{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-linear-transformation-01", "mko_id": "mko-linear-transformation", "title": { "zh-Hant": "線性映射：證明 v↦Mv 線性，且仿射式 Mv+b 線性當且僅當 b=0", "en": "Linear maps: proving v↦Mv is linear, and Mv+b is linear exactly when b=0" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-linear-transformation-01", "source": { "mko_id": "mko-linear-transformation", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "59ce4d8fac03871ea16eb62a76c9fae255470dc391ff7312ad4da120bafbf132", "english_sha256": "d95d3ac4155053ddc0d186089ae6ec2d7e4e20468aa14880d371ac260c3cd0a3" }, "mode": "theorem_proof", "audience_prerequisites": [ "distributive", "mul_assoc", "mul_comm", "add_assoc", "add_comm", "numeric_evaluation" ], "goal": { "zh-Hant": "對任意 2×2 實矩陣 M，證明 T(v)=Mv 是 ℝ²→ℝ² 的線性映射（對所有向量 u、v 與實數 c 滿足加法相容與純量相容）；證明仿射式 F(v)=Mv+b 線性當且僅當 b=0。以 M₀=[[2,−1],[3,4]]、b=(1,−2)ᵀ、u=(1,1)ᵀ、w=(2,−3)ᵀ 為例：算出 M₀u，說明 F 不線性並給出具體失效等式；以數字示範 T 的加法相容；最後以 g(x,y)=(x²,y) 說明有限測點不能證明任意映射線性。", "en": "For every 2×2 real matrix M, prove that T(v)=Mv is a linear map ℝ²→ℝ² (additivity and homogeneity for all vectors u, v and real c); prove that the affine map F(v)=Mv+b is linear exactly when b=0. For the example M₀=[[2,−1],[3,4]], b=(1,−2)ᵀ, u=(1,1)ᵀ, w=(2,−3)ᵀ: compute M₀u, show that F is not linear with an explicit failing equation, and illustrate additivity of T numerically; finally use g(x,y)=(x²,y) to show that finitely many test points cannot prove that an arbitrary map is linear." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "定義域與陪域都是全實平面 ℝ²，向量為列向量（直行），M=[[p,q],[r,s]] 為任意 2×2 實矩陣，c 為任意實數（MKO 假設 a1）。", "en": "Domain and codomain are the whole real plane ℝ², vectors are column vectors, M=[[p,q],[r,s]] is an arbitrary 2×2 real matrix and c is an arbitrary real (MKO assumption a1)." } }, { "id": "G2", "statement": { "zh-Hant": "例子：M₀=[[2,−1],[3,4]]，b=(1,−2)ᵀ，u=(1,1)ᵀ，w=(2,−3)ᵀ（整數，精確計算；MKO 假設 a2）。", "en": "Example: M₀=[[2,−1],[3,4]], b=(1,−2)ᵀ, u=(1,1)ᵀ, w=(2,−3)ᵀ (integers, computed exactly; MKO assumption a2)." } }, { "id": "G3", "statement": { "zh-Hant": "對照用映射 g:ℝ²→ℝ²，g(x,y)=(x²,y)，不是矩陣形式。", "en": "A contrast map g:ℝ²→ℝ², g(x,y)=(x²,y), which is not of matrix form." } } ], "steps": [ { "id": "S01", "uses": [ "G1" ], "before_tex": "T:\\mathbb{R}^2\\to\\mathbb{R}^2", "after_tex": "T\\ \\text{linear}\\iff\\forall u,v,c:\\ T(u+v)=T(u)+T(v)\\ \\land\\ T(cv)=cT(v)", "relation": "definition", "rule": "mko_linear_def", "conditions": [], "explanation": { "zh-Hant": "寫出 MKO 的定義。兩個條件都要對所有向量與所有實數成立；這是全稱命題，證明時要對任意的 u、v、c 論證，不能只檢查幾個點。", "en": "State the MKO's definition. Both conditions must hold for all vectors and all reals; this is a universal statement, so a proof must argue for arbitrary u, v, c rather than check a few points." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "mko_linear_def" ], "explanation": { "zh-Hant": "此步只寫出定義，不推出新事實。", "en": "This step only states the definition; it derives no new fact." } }, "leaf_basis": "mko_linear_def + def_unfold" }, { "id": "S02", "uses": [ "G1", "matrix_product_def" ], "before_tex": "Mv,\\ M=\\begin{pmatrix}p&q\\\\r&s\\end{pmatrix},\\ v=(x,y)^{\\mathsf T}", "after_tex": "Mv=(px+qy,\\ rx+sy)^{\\mathsf T}", "relation": "equivalent", "rule": "matrix_product_def", "conditions": [], "explanation": { "zh-Hant": "依矩陣乘法定義（2×2 乘 2×1）：Mv 的第 i 個分量是 M 第 i 行與 v 的對應乘積和，Σ_{k=1}^{2} M(i,k)v_k。所以第一分量 px+qy，第二分量 rx+sy。結果是一個向量，不是逐項相乘後的矩陣（MKO 第一段）。", "en": "By the definition of the matrix product (2×2 times 2×1), component i of Mv is the sum of products of row i of M with v, Σ_{k=1}^{2} M(i,k)v_k. So the first component is px+qy and the second is rx+sy. The result is a vector, not a matrix of entrywise products (MKO first paragraph)." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "matrix_product_def", "vec_ops" ], "explanation": { "zh-Hant": "這是定義展開，兩邊是同一個向量，雙向成立。", "en": "This unfolds a definition; both sides are the same vector, so it is two-way." } }, "leaf_basis": "matrix_product_def + def_unfold" }, { "id": "S03", "uses": [ "S02", "vec_ops" ], "before_tex": "M(u+v),\\ u=(u_1,u_2)^{\\mathsf T},\\ v=(v_1,v_2)^{\\mathsf T}", "after_tex": "M(u+v)=Mu+Mv\\quad\\text{for all } u,v\\in\\mathbb{R}^2", "relation": "implies", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "取任意 u、v。u+v=(u₁+v₁, u₂+v₂)ᵀ（vec_ops）。依 S02，M(u+v) 的兩個分量是 p(u₁+v₁)+q(u₂+v₂) 與 r(u₁+v₁)+s(u₂+v₂)。S03.1 把第一分量化為 Mu 與 Mv 第一分量之和，S03.2 處理第二分量並收尾。u、v 是任意的，所以對所有向量成立。", "en": "Take arbitrary u and v. u+v=(u₁+v₁, u₂+v₂)ᵀ (vec_ops). By S02 the two components of M(u+v) are p(u₁+v₁)+q(u₂+v₂) and r(u₁+v₁)+s(u₂+v₂). S03.1 turns the first component into the sum of the first components of Mu and Mv, and S03.2 handles the second component and concludes. Since u and v are arbitrary, the identity holds for all vectors." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S02", "vec_ops", "distributive", "forall_intro" ], "explanation": { "zh-Hant": "只用實數的分配律、結合律、交換律，對任意 u、v 成立；不是由有限測點推出。", "en": "It uses only distributivity, associativity and commutativity of reals and holds for arbitrary u and v; it is not inferred from finitely many test points." } }, "leaf_basis": "distributive + add_assoc + add_comm + vec_ops + forall_intro" }, { "id": "S03.1", "uses": [ "distributive", "add_assoc", "add_comm" ], "before_tex": "p(u_1+v_1)+q(u_2+v_2)", "after_tex": "(pu_1+qu_2)+(pv_1+qv_2)", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "分配律：p(u₁+v₁)=pu₁+pv₁，q(u₂+v₂)=qu₂+qv₂。四項相加，用加法結合律與交換律重排成 (pu₁+qu₂)+(pv₁+qv₂)，即 Mu 的第一分量加 Mv 的第一分量。", "en": "Distributivity: p(u₁+v₁)=pu₁+pv₁ and q(u₂+v₂)=qu₂+qv₂. Add the four terms and rearrange with associativity and commutativity of addition into (pu₁+qu₂)+(pv₁+qv₂), the first component of Mu plus the first component of Mv." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "add_assoc", "add_comm" ], "explanation": { "zh-Hant": "這是對所有實數成立的多項式恆等式（N14），雙向成立。", "en": "This is a polynomial identity valid for all reals (N14), so it is two-way." } }, "leaf_basis": "distributive + add_assoc + add_comm" }, { "id": "S03.2", "uses": [ "S03.1", "vec_ops" ], "before_tex": "r(u_1+v_1)+s(u_2+v_2)", "after_tex": "(ru_1+su_2)+(rv_1+sv_2)\\ \\Rightarrow\\ M(u+v)=Mu+Mv", "relation": "implies", "rule": "vec_ops", "conditions": [], "explanation": { "zh-Hant": "第二分量用與 S03.1 完全相同的恆等式（把 p、q 換成 r、s）。兩個分量都相等，依 vec_ops 兩向量相等：M(u+v)=Mu+Mv。", "en": "The second component uses exactly the identity of S03.1 with p, q replaced by r, s. Both components agree, so by vec_ops the vectors are equal: M(u+v)=Mu+Mv." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03.1", "vec_ops" ], "explanation": { "zh-Hant": "向量相等需要兩個分量都相等；只核對一個分量不夠。", "en": "Equality of vectors needs both components to agree; checking one component is not enough." } }, "leaf_basis": "vec_ops + distributive + add_assoc + add_comm" }, { "id": "S04", "uses": [ "S02", "vec_ops" ], "before_tex": "M(cv),\\ cv=(cx,cy)^{\\mathsf T}", "after_tex": "(p(cx)+q(cy),\\ r(cx)+s(cy))^{\\mathsf T}=c\\,(px+qy,\\ rx+sy)^{\\mathsf T}=c\\,Mv", "relation": "implies", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "取任意 c 與 v。cv=(cx,cy)ᵀ。第一分量 p(cx)+q(cy)=(pc)x+(qc)y=(cp)x+(cq)y=c(px)+c(qy)=c(px+qy)，依序用乘法結合律、交換律、結合律與分配律；第二分量同理。依 vec_ops，M(cv)=cMv。c 與 v 任意，所以對全部成立。", "en": "Take arbitrary c and v. cv=(cx,cy)ᵀ. The first component is p(cx)+q(cy)=(pc)x+(qc)y=(cp)x+(cq)y=c(px)+c(qy)=c(px+qy), using associativity, commutativity and associativity of multiplication and then distributivity; the second component is the same. By vec_ops, M(cv)=cMv. Since c and v are arbitrary, it holds for all of them." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S02", "mul_assoc", "mul_comm", "distributive", "vec_ops" ], "explanation": { "zh-Hant": "用的是對所有實數成立的恆等式（N15）；不要求 c≠0，c=0 也成立。", "en": "It uses identities valid for all reals (N15); c≠0 is not required, and c=0 works too." } }, "leaf_basis": "mul_assoc + mul_comm + distributive + vec_ops + forall_intro" }, { "id": "S05", "uses": [ "S01", "S03", "S04" ], "before_tex": "M(u+v)=Mu+Mv,\\ M(cv)=cMv\\ (\\forall u,v,c)", "after_tex": "T(v)=Mv\\ \\text{is linear for every } M\\in\\mathbb{R}^{2\\times 2}", "relation": "implies", "rule": "mko_linear_def", "conditions": [], "explanation": { "zh-Hant": "S03、S04 對任意 u、v、c 證得兩個條件，依 S01 的定義，T 是線性映射。M 也是任意的，所以結論對每個 2×2 實矩陣成立，包括零矩陣與行列式為零的矩陣：線性不要求可逆（MKO 假設 a3）。", "en": "S03 and S04 prove both conditions for arbitrary u, v, c, so by the definition in S01, T is linear. M is arbitrary as well, so the conclusion holds for every 2×2 real matrix, including the zero matrix and matrices with determinant zero: linearity does not require invertibility (MKO assumption a3)." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S03", "S04", "mko_linear_def", "forall_intro" ], "explanation": { "zh-Hant": "目標：T 線性。足夠條件：兩個全稱等式；它們由 S03、S04 正向證成，沒有把目標當作前提。", "en": "Goal: T is linear. Sufficient condition: the two universal identities; they are proved forwards in S03 and S04, and the goal is never used as a premise." } }, "leaf_basis": "mko_linear_def + forall_intro + and_intro_elim" }, { "id": "S06", "uses": [ "L_lin_zero" ], "before_tex": "F\\ \\text{linear}", "after_tex": "F(0)=0", "relation": "theorem_application", "rule": "L_lin_zero", "conditions": [], "explanation": { "zh-Hant": "任何線性映射都把零向量送到零向量（引理 L_lin_zero，在 S06.1 證明）。下面把它用在仿射式 F 上。", "en": "Every linear map sends the zero vector to the zero vector (Lemma L_lin_zero, proved in S06.1). Below it is applied to the affine map F." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_lin_zero" ], "explanation": { "zh-Hant": "只宣稱「線性 ⇒ F(0)=0」；反向不成立，例如 g(0,0)=(0,0) 但 g 不線性（S14）。", "en": "Only 'linear ⇒ F(0)=0' is claimed; the converse fails, for example g(0,0)=(0,0) but g is not linear (S14)." } }, "leaf_basis": "L_lin_zero" }, { "id": "S06.1", "uses": [ "vec_ops", "mul_zero" ], "before_tex": "T(cv)=cT(v)\\ \\forall c,v", "after_tex": "T(0)=T(0\\cdot 0)=0\\cdot T(0)=0", "relation": "implies", "rule": "vec_ops", "conditions": [], "explanation": { "zh-Hant": "零向量等於 0 乘零向量：0·(0,0)ᵀ=(0·0, 0·0)ᵀ=(0,0)ᵀ（vec_ops、mul_zero）。取 c=0、v=0 用純量相容：T(0)=T(0·0)=0·T(0)。任何向量乘以 0 是零向量（逐分量 mul_zero），所以 T(0)=0。這證得引理 L_lin_zero。", "en": "The zero vector is 0 times the zero vector: 0·(0,0)ᵀ=(0·0, 0·0)ᵀ=(0,0)ᵀ (vec_ops, mul_zero). Using homogeneity with c=0 and v=0: T(0)=T(0·0)=0·T(0). Any vector times 0 is the zero vector (mul_zero in each component), so T(0)=0. This proves Lemma L_lin_zero." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "vec_ops", "mul_zero" ], "explanation": { "zh-Hant": "只需要 c=0 這一個純量相容的實例。", "en": "Only the single instance c=0 of homogeneity is needed." } }, "leaf_basis": "vec_ops + mul_zero + forall_elim" }, { "id": "S07", "uses": [ "S05", "add_zero" ], "before_tex": "b=0", "after_tex": "F(v)=Mv+0=Mv=T(v)\\ \\Rightarrow\\ F\\ \\text{linear}", "relation": "implies", "rule": "eq_replace", "conditions": [ "b=0" ], "explanation": { "zh-Hant": "若 b 是零向量，F(v)=Mv+0=Mv（加法單位元，逐分量），所以 F 與 T 在每一點相等，是同一個映射；依 S05，F 線性。", "en": "If b is the zero vector, F(v)=Mv+0=Mv (additive identity, componentwise), so F agrees with T at every point and is the same map; by S05, F is linear." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S05", "add_zero", "function_def" ], "explanation": { "zh-Hant": "這是「b=0 ⇒ F 線性」一個方向；另一方向在 S08。", "en": "This is one direction, 'b=0 ⇒ F linear'; the other direction is S08." } }, "leaf_basis": "eq_replace + add_zero + function_def" }, { "id": "S08", "uses": [ "S02", "S06" ], "before_tex": "F\\ \\text{linear}", "after_tex": "b=F(0)=0", "relation": "implies", "rule": "L_lin_zero", "conditions": [], "explanation": { "zh-Hant": "設 F 線性。依 S06，F(0)=0。另一方面，依 S02，M·0=(p·0+q·0, r·0+s·0)ᵀ=(0,0)ᵀ，所以 F(0)=M·0+b=b。兩者合起來 b=0。", "en": "Suppose F is linear. By S06, F(0)=0. On the other hand, by S02, M·0=(p·0+q·0, r·0+s·0)ᵀ=(0,0)ᵀ, so F(0)=M·0+b=b. Together these give b=0." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S06", "S02", "mul_zero", "eq_refl_symm_trans" ], "explanation": { "zh-Hant": "只用到線性的一個後果 F(0)=0；這是「F 線性 ⇒ b=0」方向。", "en": "It uses only one consequence of linearity, F(0)=0; this is the direction 'F linear ⇒ b=0'." } }, "leaf_basis": "L_lin_zero + mul_zero + add_zero + eq_refl_symm_trans" }, { "id": "S09", "uses": [ "S07", "S08" ], "before_tex": "b=0\\Rightarrow F\\ \\text{linear},\\ F\\ \\text{linear}\\Rightarrow b=0", "after_tex": "F\\ \\text{linear}\\iff b=0", "relation": "implies", "rule": "iff_intro", "conditions": [], "explanation": { "zh-Hant": "兩個方向都已證明，依 iff_intro 得到雙條件。這正是 MKO 對仿射式的判準；它對每個 M 都成立，不需要改動 M 的元素。", "en": "Both directions are proved, so iff_intro gives the biconditional. This is exactly the MKO's criterion for affine maps; it holds for every M, without changing any entry of M." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S07", "S08", "iff_intro" ], "explanation": { "zh-Hant": "雙條件需要兩個方向各有證明：S07 與 S08。", "en": "The biconditional needs a proof of each direction: S07 and S08." } }, "leaf_basis": "iff_intro" }, { "id": "S10", "uses": [ "S02", "G2" ], "before_tex": "M_0u,\\ M_0=\\begin{pmatrix}2&-1\\\\3&4\\end{pmatrix},\\ u=(1,1)^{\\mathsf T}", "after_tex": "M_0u=(2\\cdot 1+(-1)\\cdot 1,\\ 3\\cdot 1+4\\cdot 1)^{\\mathsf T}=(1,7)^{\\mathsf T}", "relation": "equivalent", "rule": "matrix_product_def", "conditions": [], "explanation": { "zh-Hant": "依 S02（p=2、q=−1、r=3、s=4、x=y=1）：第一分量 2−1=1，第二分量 3+4=7。每個分量是一行與 u 的乘積和。", "en": "By S02 (p=2, q=−1, r=3, s=4, x=y=1): the first component is 2−1=1 and the second is 3+4=7. Each component is a sum of products of one row with u." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "matrix_product_def", "numeric_evaluation" ], "explanation": { "zh-Hant": "精確計算（N1、N2），雙向成立。", "en": "Exact computation (N1, N2), two-way." } }, "leaf_basis": "matrix_product_def + numeric_evaluation" }, { "id": "S11", "uses": [ "S09", "G2" ], "before_tex": "F(v)=M_0v+(1,-2)^{\\mathsf T}", "after_tex": "F(0)=(1,-2)^{\\mathsf T}\\neq 0\\ \\Rightarrow\\ F\\ \\text{not linear}", "relation": "implies", "rule": "contrapositive", "conditions": [], "explanation": { "zh-Hant": "b=(1,−2)ᵀ 不是零向量（第一分量 1≠0），依 S09 的逆否形式，F 不是線性映射。等價地，F(0)=b≠0 違反 S06。", "en": "b=(1,−2)ᵀ is not the zero vector (its first component is 1≠0), so by the contrapositive of S09, F is not linear. Equivalently, F(0)=b≠0 violates S06." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S09", "contrapositive", "vec_ops" ], "explanation": { "zh-Hant": "只需要 b 有一個非零分量。", "en": "It needs only one nonzero component of b." } }, "leaf_basis": "contrapositive + vec_ops" }, { "id": "S12", "uses": [ "S10", "G2", "S02" ], "before_tex": "F(2u)\\ \\text{vs}\\ 2F(u)", "after_tex": "F(2u)=(2,14)^{\\mathsf T}+(1,-2)^{\\mathsf T}=(3,12)^{\\mathsf T}\\neq(4,10)^{\\mathsf T}=2F(u)", "relation": "implies", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "具體失效：F(u)=M₀u+b=(1,7)ᵀ+(1,−2)ᵀ=(2,5)ᵀ，所以 2F(u)=(4,10)ᵀ。另一方面 2u=(2,2)ᵀ，M₀(2u)=(4−2, 6+8)ᵀ=(2,14)ᵀ，F(2u)=(3,12)ᵀ。第一分量 3≠4，純量相容在 c=2 失效；因為 F(u+u)=F(2u)，加法相容在 u、u 也失效。常數項 b 在 F(2u) 中只加一次，在 2F(u) 中加了兩次。", "en": "An explicit failure: F(u)=M₀u+b=(1,7)ᵀ+(1,−2)ᵀ=(2,5)ᵀ, so 2F(u)=(4,10)ᵀ. On the other hand 2u=(2,2)ᵀ, M₀(2u)=(4−2, 6+8)ᵀ=(2,14)ᵀ and F(2u)=(3,12)ᵀ. The first components differ, 3≠4, so homogeneity fails at c=2; since F(u+u)=F(2u), additivity fails at u, u as well. The constant b is added once in F(2u) but twice in 2F(u)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S10", "numeric_evaluation", "quantifier_negation" ], "explanation": { "zh-Hant": "一個失效的實例就足以否定全稱條件（quantifier_negation）；這與 S11 的結論一致。", "en": "One failing instance suffices to refute a universal condition (quantifier_negation); this agrees with the conclusion of S11." } }, "leaf_basis": "numeric_evaluation + quantifier_negation + vec_ops" }, { "id": "S13", "uses": [ "S03", "S10", "G2" ], "before_tex": "M_0(u+w)\\ \\text{vs}\\ M_0u+M_0w", "after_tex": "M_0(3,-2)^{\\mathsf T}=(8,1)^{\\mathsf T}=(1,7)^{\\mathsf T}+(7,-6)^{\\mathsf T}", "relation": "implies", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "數值示範 T 的加法相容：u+w=(3,−2)ᵀ，M₀(u+w)=(2·3+(−1)(−2), 3·3+4·(−2))ᵀ=(8,1)ᵀ；M₀w=(2·2+(−1)(−3), 3·2+4·(−3))ᵀ=(7,−6)ᵀ，M₀u+M₀w=(1+7, 7+(−6))ᵀ=(8,1)ᵀ。兩邊相等，與 S03 一致。這只是一個實例，一般結論來自 S03 的符號證明（MKO 第二段）。", "en": "A numeric illustration of additivity of T: u+w=(3,−2)ᵀ, M₀(u+w)=(2·3+(−1)(−2), 3·3+4·(−2))ᵀ=(8,1)ᵀ; M₀w=(2·2+(−1)(−3), 3·2+4·(−3))ᵀ=(7,−6)ᵀ and M₀u+M₀w=(1+7, 7+(−6))ᵀ=(8,1)ᵀ. Both sides agree, consistent with S03. This is one instance only; the general result comes from the symbolic proof in S03 (MKO second paragraph)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "numeric_evaluation" ], "explanation": { "zh-Hant": "有限計算不被包裝成普遍證明；它只是 S03 的一致性核對（N8 到 N13）。", "en": "The finite computation is not presented as a universal proof; it is only a consistency check of S03 (N8 to N13)." } }, "leaf_basis": "numeric_evaluation + matrix_product_def" }, { "id": "S14", "uses": [ "G3", "S01" ], "before_tex": "g(x,y)=(x^2,y)", "after_tex": "g((0,1)+(0,2))=g(0,1)+g(0,2),\\ \\text{but}\\ g((1,0)+(1,0))=(4,0)\\neq(2,0)", "relation": "implies", "rule": "quantifier_negation", "conditions": [], "explanation": { "zh-Hant": "對 g：g(0,3)=(0,3)=(0,1)+(0,2)=g(0,1)+g(0,2)，在這組測點加法相容「看起來成立」；g(0,5)=(0,5)=5·g(0,1) 也通過純量測試。但 g(2,0)=(4,0)，而 g(1,0)+g(1,0)=(1,0)+(1,0)=(2,0)，4≠2。所以 g 不線性；通過有限個測點不能推出任意映射線性（MKO 假設 a1 與誤解）。對矩陣映射，線性來自 S03、S04 對任意輸入的證明。", "en": "For g: g(0,3)=(0,3)=(0,1)+(0,2)=g(0,1)+g(0,2), so additivity 'seems to hold' at these test points; g(0,5)=(0,5)=5·g(0,1) also passes a homogeneity test. But g(2,0)=(4,0), while g(1,0)+g(1,0)=(1,0)+(1,0)=(2,0), and 4≠2. So g is not linear; passing finitely many test points cannot establish linearity of an arbitrary map (MKO assumption a1 and misconception). For matrix maps, linearity comes from the proofs in S03 and S04 for arbitrary inputs." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S01", "quantifier_negation", "numeric_evaluation" ], "explanation": { "zh-Hant": "否定全稱只需一個反例；通過的測點不構成證明。", "en": "Refuting a universal statement needs one counterexample; the passing test points do not constitute a proof of linearity." } }, "leaf_basis": "quantifier_negation + numeric_evaluation" }, { "id": "S15", "uses": [ "S05", "S09", "S11", "S12", "S13", "S14" ], "before_tex": "\\text{tasks of the goal}", "after_tex": "v\\mapsto Mv\\ \\text{linear};\\ Mv+b\\ \\text{linear}\\iff b=0;\\ F\\ \\text{not linear};\\ g\\ \\text{not linear}", "relation": "implies", "rule": "and_intro_elim", "conditions": [], "explanation": { "zh-Hant": "合併：每個 2×2 實矩陣給出線性映射（S05）；仿射式線性當且僅當 b=0（S09）；例中的 F 不線性，失效式為 F(2u)=(3,12)ᵀ≠(4,10)ᵀ=2F(u)（S11、S12）；T 的加法相容有數值示範（S13）；g 說明有限測點不足（S14）。", "en": "Combine: every 2×2 real matrix gives a linear map (S05); an affine map is linear exactly when b=0 (S09); the example F is not linear, failing as F(2u)=(3,12)ᵀ≠(4,10)ᵀ=2F(u) (S11, S12); additivity of T is illustrated numerically (S13); g shows that finitely many test points are not enough (S14)." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S05", "S09", "S11", "S12", "S13", "S14" ], "explanation": { "zh-Hant": "目標 G：題目要求的各項。足夠條件 H：S05 到 S14 的結論；H⇒G 只用合取。H 由 G1 到 G3 正向推出，沒有把 G 當作前提。", "en": "Goal G: the items asked for. Sufficient condition H: the conclusions of S05 to S14; H⇒G uses only conjunction. H is derived forwards from G1 to G3; G is never used as a premise." } }, "leaf_basis": "and_intro_elim" } ], "result": { "zh-Hant": "對每個 2×2 實矩陣 M，T(v)=Mv 滿足 T(u+v)=T(u)+T(v) 與 T(cv)=cT(v)，所以線性（不需可逆）。仿射式 F(v)=Mv+b 線性當且僅當 b=0。例中 M₀u=(1,7)ᵀ；F(v)=M₀v+(1,−2)ᵀ 不線性，因 F(0)=(1,−2)ᵀ≠0，且 F(2u)=(3,12)ᵀ≠(4,10)ᵀ=2F(u)。g(x,y)=(x²,y) 通過部分測點但不線性。", "en": "For every 2×2 real matrix M, T(v)=Mv satisfies T(u+v)=T(u)+T(v) and T(cv)=cT(v), so it is linear (invertibility is not needed). The affine map F(v)=Mv+b is linear exactly when b=0. In the example M₀u=(1,7)ᵀ; F(v)=M₀v+(1,−2)ᵀ is not linear, since F(0)=(1,−2)ᵀ≠0 and F(2u)=(3,12)ᵀ≠(4,10)ᵀ=2F(u). g(x,y)=(x²,y) passes some test points but is not linear." }, "negative_controls": [ { "claim": { "zh-Hant": "F(v)=M₀v+(1,−2)ᵀ 是線性的，因為每個分量都是 x、y 的一次式", "en": "F(v)=M₀v+(1,−2)ᵀ is linear because each component is a first-degree expression in x and y" }, "expected": "reject", "reason": { "zh-Hant": "F(0)=(1,−2)ᵀ≠0；且 F(2u)=(3,12)ᵀ≠(4,10)ᵀ=2F(u)。", "en": "F(0)=(1,−2)ᵀ≠0, and F(2u)=(3,12)ᵀ≠(4,10)ᵀ=2F(u)." } }, { "claim": { "zh-Hant": "g(x,y)=(x²,y) 線性，因為 g((0,1)+(0,2))=g(0,1)+g(0,2)", "en": "g(x,y)=(x²,y) is linear because g((0,1)+(0,2))=g(0,1)+g(0,2)" }, "expected": "reject", "reason": { "zh-Hant": "g(2,0)=(4,0)≠(2,0)=g(1,0)+g(1,0)。", "en": "g(2,0)=(4,0)≠(2,0)=g(1,0)+g(1,0)." } }, { "claim": { "zh-Hant": "零矩陣給出的映射不線性，因為它不可逆", "en": "The map given by the zero matrix is not linear because it is not invertible" }, "expected": "reject", "reason": { "zh-Hant": "0(u+v)=0=0u+0v、0(cv)=0=c·0v，兩個條件都成立；線性不要求可逆（MKO 假設 a3）。", "en": "0(u+v)=0=0u+0v and 0(cv)=0=c·0v, so both conditions hold; linearity does not require invertibility (MKO assumption a3)." } }, { "claim": { "zh-Hant": "M₀u 是把 M₀ 的每個元素乘以 1，得到矩陣 [[2,−1],[3,4]]", "en": "M₀u is obtained by multiplying each entry of M₀ by 1, giving the matrix [[2,−1],[3,4]]" }, "expected": "reject", "reason": { "zh-Hant": "Mv 是行與向量的乘積和，結果是向量 (1,7)ᵀ。", "en": "Mv is formed from sums of products of rows with the vector; the result is the vector (1,7)ᵀ." } } ], "status": "candidate", "gaps": [], "verification": { "structure": "checked_local_script", "mathematics": "candidate_argument", "formalization": "not_formalized" }, "rule_glossary": [ { "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": "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": "add_assoc", "kind": "prerequisite", "statement": { "zh-Hant": "加法結合律：(u+v)+w=u+(v+w)。", "en": "Associativity of addition: (u+v)+w=u+(v+w)." } }, { "id": "add_comm", "kind": "prerequisite", "statement": { "zh-Hant": "加法交換律：u+v=v+u。", "en": "Commutativity of addition: u+v=v+u." } }, { "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": "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": "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": "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": "mul_zero", "kind": "prerequisite", "statement": { "zh-Hant": "乘以零：u·0=0。理由：u·0=u·(0+0)=u·0+u·0，兩邊同減 u·0 得 0=u·0。", "en": "Multiplying by zero: u·0=0. Reason: u·0=u·(0+0)=u·0+u·0; subtracting u·0 from both sides gives 0=u·0." } }, { "id": "forall_elim", "kind": "prerequisite", "statement": { "zh-Hant": "∀消去（代入特例）：若 ∀x∈D, P(x) 成立且 a∈D，則 P(a) 成立。", "en": "∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds." } }, { "id": "add_zero", "kind": "prerequisite", "statement": { "zh-Hant": "加法單位元：u+0=u。", "en": "Additive identity: u+0=u." } }, { "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": "function_def", "kind": "prerequisite", "statement": { "zh-Hant": "函數：f:A→B 對每個 a∈A 指定恰好一個 f(a)∈B。兩函數相等 ⇔ 定義域、對應域相同且對每個 a 都有相同值。", "en": "Function: f:A→B assigns to each a∈A exactly one f(a)∈B. Two functions are equal ⇔ they have the same domain and codomain and agree at every a." } }, { "id": "eq_refl_symm_trans", "kind": "prerequisite", "statement": { "zh-Hant": "等號的性質：u=u；若 u=v 則 v=u；若 u=v 且 v=w 則 u=w。", "en": "Properties of equality: u=u; if u=v then v=u; if u=v and v=w then u=w." } }, { "id": "iff_intro", "kind": "prerequisite", "statement": { "zh-Hant": "雙條件：P⇒Q 與 Q⇒P 都成立時，P⇔Q 成立；反之亦然。", "en": "Biconditional: P⇔Q holds exactly when both P⇒Q and Q⇒P hold." } }, { "id": "contrapositive", "kind": "prerequisite", "statement": { "zh-Hant": "逆否命題：P⇒Q 與 ¬Q⇒¬P 等價。逆命題 Q⇒P 一般不與 P⇒Q 等價。", "en": "Contrapositive: P⇒Q is equivalent to ¬Q⇒¬P. The converse Q⇒P is in general not equivalent to P⇒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": "mko_linear_def", "kind": "cited", "source": { "mko_id": "mko-linear-transformation", "sha256": "59ce4d8fac03871ea16eb62a76c9fae255470dc391ff7312ad4da120bafbf132", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "映射 T:R²→R² 為線性映射，當且僅當對所有向量 u、v 與實純量 c，T(u+v)=T(u)+T(v)、T(cv)=cT(v)。本輪 T(v)=Mv，M 是2×2實矩陣，向量以列向量約定。仿射式 F(v)=Mv+b 在全實平面上為線性映射，當且僅當 b 是零向量。", "en": "A map T:R²→R² is linear exactly when T(u+v)=T(u)+T(v) and T(cv)=cT(v) for all vectors u,v and real scalars c. Here T(v)=Mv, with a 2×2 real matrix M and the column-vector convention. The affine map F(v)=Mv+b on the whole real plane is linear exactly when b is the zero vector." } }, { "id": "vec_ops", "kind": "prerequisite", "statement": { "zh-Hant": "ℝ² 的列向量逐分量相加與數乘：(x,y)ᵀ+(x′,y′)ᵀ=(x+x′,y+y′)ᵀ，c(x,y)ᵀ=(cx,cy)ᵀ；兩向量相等 ⇔ 兩個分量分別相等。", "en": "Column vectors in ℝ² add and scale componentwise: (x,y)ᵀ+(x′,y′)ᵀ=(x+x′,y+y′)ᵀ and c(x,y)ᵀ=(cx,cy)ᵀ; two vectors are equal ⇔ both components are equal." } }, { "id": "L_lin_zero", "kind": "lemma", "proved_in": [ "S06.1" ], "statement": { "zh-Hant": "若映射 T:ℝ²→ℝ² 滿足純量相容 T(cv)=cT(v)（對所有 c、v），則 T(0)=0。", "en": "If a map T:ℝ²→ℝ² satisfies homogeneity T(cv)=cT(v) for all c and v, then T(0)=0." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S10", "lhs": "2*1+(-1)*1", "rhs": "1" }, { "id": "N2", "step": "S10", "lhs": "3*1+4*1", "rhs": "7" }, { "id": "N3", "step": "S12", "lhs": "2*2+(-1)*2+1", "rhs": "3" }, { "id": "N4", "step": "S12", "lhs": "3*2+4*2+(-2)", "rhs": "12" }, { "id": "N5", "step": "S12", "lhs": "2*(1+1)", "rhs": "4" }, { "id": "N6", "step": "S12", "lhs": "2*(7+(-2))", "rhs": "10" }, { "id": "N7", "step": "S12", "lhs": "3", "relation": "!=", "rhs": "4" }, { "id": "N8", "step": "S13", "lhs": "2*3+(-1)*(-2)", "rhs": "8" }, { "id": "N9", "step": "S13", "lhs": "3*3+4*(-2)", "rhs": "1" }, { "id": "N10", "step": "S13", "lhs": "2*2+(-1)*(-3)", "rhs": "7" }, { "id": "N11", "step": "S13", "lhs": "3*2+4*(-3)", "rhs": "-6" }, { "id": "N12", "step": "S13", "lhs": "1+7", "rhs": "8" }, { "id": "N13", "step": "S13", "lhs": "7+(-6)", "rhs": "1" }, { "id": "N14", "step": "S03.1", "vars": [ "p", "q", "u1", "u2", "v1", "v2" ], "degree": 2, "lhs": "p*(u1+v1)+q*(u2+v2)", "rhs": "(p*u1+q*u2)+(p*v1+q*v2)" }, { "id": "N15", "step": "S04", "vars": [ "p", "q", "c", "x", "y" ], "degree": 2, "lhs": "p*(c*x)+q*(c*y)", "rhs": "c*(p*x+q*y)" }, { "id": "N16", "step": "S14", "lhs": "(1+1)**2", "relation": "!=", "rhs": "1**2+1**2" }, { "id": "N17", "step": "S08", "vars": [ "p", "q" ], "degree": 1, "lhs": "p*0+q*0", "rhs": "0" } ] }, "presentation": { "md_scope": { "zh-Hant": "只在 MKO 的範圍內：全實平面為定義域與陪域、列向量約定、2×2 實矩陣。完整證明矩陣映射的線性與仿射式的判準；不處理其他維度、可逆性、單射或等距，也不做 Lean 的 Matrix.mulVecLin 橋接；不改動 canonical MKO 或英文 edition。", "en": "Stays inside the MKO's scope: the whole real plane as domain and codomain, the column-vector convention, 2×2 real matrices. It fully proves linearity of matrix maps and the criterion for affine maps; it does not treat other dimensions, invertibility, injectivity or isometry, and does not build the Lean Matrix.mulVecLin bridge; it does not change the canonical MKO or its English edition." }, "md_symbols": [ [ "T(v)=Mv", "矩陣 M 決定的映射", "the map determined by the matrix M" ], [ "M=\\begin{pmatrix}p&q\\\\r&s\\end{pmatrix}", "一般 2×2 實矩陣", "a general 2×2 real matrix" ], [ "v=(x,y)^{\\mathsf T}", "列向量（直行向量）", "a column vector" ], [ "c", "實純量", "a real scalar" ], [ "F(v)=Mv+b", "仿射映射", "an affine map" ] ], "md_error_controls": [ { "title": "非零截距破壞線性 / A nonzero intercept breaks linearity", "zh": "MKO 的誤解之一是把圖上的直線或仿射式當成線性。一維的 f(x)=2x+1 圖形是直線，但 f(0)=1≠0，不滿足 f(cx)=cf(x)。同理 F(v)=M₀v+(1,−2)ᵀ 的每個分量都是一次式加常數，但 F(0)=b≠0，所以不線性（S11、S12）。", "en": "One misconception the MKO names is calling a straight-line graph or an affine formula linear. In one dimension f(x)=2x+1 has a straight-line graph, but f(0)=1≠0, so f(cx)=cf(x) fails. Likewise each component of F(v)=M₀v+(1,−2)ᵀ is a first-degree expression plus a constant, but F(0)=b≠0, so F is not linear (S11, S12)." }, { "title": "有限測點不是證明 / Finitely many test points are not a proof", "zh": "g(x,y)=(x²,y) 在 (0,1)、(0,2) 上通過加法測試，在 c=5、v=(0,1) 上通過純量測試，卻在 (1,0)+(1,0) 失效（S14）。線性是全稱條件，對任意映射只能由對任意輸入的證明（如 S03、S04）確立，有限測點只能否定，不能肯定。", "en": "g(x,y)=(x²,y) passes the additivity test at (0,1), (0,2) and the homogeneity test at c=5, v=(0,1), yet fails at (1,0)+(1,0) (S14). Linearity is a universal condition; for an arbitrary map it can be established only by a proof for arbitrary inputs (as in S03, S04). Finitely many test points can refute it but never confirm it." }, { "title": "Mv 的各分量是乘積和，不是逐項乘 / Each component of Mv is a sum of products, not an entrywise product", "zh": "M₀u 的第一分量是 2·1+(−1)·1=1，是第一行與 u 的乘積和。逐項相乘會得到 [[2,−1],[3,4]] 這樣的矩陣，而不是向量，也不是 Mv（S02、S10）。", "en": "The first component of M₀u is 2·1+(−1)·1=1, the sum of products of the first row with u. Multiplying entry by entry would give a matrix such as [[2,−1],[3,4]], not a vector and not Mv (S02, S10)." } ], "md_open_items": [ { "zh-Hant": "本稿不處理可逆、單射、等距等另需條件的性質，也不處理 ℝ² 以外的維度。", "en": "This guide does not treat properties that need further conditions, such as invertibility, injectivity or isometry, nor dimensions other than ℝ²." }, { "zh-Hant": "MKO 的 formal next（Matrix.mulVecLin 與標準基底的橋接）未做。", "en": "The MKO's formal next step (bridging Matrix.mulVecLin with the standard basis) is not done." } ] }, "revision": { "id": "production-20261019-adopted-96", "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "adopted_source_sha256": "38b64f81831e2d28908dc118d70f5a88e7a287d2c9e36e7095f6fc17b8af399b", "original_source_sha256": "38b64f81831e2d28908dc118d70f5a88e7a287d2c9e36e7095f6fc17b8af399b", "owner_adopted": true, "bilingual_sync": "complete_source_bound", "formal_Lean": "NOT_RUN", "independent_AI_concurrence": "NOT_MEASURED", "issue_ids": [], "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion." } }