2³−x2)y2 looks additive on samples with x2∈{−1,0,1}, but f((0,2),(0,1))=6≠0, so it is not bilinear." }, "negative_controls": [ { "claim": { "zh-Hant": "g(x,y)=(x1y1)² 對稱，所以是雙線性形式", "en": "g(x,y)=(x1y1)² is symmetric, so it is a bilinear form" }, "expected": "reject", "reason": { "zh-Hant": "g(2e1,e1)=4，而 2g(e1,e1)=2。", "en": "g(2e1,e1)=4 while 2g(e1,e1)=2." } }, { "claim": { "zh-Hant": "因為 B(x,y)=7≠−20=B(y,x)，所以 x^T A y 不是雙線性形式", "en": "Since B(x,y)=7≠−20=B(y,x), x^T A y is not a bilinear form" }, "expected": "reject", "reason": { "zh-Hant": "雙線性不要求對稱；S04 到 S08 已證 B 雙線性。", "en": "Bilinearity does not require symmetry; S04 to S08 prove B bilinear." } }, { "claim": { "zh-Hant": "雙線性表示 B(x+x′,y+y′)=B(x,y)+B(x′,y′)", "en": "Bilinear means B(x+x′,y+y′)=B(x,y)+B(x′,y′)" }, "expected": "reject", "reason": { "zh-Hant": "x=x′=y=y′=e1 時左邊 4、右邊 2；漏了交叉項。", "en": "With x=x′=y=y′=e1 the left side is 4 and the right side 2; the cross terms are missing." } }, { "claim": { "zh-Hant": "f 通過了 x=(1,1)、x′=(2,−1) 等多組可加性測試，所以 f 雙線性", "en": "f passed several additivity tests such as x=(1,1), x′=(2,−1), so f is bilinear" }, "expected": "reject", "reason": { "zh-Hant": "f((0,2),(0,1))=6，而 f((0,1),(0,1))+f((0,1),(0,1))=0。", "en": "f((0,2),(0,1))=6 while f((0,1),(0,1))+f((0,1),(0,1))=0." } } ], "status": "candidate", "gaps": [], "verification": { "structure": "checked_local_script", "mathematics": "candidate_argument", "formalization": "not_formalized" }, "rule_glossary": [ { "id": "matrix_product_def", "kind": "prerequisite", "statement": { "zh-Hant": "矩陣乘法：若 A 為 m×n、B 為 n×p，則 (AB)_{ij}=Σ_{k=1}^{n} A_{ik}B_{kj}。矩陣乘法一般不可交換。", "en": "Matrix product: for A of size m×n and B of size n×p, (AB)_{ij}=Σ_{k=1}^{n} A_{ik}B_{kj}. Matrix multiplication is in general not commutative." } }, { "id": "distributive", "kind": "prerequisite", "statement": { "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。", "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu." } }, { "id": "finite_sum_rules", "kind": "prerequisite", "statement": { "zh-Hant": "有限和：Σ_{i=1}^{0} a_i=0、Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}；Σ(a_i+b_i)=Σa_i+Σb_i；Σ c·a_i=c·Σa_i。", "en": "Finite sums: Σ_{i=1}^{0} a_i=0 and Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}; Σ(a_i+b_i)=Σa_i+Σb_i; Σ c·a_i=c·Σa_i." } }, { "id": "vector_space_axioms", "kind": "prerequisite", "statement": { "zh-Hant": "向量空間公理（實數或其他域 F 上）：加法成交換群；純量乘法滿足 a(u+v)=au+av、(a+b)u=au+bu、(ab)u=a(bu)、1u=u。", "en": "Vector space axioms (over ℝ or another field F): vectors form a commutative group under addition; scalar multiplication satisfies a(u+v)=au+av, (a+b)u=au+bu, (ab)u=a(bu) and 1u=u." } }, { "id": "def_unfold", "kind": "prerequisite", "statement": { "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。", "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true." } }, { "id": "numeric_evaluation", "kind": "prerequisite", "statement": { "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。", "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable." } }, { "id": "eq_replace", "kind": "prerequisite", "statement": { "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。", "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true." } }, { "id": "mul_assoc", "kind": "prerequisite", "statement": { "zh-Hant": "乘法結合律：(uv)w=u(vw)。", "en": "Associativity of multiplication: (uv)w=u(vw)." } }, { "id": "mul_comm", "kind": "prerequisite", "statement": { "zh-Hant": "乘法交換律：uv=vu（在交換的數系中，例如 ℤ、ℚ、ℝ、ℂ）。", "en": "Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ)." } }, { "id": "forall_intro", "kind": "prerequisite", "statement": { "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。", "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain." } }, { "id": "and_intro_elim", "kind": "prerequisite", "statement": { "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。", "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q." } }, { "id": "quantifier_negation", "kind": "prerequisite", "statement": { "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。", "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)." } }, { "id": "contradiction", "kind": "prerequisite", "statement": { "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。", "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds." } }, { "id": "power_laws", "kind": "prerequisite", "statement": { "zh-Hant": "整數指數律（底數非零或指數為正整數）：a^m·a^n=a^{m+n}，(a^m)^n=a^{mn}，(ab)^n=a^n b^n，a^0=1。", "en": "Integer exponent laws (base nonzero, or exponents positive integers): a^m·a^n=a^{m+n}, (a^m)^n=a^{mn}, (ab)^n=a^n b^n, a^0=1." } }, { "id": "mko_bilinear_def", "kind": "cited", "source": { "mko_id": "mko-real-bilinear-form", "sha256": "fa2ccad4a26e9cf94b590f02594c9e1c0db0ba86fc50c60ee410e2b934d4c7d5", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "B:V乘V到R為雙線性，若固定任一輸入後，另一輸入上的映射皆為線性。本輪用整係數矩陣公式B(x,y)=x^T A y作全稱座標證書並精確求值。", "en": "B:V×V→R is bilinear when fixing either input leaves a linear map in the other. This slice uses the integer-coefficient matrix formula B(x,y)=x^T A y as a universal coordinate certificate and for exact evaluation." } }, { "id": "linear_map_def", "kind": "prerequisite", "statement": { "zh-Hant": "映射 L:ℝ²→ℝ 是線性的，若對所有 u、u′∈ℝ² 與 α、β∈ℝ，L(αu+βu′)=αL(u)+βL(u′)。", "en": "A map L:ℝ²→ℝ is linear if L(αu+βu′)=αL(u)+βL(u′) for all u, u′∈ℝ² and α, β∈ℝ." } }, { "id": "L_slot2", "kind": "lemma", "proved_in": [ "S07.1", "S07.2", "S07.3" ], "statement": { "zh-Hant": "對每個固定的 x∈ℝ²，y↦B(x,y)=Σ_iΣ_j x_i a_ij y_j 是線性的：B(x,αy+βy′)=αB(x,y)+βB(x,y′)。", "en": "For every fixed x∈ℝ², y↦B(x,y)=Σ_iΣ_j x_i a_ij y_j is linear: B(x,αy+βy′)=αB(x,y)+βB(x,y′)." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S03", "vars": [ "x1", "x2", "y1", "y2" ], "degree": 1, "lhs": "x1*(1*y1+2*y2)+x2*(-1*y1+3*y2)", "rhs": "x1*y1+2*x1*y2-x2*y1+3*x2*y2" }, { "id": "N2", "step": "S06", "vars": [ "a", "b", "u1", "u2", "v1", "v2", "y1", "y2" ], "degree": 1, "lhs": "(a*u1+b*v1)*y1+2*(a*u1+b*v1)*y2-(a*u2+b*v2)*y1+3*(a*u2+b*v2)*y2", "rhs": "a*(u1*y1+2*u1*y2-u2*y1+3*u2*y2)+b*(v1*y1+2*v1*y2-v2*y1+3*v2*y2)" }, { "id": "N3", "step": "S07.3", "vars": [ "a", "b", "x1", "x2", "u1", "u2", "v1", "v2" ], "degree": 1, "lhs": "x1*(a*u1+b*v1)+2*x1*(a*u2+b*v2)-x2*(a*u1+b*v1)+3*x2*(a*u2+b*v2)", "rhs": "a*(x1*u1+2*x1*u2-x2*u1+3*x2*u2)+b*(x1*v1+2*x1*v2-x2*v1+3*x2*v2)" }, { "id": "N4", "step": "S09.1", "lhs": "1*1+2*4", "rhs": "9" }, { "id": "N5", "step": "S09.1", "lhs": "-1*1+3*4", "rhs": "11" }, { "id": "N6", "step": "S09.2", "lhs": "2*9+(-1)*11", "rhs": "7" }, { "id": "N7", "step": "S10", "lhs": "1*(1*2+2*(-1))+4*(-1*2+3*(-1))", "rhs": "-20" }, { "id": "N8", "step": "S11", "lhs": "7", "relation": "!=", "rhs": "-20" }, { "id": "N9", "step": "S12", "lhs": "4*9+(-5)*11", "rhs": "3*7+(-2)*(1*9+1*11)" }, { "id": "N10", "step": "S13", "lhs": "0*0+(2**3-2)*1", "rhs": "6" }, { "id": "N11", "step": "S13", "lhs": "2*(0*0+(1**3-1)*1)", "rhs": "0" }, { "id": "N12", "step": "S14", "vars": [ "t" ], "range": [ -1, 1 ], "lhs": "t**3-t", "rhs": "0" }, { "id": "N13", "step": "S09", "lhs": "2*1+2*2*4-(-1)*1+3*(-1)*4", "rhs": "7" }, { "id": "N14", "vars": [ "x1", "x2" ], "degree": 2, "lhs": "x1*x1+2*x1*x2-x2*x1+3*x2*x2", "rhs": "(x1+x2/2)**2+(11/4)*x2**2" }, { "id": "N15", "lhs": "2*2*1", "relation": "!=", "rhs": "1+1" } ] }, "presentation": { "md_scope": { "zh-Hant": "只在 MKO 的範圍內：具體數值核對使用整係數矩陣、相容維度與安全整數座標（假設 a1）；一般雙線性證書對任意實數座標與純量成立，固定實矩陣也可；矩陣展開是雙線性的全稱代數證書，有限樣本不足以證任意黑箱函數雙線性（假設 a2）。對稱、正定、非退化與內積資格只作對照，不在本稿證明。不改動 canonical MKO 或英文 edition。", "en": "Stays inside the MKO: the numeric checks use an integer-coefficient matrix, compatible dimensions and safe integer coordinates (assumption a1); the general bilinearity certificate permits arbitrary real coordinates and scalars, and a fixed real matrix; matrix expansion is a universal algebraic certificate of bilinearity, and finite samples cannot prove that an arbitrary black-box function is bilinear (assumption a2). Symmetry, positive definiteness, nondegeneracy and inner products appear only as contrasts and are not proved here. The canonical MKO and its English edition are not changed." }, "md_symbols": [ [ "A=(a_{ij})", "2×2 整係數矩陣 [[1,2],[−1,3]]", "the 2×2 integer matrix [[1,2],[−1,3]]" ], [ "B(x,y)=x^{T}Ay", "由 A 給出的雙線性形式", "the bilinear form given by A" ], [ "x,\\ x',\\ y,\\ y'", "ℝ² 中的向量", "vectors in ℝ²" ], [ "\\alpha,\\ \\beta", "實數係數", "real coefficients" ], [ "f", "對照用的黑箱函數", "the black-box function used for contrast" ] ], "md_error_controls": [ { "title": "對稱不是雙線性 / Symmetry is not bilinearity", "zh": "MKO 指出的誤解是只核 B(x,y)=B(y,x) 就宣稱雙線性或內積。g(x,y)=(x1y1)² 對稱，但 g(2e1,e1)=4≠2=2g(e1,e1)，不是雙線性。反過來，本稿的 B 雙線性卻不對稱（7≠−20）。B(x,x)=x1²+x1x2+3x2²=(x1+x2/2)²+(11/4)x2² 對 x≠0 為正，但因不對稱仍不是內積。", "en": "The mistake named by the MKO is to declare bilinearity or an inner product after checking only B(x,y)=B(y,x). g(x,y)=(x1y1)² is symmetric, but g(2e1,e1)=4≠2=2g(e1,e1), so it is not bilinear. Conversely B here is bilinear but not symmetric (7≠−20). B(x,x)=x1²+x1x2+3x2²=(x1+x2/2)²+(11/4)x2² is positive for x≠0, yet B is not an inner product because it is not symmetric." }, { "title": "兩個槽分開核；雙線性不是對 (x,y) 線性 / Check each slot separately; bilinear is not linear in the pair", "zh": "A 不對稱，所以第二槽不能由第一槽交換得到（S07 另證）。雙線性也不是 B(x+x′,y+y′)=B(x,y)+B(x′,y′)：正確展開還有交叉項 B(x,y′)+B(x′,y)。例如 x=x′=y=y′=e1：B(2e1,2e1)=4，而 B(e1,e1)+B(e1,e1)=2。", "en": "A is not symmetric, so the second slot cannot be obtained from the first by swapping (S07 proves it separately). Bilinear also does not mean B(x+x′,y+y′)=B(x,y)+B(x′,y′): the correct expansion has the cross terms B(x,y′)+B(x′,y). For example x=x′=y=y′=e1: B(2e1,2e1)=4 while B(e1,e1)+B(e1,e1)=2." }, { "title": "黑箱樣本不足 / Black-box samples are not enough", "zh": "f 在所有 x2、x′2、x2+x′2∈{−1,0,1} 的樣本上都通過可加性（S14），卻在 (0,1)+(0,1) 失敗（S13）。MKO 假設 a2：矩陣展開是全稱證書，有限樣本不能證任意黑箱函數雙線性。", "en": "f passes additivity on every sample with x2, x′2, x2+x′2∈{−1,0,1} (S14), yet fails at (0,1)+(0,1) (S13). MKO assumption a2: matrix expansion is a universal certificate, and finite samples cannot prove an arbitrary black-box function bilinear." } ], "md_open_items": [ { "zh-Hant": "錯誤控制中 B(x,x) 的配方式由 N14 以多項式恆等式核對；本稿不處理正定性或內積的一般理論。", "en": "The completed square for B(x,x) in the error controls is checked as a polynomial identity by N14; the general theory of positive definiteness and inner products is not treated." }, { "zh-Hant": "N2、N3 是每個變數的次數分別至多 1（總次數可大於 1）的多線性恆等式格點核對，決定該恆等式；N9 只是一個實例。", "en": "N2 and N3 are grid checks of multilinear identities of degree at most 1 per variable (not necessarily total degree at most 1), which decide those identities; N9 is only one instance." } ] }, "revision": { "id": "production-20261019-adopted-96", "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "adopted_source_sha256": "e1078dee19a7ff2ca3fc623044adf902aaf5c6edb8681d13aaaa972aa31c4f49", "original_source_sha256": "1412fb7e13e4018d6967accecdadadcf60832aad360e17ead14a7174bc1399a7", "owner_adopted": true, "bilingual_sync": "complete_source_bound", "formal_Lean": "NOT_RUN", "independent_AI_concurrence": "NOT_MEASURED", "issue_ids": [ "BR-M073", "BR-M074" ], "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion." } }