{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-eigenvalue-01", "mko_id": "mko-eigenvalue", "title": { "zh-Hant": "二階實矩陣的特徵值：特徵多項式、判別式與非零特徵向量", "en": "Eigenvalues of 2×2 real matrices: characteristic polynomial, discriminant and nonzero eigenvectors" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-eigenvalue-01", "source": { "mko_id": "mko-eigenvalue", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "5e1e1186120be48f1e86332bf1d679143e44074e2b3e9f59c5c219cb76a99e36", "english_sha256": "e48363ac57349147d31e8427dc2281ab02ae34f1abaea817c89e3e0f431fe5ed" }, "mode": "theorem_proof", "audience_prerequisites": [ "matrix_product_def", "distributive", "square_nonneg", "sqrt_def", "zero_product", "numeric_evaluation" ], "goal": { "zh-Hant": "對 2×2 實矩陣證明：λ 是實特徵值 ⇔ det(M−λI)=0 ⇔ λ²−tr(M)λ+det(M)=0，且互異實特徵值的個數由 Δ=tr(M)²−4det(M) 的正負決定（負、零、正分別為 0、1、2）。以 M=[[4,1],[2,3]] 求出全部互異實特徵值與各自的非零特徵向量，說明候選 λ=3 的核 K₃={0} 而 3 不是特徵值，並拒絕候選特徵對 (3,0)；再以旋轉矩陣 R=[[0,−1],[1,0]]（Δ<0）與 J=[[3,1],[0,3]]（Δ=0）對照。", "en": "For 2×2 real matrices prove: λ is a real eigenvalue ⇔ det(M−λI)=0 ⇔ λ²−tr(M)λ+det(M)=0, and the number of distinct real eigenvalues is decided by the sign of Δ=tr(M)²−4det(M) (negative, zero, positive give 0, 1, 2). For M=[[4,1],[2,3]] find all distinct real eigenvalues with nonzero eigenvectors, show that for the candidate λ=3 the kernel is K₃={0} and 3 is not an eigenvalue, and reject the candidate eigenpair (3,0); then contrast with the rotation R=[[0,−1],[1,0]] (Δ<0) and J=[[3,1],[0,3]] (Δ=0)." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "全程在實數域，M=[[a,b],[c,d]] 為 2×2 實矩陣，向量為 ℝ² 的列向量；I=[[1,0],[0,1]]（MKO 假設 a1）。", "en": "Throughout we work over ℝ; M=[[a,b],[c,d]] is a 2×2 real matrix, vectors are column vectors in ℝ², and I=[[1,0],[0,1]] (MKO assumption a1)." } }, { "id": "G2", "statement": { "zh-Hant": "例子：M=[[4,1],[2,3]]；候選值 λ=3；候選特徵對 (3,0)。", "en": "Example: M=[[4,1],[2,3]]; candidate value λ=3; candidate eigenpair (3,0)." } }, { "id": "G3", "statement": { "zh-Hant": "對照：R=[[0,−1],[1,0]]，J=[[3,1],[0,3]]。", "en": "Contrasts: R=[[0,−1],[1,0]] and J=[[3,1],[0,3]]." } } ], "steps": [ { "id": "S01", "uses": [ "G1" ], "before_tex": "\\lambda\\in\\mathbb{R}", "after_tex": "\\lambda\\ \\text{eigenvalue}\\iff\\exists v\\neq 0:\\ Mv=\\lambda v;\\quad K_\\lambda=\\{v\\mid Mv=\\lambda v\\}", "relation": "definition", "rule": "mko_eig_def", "conditions": [], "explanation": { "zh-Hant": "寫出 MKO 的定義與非零特徵向量慣例。核 K_λ 是所有滿足 Mv=λv 的向量（包括零向量）；特徵值則要求核中有非零向量。兩者必須分開：因為 M0=0=λ0 對每個 λ 都成立，零向量本身不能證明任何 λ 是特徵值。", "en": "State the MKO's definition and the nonzero-eigenvector convention. The kernel K_λ consists of all vectors with Mv=λv (the zero vector included); being an eigenvalue requires a nonzero vector in the kernel. The two must be kept apart: since M0=0=λ0 for every λ, the zero vector alone cannot make any λ an eigenvalue." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "mko_eig_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_eig_def + def_unfold" }, { "id": "S02", "uses": [ "S01", "matrix_product_def", "vec_ops" ], "before_tex": "Mv=\\lambda v", "after_tex": "(M-\\lambda I)v=0", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "設 v=(x,y)ᵀ。(M−λI)v=((a−λ)x+by, cx+(d−λ)y)ᵀ，依分配律等於 (ax+by−λx, cx+dy−λy)ᵀ=Mv−λv。所以 Mv=λv ⇔ Mv−λv=0 ⇔ (M−λI)v=0。因此 K_λ 就是 M−λI 的零空間。", "en": "Let v=(x,y)ᵀ. (M−λI)v=((a−λ)x+by, cx+(d−λ)y)ᵀ, which by distributivity equals (ax+by−λx, cx+dy−λy)ᵀ=Mv−λv. So Mv=λv ⇔ Mv−λv=0 ⇔ (M−λI)v=0. Hence K_λ is the null space of M−λI." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "vec_ops", "add_inverse" ], "explanation": { "zh-Hant": "兩邊同減 λv 或同加 λv，互為反運算，雙向成立。", "en": "Subtracting λv from both sides and adding it back undo each other, so it is two-way." } }, "leaf_basis": "distributive + matrix_product_def + vec_ops + add_inverse" }, { "id": "S03", "uses": [ "S01", "S02", "L_kernel_det" ], "before_tex": "\\lambda\\ \\text{eigenvalue}", "after_tex": "\\det(M-\\lambda I)=0", "relation": "equivalent", "rule": "L_kernel_det", "conditions": [], "explanation": { "zh-Hant": "依 S01、S02，λ 是特徵值 ⇔ 存在非零 v 使 (M−λI)v=0。對 N=M−λI 用引理 L_kernel_det，這等價於 det(M−λI)=0。引理在 S03.1 到 S03.3 證明。", "en": "By S01 and S02, λ is an eigenvalue ⇔ some nonzero v satisfies (M−λI)v=0. Applying Lemma L_kernel_det to N=M−λI, this is equivalent to det(M−λI)=0. The lemma is proved in S03.1 to S03.3." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_kernel_det", "S02" ], "explanation": { "zh-Hant": "引理兩個方向都有證明（S03.1、S03.2 為 ⇐，S03.3 為 ⇒）。", "en": "Both directions of the lemma are proved (S03.1 and S03.2 for ⇐, S03.3 for ⇒)." } }, "leaf_basis": "L_kernel_det + logic_chain" }, { "id": "S03.1", "uses": [ "matrix_product_def", "vec_ops" ], "before_tex": "\\det N=ps-qr=0,\\ (p,q)\\neq(0,0)", "after_tex": "v=(q,-p)^{\\mathsf T}\\neq 0,\\ Nv=(pq-qp,\\ rq-sp)^{\\mathsf T}=(0,\\ -\\det N)^{\\mathsf T}=0", "relation": "implies", "rule": "matrix_product_def", "conditions": [ "\\det N=0", "(p,q)\\neq(0,0)" ], "explanation": { "zh-Hant": "情況一：第一行不全為零。取 v=(q,−p)ᵀ；因 p、q 至少一個非零，v≠0（vec_ops）。計算 Nv：第一分量 pq+q(−p)=0，第二分量 rq+s(−p)=−(ps−qr)=−det N=0。所以 v 是非零核向量。", "en": "Case one: the first row is not all zero. Take v=(q,−p)ᵀ; since at least one of p, q is nonzero, v≠0 (vec_ops). Compute Nv: the first component is pq+q(−p)=0 and the second is rq+s(−p)=−(ps−qr)=−det N=0. So v is a nonzero kernel vector." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "matrix_product_def", "vec_ops", "neg_mul" ], "explanation": { "zh-Hant": "第二分量為零正是用到 det N=0；第一行不全為零保證 v≠0。", "en": "The second component vanishes exactly because det N=0; the first row being nonzero guarantees v≠0." } }, "leaf_basis": "matrix_product_def + vec_ops + neg_mul + mul_comm" }, { "id": "S03.2", "uses": [ "S03.1", "matrix_product_def", "vec_ops" ], "before_tex": "p=q=0", "after_tex": "v=(s,-r)^{\\mathsf T}\\ \\text{if}\\ (r,s)\\neq(0,0),\\ \\text{else}\\ v=(1,0)^{\\mathsf T};\\ \\ v\\neq 0,\\ Nv=0", "relation": "implies", "rule": "case_analysis", "conditions": [ "p=q=0" ], "explanation": { "zh-Hant": "情況二：p=q=0（此時 det N=0·s−0·r=0 自動成立）。若 (r,s)≠(0,0)，取 v=(s,−r)ᵀ≠0：第一分量 0·s+0·(−r)=0，第二分量 rs+s(−r)=0。若 r=s=0，N 是零矩陣，取 v=(1,0)ᵀ，Nv=0。兩種情況與 S03.1 合起來涵蓋所有 det N=0 的 N，所以「det N=0 ⇒ 有非零核向量」成立。", "en": "Case two: p=q=0 (then det N=0·s−0·r=0 automatically). If (r,s)≠(0,0), take v=(s,−r)ᵀ≠0: the first component is 0·s+0·(−r)=0 and the second is rs+s(−r)=0. If r=s=0, N is the zero matrix; take v=(1,0)ᵀ, so Nv=0. These cases together with S03.1 cover every N with det N=0, so 'det N=0 ⇒ some nonzero kernel vector' holds." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03.1", "case_analysis", "matrix_product_def" ], "explanation": { "zh-Hant": "三種情況（第一行非零、第一行為零但第二行非零、全零）窮盡所有可能。", "en": "The three cases (first row nonzero; first row zero but second row nonzero; all zero) exhaust every possibility." } }, "leaf_basis": "case_analysis + matrix_product_def + mul_zero + vec_ops" }, { "id": "S03.3", "uses": [ "mko_det_inv", "vec_ops" ], "before_tex": "\\det N\\neq 0,\\ Nv=0", "after_tex": "v=N^{-1}(Nv)=N^{-1}0=0", "relation": "implies", "rule": "contrapositive", "conditions": [ "\\det N\\neq 0" ], "explanation": { "zh-Hant": "以逆否證「有非零核向量 ⇒ det N=0」：設 det N≠0。依行列式條目（cited），N 在 ℝ 上可逆。若 Nv=0，兩邊左乘 N⁻¹：v=N⁻¹(Nv)=N⁻¹0=0。所以核中只有零向量。S03.1 到 S03.3 合起來證得引理 L_kernel_det。", "en": "Prove 'some nonzero kernel vector ⇒ det N=0' by contraposition: suppose det N≠0. By the determinant entry (cited), N is invertible over ℝ. If Nv=0, multiply both sides on the left by N⁻¹: v=N⁻¹(Nv)=N⁻¹0=0. So the kernel contains only the zero vector. S03.1 to S03.3 together prove Lemma L_kernel_det." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "mko_det_inv", "contrapositive", "vec_ops" ], "explanation": { "zh-Hant": "依賴行列式條目的實數域可逆判準與矩陣乘法結合律；N⁻¹0=0 是每個分量乘零之和。", "en": "It relies on the real-field invertibility criterion of the determinant entry and associativity of matrix multiplication; N⁻¹0=0 because each component is a sum of products with zero." } }, "leaf_basis": "mko_det_inv + contrapositive + mul_zero" }, { "id": "S04", "uses": [ "S03", "G1" ], "before_tex": "\\det(M-\\lambda I)=(a-\\lambda)(d-\\lambda)-bc", "after_tex": "\\lambda^2-(a+d)\\lambda+(ad-bc)=\\lambda^2-\\operatorname{tr}(M)\\lambda+\\det(M)", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "M−λI=[[a−λ,b],[c,d−λ]]，依行列式定義得左式。展開 (a−λ)(d−λ)=ad−aλ−dλ+λ²，再減 bc，整理得 λ²−(a+d)λ+(ad−bc)。對例子 M：tr(M)=4+3=7，det(M)=4·3−1·2=10，特徵多項式 λ²−7λ+10。", "en": "M−λI=[[a−λ,b],[c,d−λ]], and the definition of the determinant gives the left side. Expand (a−λ)(d−λ)=ad−aλ−dλ+λ², subtract bc and collect terms: λ²−(a+d)λ+(ad−bc). For the example M: tr(M)=4+3=7 and det(M)=4·3−1·2=10, so the characteristic polynomial is λ²−7λ+10." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "neg_mul", "add_comm" ], "explanation": { "zh-Hant": "多項式恆等式（N1），雙向成立。", "en": "A polynomial identity (N1), two-way." } }, "leaf_basis": "distributive + neg_mul + add_comm + mko_det_inv" }, { "id": "S05", "uses": [ "S04", "L_disc" ], "before_tex": "\\Delta=\\operatorname{tr}(M)^2-4\\det(M)=49-40", "after_tex": "\\Delta=9>0\\ \\Rightarrow\\ \\text{two distinct real eigenvalues}", "relation": "theorem_application", "rule": "L_disc", "conditions": [], "explanation": { "zh-Hant": "依 S03、S04，互異實特徵值恰是 λ²−7λ+10=0 的互異實根。Δ=7²−4·10=9>0，依引理 L_disc（T=7、D=10），恰有兩個互異實根。引理在 S05.1 到 S05.4 證明，涵蓋 MKO 假設 a3 的三種情形。", "en": "By S03 and S04 the distinct real eigenvalues are exactly the distinct real roots of λ²−7λ+10=0. Δ=7²−4·10=9>0, so by Lemma L_disc (T=7, D=10) there are exactly two distinct real roots. The lemma is proved in S05.1 to S05.4, covering the three cases of MKO assumption a3." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "S04", "L_disc" ], "explanation": { "zh-Hant": "需要 S03 的等價把特徵值問題化為求根問題；引理只計互異根，不計重數。", "en": "It needs the equivalence of S03 to turn the eigenvalue question into a root question; the lemma counts distinct roots, not multiplicities." } }, "leaf_basis": "L_disc + numeric_evaluation" }, { "id": "S05.1", "uses": [ "distributive", "frac_rules" ], "before_tex": "\\lambda^2-T\\lambda+D", "after_tex": "\\left(\\lambda-\\tfrac{T}{2}\\right)^2-\\tfrac{\\Delta}{4},\\quad \\Delta=T^2-4D", "relation": "equivalent", "rule": "distributive", "conditions": [], "explanation": { "zh-Hant": "配方：(λ−T/2)²=λ²−Tλ+T²/4（分配律，2≠0），所以 (λ−T/2)²−Δ/4=λ²−Tλ+T²/4−(T²−4D)/4=λ²−Tλ+D。", "en": "Complete the square: (λ−T/2)²=λ²−Tλ+T²/4 (distributivity, with 2≠0), so (λ−T/2)²−Δ/4=λ²−Tλ+T²/4−(T²−4D)/4=λ²−Tλ+D." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "distributive", "frac_rules", "two_nonzero" ], "explanation": { "zh-Hant": "對所有實數 λ、T、D 成立的恆等式（N4），雙向成立。", "en": "An identity for all reals λ, T, D (N4), two-way." } }, "leaf_basis": "distributive + frac_rules + two_nonzero" }, { "id": "S05.2", "uses": [ "S05.1", "square_nonneg" ], "before_tex": "\\Delta<0", "after_tex": "\\left(\\lambda-\\tfrac{T}{2}\\right)^2-\\tfrac{\\Delta}{4}>0\\ \\text{for all }\\lambda:\\ \\text{no real root}", "relation": "implies", "rule": "square_nonneg", "conditions": [ "\\Delta<0" ], "explanation": { "zh-Hant": "Δ<0 時 −Δ/4>0；又 (λ−T/2)²≥0（square_nonneg），兩者相加得正數。所以依 S05.1，λ²−Tλ+D 對每個實數 λ 都大於 0，沒有實根。", "en": "If Δ<0 then −Δ/4>0; also (λ−T/2)²≥0 (square_nonneg), and the sum is positive. So by S05.1, λ²−Tλ+D is greater than 0 for every real λ, and there is no real root." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S05.1", "square_nonneg", "order_basics" ], "explanation": { "zh-Hant": "結論是對所有實數 λ；論證沒有對 λ 加任何假設。", "en": "The conclusion is for every real λ; the argument makes no assumption about λ." } }, "leaf_basis": "square_nonneg + order_basics + forall_intro" }, { "id": "S05.3", "uses": [ "S05.1", "square_nonneg" ], "before_tex": "\\Delta=0", "after_tex": "\\left(\\lambda-\\tfrac{T}{2}\\right)^2=0\\iff\\lambda=\\tfrac{T}{2}:\\ \\text{exactly one root}", "relation": "implies", "rule": "square_nonneg", "conditions": [ "\\Delta=0" ], "explanation": { "zh-Hant": "Δ=0 時方程變成 (λ−T/2)²=0；依 square_nonneg（u²=0 ⇔ u=0），等價於 λ=T/2。恰有一個實根；它只列一次，不按重數算兩個（MKO 假設 a3）。", "en": "If Δ=0 the equation becomes (λ−T/2)²=0; by square_nonneg (u²=0 ⇔ u=0) this is equivalent to λ=T/2. There is exactly one real root; it is listed once and not counted twice by multiplicity (MKO assumption a3)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S05.1", "square_nonneg" ], "explanation": { "zh-Hant": "u²=0 ⇔ u=0 兩向都成立，所以根存在且唯一。", "en": "u²=0 ⇔ u=0 holds both ways, so the root exists and is unique." } }, "leaf_basis": "square_nonneg + eq_congruence" }, { "id": "S05.4", "uses": [ "S05.1", "sqrt_def", "zero_product" ], "before_tex": "\\Delta>0", "after_tex": "\\lambda=\\tfrac{T+\\sqrt{\\Delta}}{2}\\ \\lor\\ \\lambda=\\tfrac{T-\\sqrt{\\Delta}}{2},\\ \\text{two distinct roots}", "relation": "implies", "rule": "zero_product", "conditions": [ "\\Delta>0" ], "explanation": { "zh-Hant": "Δ>0 時令 k=√Δ/2；依 sqrt_def，k≥0 且 k²=Δ/4，又 Δ≠0 所以 k>0。方程等價於 u²−k²=0（u=λ−T/2），即 (u−k)(u+k)=0；依零乘積性質，u=k 或 u=−k，即 λ=(T+√Δ)/2 或 λ=(T−√Δ)/2。兩者相差 2k>0，所以互異；沒有其他實根。S05.1 到 S05.4 合起來證得引理 L_disc。", "en": "If Δ>0 let k=√Δ/2; by sqrt_def k≥0 and k²=Δ/4, and since Δ≠0, k>0. The equation is equivalent to u²−k²=0 with u=λ−T/2, that is (u−k)(u+k)=0; by the zero-product property u=k or u=−k, so λ=(T+√Δ)/2 or λ=(T−√Δ)/2. They differ by 2k>0, so they are distinct, and there are no other real roots. S05.1 to S05.4 together prove Lemma L_disc." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S05.1", "sqrt_def", "zero_product", "distributive" ], "explanation": { "zh-Hant": "零乘積性質是雙向的，所以這兩個值確實是根，且只有它們；不要求 Δ 是完全平方（MKO 第三段）。", "en": "The zero-product property is two-way, so these two values are roots and the only ones; Δ need not be a perfect square (MKO third paragraph)." } }, "leaf_basis": "zero_product + sqrt_def + distributive + order_basics" }, { "id": "S06", "uses": [ "S05", "sqrt_def" ], "before_tex": "\\lambda=\\tfrac{7\\pm\\sqrt{9}}{2}", "after_tex": "\\lambda\\in\\{2,5\\},\\quad \\lambda^2-7\\lambda+10=(\\lambda-2)(\\lambda-5)", "relation": "equivalent", "rule": "sqrt_def", "conditions": [], "explanation": { "zh-Hant": "3≥0 且 3²=9，依 sqrt_def √9=3，所以兩根為 (7+3)/2=5 與 (7−3)/2=2。交叉核對：(λ−2)(λ−5)=λ²−7λ+10。依 S03，M 的互異實特徵值集合是 {2,5}。", "en": "3≥0 and 3²=9, so √9=3 by sqrt_def, and the two roots are (7+3)/2=5 and (7−3)/2=2. Cross-check: (λ−2)(λ−5)=λ²−7λ+10. By S03 the set of distinct real eigenvalues of M is {2,5}." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "sqrt_def", "zero_product", "numeric_evaluation" ], "explanation": { "zh-Hant": "分解式與零乘積性質雙向給出同樣兩根（N7 到 N10）。", "en": "The factorization and the zero-product property give the same two roots in both directions (N7 to N10)." } }, "leaf_basis": "sqrt_def + zero_product + numeric_evaluation" }, { "id": "S07", "uses": [ "S02", "S06", "G2" ], "before_tex": "(M-5I)v=\\begin{pmatrix}-1&1\\\\2&-2\\end{pmatrix}\\begin{pmatrix}x\\\\y\\end{pmatrix}=0", "after_tex": "K_5=\\{(t,t)^{\\mathsf T}\\mid t\\in\\mathbb{R}\\},\\ v=(1,1)^{\\mathsf T}:\\ Mv=(5,5)^{\\mathsf T}=5v", "relation": "implies", "rule": "exists_intro", "conditions": [], "explanation": { "zh-Hant": "(M−5I)v=(−x+y, 2x−2y)ᵀ=0 ⇔ y=x（第二式是第一式的 −2 倍）。所以 K₅ 是直線 {(t,t)ᵀ}，包含 t=0 的零向量。取 t=1 得非零向量 v=(1,1)ᵀ，Mv=(4+1, 2+3)ᵀ=(5,5)ᵀ=5v，所以 5 是特徵值、v 是特徵向量。", "en": "(M−5I)v=(−x+y, 2x−2y)ᵀ=0 ⇔ y=x (the second equation is −2 times the first). So K₅ is the line {(t,t)ᵀ}, which contains the zero vector at t=0. Taking t=1 gives the nonzero vector v=(1,1)ᵀ with Mv=(4+1, 2+3)ᵀ=(5,5)ᵀ=5v, so 5 is an eigenvalue and v is an eigenvector." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S02", "exists_intro", "numeric_evaluation" ], "explanation": { "zh-Hant": "特徵值只需找到一個非零向量；直接計算 Mv（N11、N12）獨立確認。", "en": "Being an eigenvalue needs only one nonzero vector; computing Mv directly (N11, N12) confirms it independently." } }, "leaf_basis": "exists_intro + matrix_product_def + numeric_evaluation" }, { "id": "S08", "uses": [ "S02", "S06", "G2" ], "before_tex": "(M-2I)v=\\begin{pmatrix}2&1\\\\2&1\\end{pmatrix}\\begin{pmatrix}x\\\\y\\end{pmatrix}=0", "after_tex": "K_2=\\{(t,-2t)^{\\mathsf T}\\mid t\\in\\mathbb{R}\\},\\ v=(1,-2)^{\\mathsf T}:\\ Mv=(2,-4)^{\\mathsf T}=2v", "relation": "implies", "rule": "exists_intro", "conditions": [], "explanation": { "zh-Hant": "(M−2I)v=(2x+y, 2x+y)ᵀ=0 ⇔ y=−2x。取 x=1 得 v=(1,−2)ᵀ≠0，Mv=(4−2, 2−6)ᵀ=(2,−4)ᵀ=2v，所以 2 是特徵值。", "en": "(M−2I)v=(2x+y, 2x+y)ᵀ=0 ⇔ y=−2x. Taking x=1 gives v=(1,−2)ᵀ≠0 with Mv=(4−2, 2−6)ᵀ=(2,−4)ᵀ=2v, so 2 is an eigenvalue." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S02", "exists_intro", "numeric_evaluation" ], "explanation": { "zh-Hant": "直接計算（N13 到 N15）確認；特徵向量不唯一，任何非零倍數也可以。", "en": "Direct computation (N13 to N15) confirms it; the eigenvector is not unique, and any nonzero multiple also works." } }, "leaf_basis": "exists_intro + matrix_product_def + numeric_evaluation" }, { "id": "S09", "uses": [ "S02", "S03", "S04", "G2" ], "before_tex": "(M-3I)v=\\begin{pmatrix}1&1\\\\2&0\\end{pmatrix}\\begin{pmatrix}x\\\\y\\end{pmatrix}=0", "after_tex": "K_3=\\{0\\},\\ 3\\ \\text{not an eigenvalue}", "relation": "implies", "rule": "eq_congruence", "conditions": [], "explanation": { "zh-Hant": "(M−3I)v=(x+y, 2x)ᵀ=0：由 2x=0 得 x=0，再由 x+y=0 得 y=0。所以 K₃={0}：零向量在核中，但沒有非零向量，3 不是特徵值。交叉核對：det(M−3I)=1·0−1·2=−2≠0（S03），特徵多項式在 3 的值 9−21+10=−2≠0（S04）。", "en": "(M−3I)v=(x+y, 2x)ᵀ=0: from 2x=0 we get x=0, and then x+y=0 gives y=0. So K₃={0}: the zero vector is in the kernel but no nonzero vector is, and 3 is not an eigenvalue. Cross-check: det(M−3I)=1·0−1·2=−2≠0 (S03), and the characteristic polynomial at 3 is 9−21+10=−2≠0 (S04)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S02", "S03", "eq_congruence", "mul_inverse_nonzero" ], "explanation": { "zh-Hant": "直接解方程與行列式判準兩種方法一致（N16、N17）。核中有零向量不構成特徵值。", "en": "Solving the equations directly and the determinant criterion agree (N16, N17). The zero vector in the kernel does not create an eigenvalue." } }, "leaf_basis": "eq_congruence + mul_inverse_nonzero + two_nonzero + numeric_evaluation" }, { "id": "S10", "uses": [ "S01", "S09", "G2" ], "before_tex": "(\\lambda,v)=(3,0)", "after_tex": "M0=0=3\\cdot 0\\ \\text{but}\\ v=0\\ \\Rightarrow\\ \\text{reject as eigenpair}", "relation": "implies", "rule": "mko_eig_def", "conditions": [], "explanation": { "zh-Hant": "候選特徵對 (3,0) 滿足 M0=0=3·0，所以 0∈K₃（核會員題接受）；但非零特徵向量慣例要求 v≠0，所以作為特徵對必須拒絕（MKO 假設 a2）。同理 (5,0) 也不是特徵對，即使 5 是特徵值。", "en": "The candidate eigenpair (3,0) satisfies M0=0=3·0, so 0∈K₃ (a kernel-membership question accepts it); but the nonzero-eigenvector convention requires v≠0, so it must be rejected as an eigenpair (MKO assumption a2). Likewise (5,0) is not an eigenpair, even though 5 is an eigenvalue." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S01", "mko_eig_def" ], "explanation": { "zh-Hant": "拒絕的理由只是 v=0；核會員與特徵對是兩個不同的問題。", "en": "The only reason for rejection is v=0; kernel membership and being an eigenpair are different questions." } }, "leaf_basis": "mko_eig_def + mul_zero" }, { "id": "S11", "uses": [ "S03", "S04", "S05", "G3" ], "before_tex": "R=\\begin{pmatrix}0&-1\\\\1&0\\end{pmatrix}", "after_tex": "\\lambda^2+1,\\ \\Delta=0^2-4\\cdot 1=-4<0\\ \\Rightarrow\\ \\text{no real eigenvalue}", "relation": "implies", "rule": "L_disc", "conditions": [], "explanation": { "zh-Hant": "tr(R)=0，det(R)=0·0−(−1)·1=1，特徵多項式 λ²+1，Δ=−4<0。依 L_disc（Δ<0 情形），沒有實根；直接看也是 λ²+1≥1>0。依 S03，R 沒有實特徵值：不是每個實矩陣都有實特徵值（MKO 假設 a1）。", "en": "tr(R)=0 and det(R)=0·0−(−1)·1=1, so the characteristic polynomial is λ²+1 and Δ=−4<0. By L_disc (the case Δ<0) there is no real root; directly, λ²+1≥1>0. By S03, R has no real eigenvalue: not every real matrix has a real eigenvalue (MKO assumption a1)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "L_disc", "square_nonneg" ], "explanation": { "zh-Hant": "本稿只在實數域工作，不討論複數特徵值。", "en": "This guide works over ℝ only and does not discuss complex eigenvalues." } }, "leaf_basis": "L_disc + square_nonneg + numeric_evaluation" }, { "id": "S12", "uses": [ "S03", "S04", "S05", "G3" ], "before_tex": "J=\\begin{pmatrix}3&1\\\\0&3\\end{pmatrix}", "after_tex": "(\\lambda-3)^2,\\ \\Delta=36-36=0\\ \\Rightarrow\\ E(J)=\\{3\\}", "relation": "implies", "rule": "L_disc", "conditions": [], "explanation": { "zh-Hant": "tr(J)=6，det(J)=9，特徵多項式 λ²−6λ+9=(λ−3)²，Δ=36−36=0。依 L_disc（Δ=0 情形），恰有一個互異實特徵值 3；它是二重根，但互異特徵值集合只列一次 {3}，不算成兩個。非零特徵向量例如 (1,0)ᵀ：J(1,0)ᵀ=(3,0)ᵀ。", "en": "tr(J)=6 and det(J)=9, so the characteristic polynomial is λ²−6λ+9=(λ−3)² and Δ=36−36=0. By L_disc (the case Δ=0) there is exactly one distinct real eigenvalue, 3; it is a double root, but the set of distinct eigenvalues lists it once, {3}, and does not count it twice. A nonzero eigenvector is for example (1,0)ᵀ: J(1,0)ᵀ=(3,0)ᵀ." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "L_disc", "numeric_evaluation" ], "explanation": { "zh-Hant": "只宣稱互異實特徵值的個數為 1；不建立代數重數或特徵基底。", "en": "Only the number of distinct real eigenvalues, 1, is claimed; no algebraic multiplicity or eigenbasis is established." } }, "leaf_basis": "L_disc + numeric_evaluation + exists_intro" }, { "id": "S13", "uses": [ "S03", "S05", "S06", "S07", "S08", "S09", "S10", "S11", "S12" ], "before_tex": "\\text{tasks of the goal}", "after_tex": "E(M)=\\{2,5\\},\\ K_3=\\{0\\},\\ (3,0)\\ \\text{rejected},\\ E(R)=\\varnothing,\\ E(J)=\\{3\\}", "relation": "implies", "rule": "and_intro_elim", "conditions": [], "explanation": { "zh-Hant": "合併：一般結論 S03、S05；M 的互異實特徵值為 {2,5}，2 的特徵向量可取 (1,−2)ᵀ，5 的特徵向量可取 (1,1)ᵀ；候選 3 的核 {0}，3 不是特徵值；(3,0) 不是特徵對；R 無實特徵值；J 只有一個互異實特徵值 3。", "en": "Combine: the general results S03 and S05; the distinct real eigenvalues of M are {2,5}, with (1,−2)ᵀ an eigenvector for 2 and (1,1)ᵀ an eigenvector for 5; the candidate 3 has kernel {0} and is not an eigenvalue; (3,0) is not an eigenpair; R has no real eigenvalue; J has exactly one distinct real eigenvalue, 3." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S03", "S05", "S06", "S07", "S08", "S09", "S10", "S11", "S12" ], "explanation": { "zh-Hant": "目標 G：題目要求的各項。足夠條件 H：S03 到 S12 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 S03 to S12; 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 2×2 實矩陣，λ 是實特徵值 ⇔ det(M−λI)=0 ⇔ λ²−tr(M)λ+det(M)=0；互異實特徵值個數在 Δ<0、Δ=0、Δ>0 時分別為 0、1、2。M=[[4,1],[2,3]] 的互異實特徵值為 {2,5}，特徵向量 (1,−2)ᵀ、(1,1)ᵀ；K₃={0}，3 不是特徵值，(3,0) 不是特徵對。R 無實特徵值（Δ=−4），J 只有 3（Δ=0）。", "en": "For 2×2 real matrices, λ is a real eigenvalue ⇔ det(M−λI)=0 ⇔ λ²−tr(M)λ+det(M)=0; the number of distinct real eigenvalues is 0, 1, 2 for Δ<0, Δ=0, Δ>0. The distinct real eigenvalues of M=[[4,1],[2,3]] are {2,5}, with eigenvectors (1,−2)ᵀ and (1,1)ᵀ; K₃={0}, so 3 is not an eigenvalue and (3,0) is not an eigenpair. R has no real eigenvalue (Δ=−4), and J has only 3 (Δ=0)." }, "negative_controls": [ { "claim": { "zh-Hant": "M·0=3·0，所以 3 是 M 的特徵值", "en": "M·0=3·0, so 3 is an eigenvalue of M" }, "expected": "reject", "reason": { "zh-Hant": "需要非零向量；K₃={0}，det(M−3I)=−2≠0。", "en": "A nonzero vector is needed; K₃={0} and det(M−3I)=−2≠0." } }, { "claim": { "zh-Hant": "K₃ 非空，所以 3 是特徵值", "en": "K₃ is nonempty, so 3 is an eigenvalue" }, "expected": "reject", "reason": { "zh-Hant": "K₃ 只含零向量；每個 λ 的核都含零向量。", "en": "K₃ contains only the zero vector; every kernel K_λ contains the zero vector." } }, { "claim": { "zh-Hant": "J=[[3,1],[0,3]] 有兩個互異實特徵值，因為特徵多項式是二次", "en": "J=[[3,1],[0,3]] has two distinct real eigenvalues because its characteristic polynomial is quadratic" }, "expected": "reject", "reason": { "zh-Hant": "Δ=0，唯一的根是 3；互異值只有一個。", "en": "Δ=0 and the only root is 3; there is one distinct value." } }, { "claim": { "zh-Hant": "每個 2×2 實矩陣都至少有一個實特徵值", "en": "Every 2×2 real matrix has at least one real eigenvalue" }, "expected": "reject", "reason": { "zh-Hant": "R=[[0,−1],[1,0]] 的特徵多項式 λ²+1 沒有實根。", "en": "The characteristic polynomial λ²+1 of R=[[0,−1],[1,0]] has no real root." } } ], "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": "square_nonneg", "kind": "prerequisite", "statement": { "zh-Hant": "實數平方非負：對所有實數 u，u²≥0，且 u²=0 ⇔ u=0。", "en": "Squares are nonnegative: for every real u, u²≥0, and u²=0 ⇔ u=0." } }, { "id": "sqrt_def", "kind": "prerequisite", "statement": { "zh-Hant": "主平方根：對 a≥0，√a 是唯一滿足 b≥0 且 b²=a 的實數 b。負數在實數中沒有平方根。", "en": "Principal square root: for a≥0, √a is the unique real b with b≥0 and b²=a. Negative numbers have no real square root." } }, { "id": "zero_product", "kind": "prerequisite", "statement": { "zh-Hant": "零乘積性質：在 ℤ、ℚ、ℝ、ℂ（或任何整域）中，uv=0 ⇔ u=0 或 v=0。", "en": "Zero-product property: in ℤ, ℚ, ℝ, ℂ (or any integral domain), uv=0 ⇔ u=0 or v=0." } }, { "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": "add_inverse", "kind": "prerequisite", "statement": { "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。", "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)." } }, { "id": "logic_chain", "kind": "prerequisite", "statement": { "zh-Hant": "推論的傳遞：由 P⇒Q 與 Q⇒R 得 P⇒R；⇔ 同理。", "en": "Chaining: from P⇒Q and Q⇒R infer P⇒R; likewise for ⇔." } }, { "id": "neg_mul", "kind": "prerequisite", "statement": { "zh-Hant": "負號與乘法：(−u)v=−(uv)，(−u)(−v)=uv。", "en": "Signs and multiplication: (−u)v=−(uv) and (−u)(−v)=uv." } }, { "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": "case_analysis", "kind": "prerequisite", "statement": { "zh-Hant": "分情況論證：若 P∨Q 成立，且 P⇒R、Q⇒R 都已證，則 R 成立。所有情況必須涵蓋全部可能。", "en": "Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility." } }, { "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": "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": "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": "frac_rules", "kind": "prerequisite", "statement": { "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。", "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0." } }, { "id": "two_nonzero", "kind": "prerequisite", "statement": { "zh-Hant": "實數中 0<1，故 0<1+1=2，所以 2≠0。", "en": "In the reals 0<1, hence 0<1+1=2, so 2≠0." } }, { "id": "order_basics", "kind": "prerequisite", "statement": { "zh-Hant": "實數的序：任兩數恰有 uv 之一；u0 推出 uwvw。", "en": "Order of the reals: exactly one of uv holds; u0 give uwvw." } }, { "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": "eq_congruence", "kind": "prerequisite", "statement": { "zh-Hant": "等式同用：若 u=v，對兩邊做同一個有定義的運算 f，得 f(u)=f(v)。只對一邊做運算不屬於此規則。", "en": "Congruence of equality: if u=v, applying the same defined operation f to both sides gives f(u)=f(v). Operating on one side only is not an instance of this rule." } }, { "id": "exists_intro", "kind": "prerequisite", "statement": { "zh-Hant": "∃引入：若找到 a∈D 使 P(a) 成立，則 ∃x∈D, P(x)。", "en": "∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x)." } }, { "id": "mul_inverse_nonzero", "kind": "prerequisite", "statement": { "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。", "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined." } }, { "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": "mko_eig_def", "kind": "cited", "source": { "mko_id": "mko-eigenvalue", "sha256": "5e1e1186120be48f1e86332bf1d679143e44074e2b3e9f59c5c219cb76a99e36", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "對2×2實矩陣 M，實數 λ 是特徵值當且僅當存在非零 v∈R² 使 Mv=λv。本輪採非零特徵向量慣例；候選 λ 的核 Kλ={v:Mv=λv} 仍包含零向量，即使 λ 並非特徵值。互異實特徵值是 λ²−tr(M)λ+det(M)=0 的互異實根。", "en": "For a 2×2 real matrix M, a real λ is an eigenvalue exactly when some nonzero v∈R² satisfies Mv=λv. This pulse uses the nonzero-eigenvector convention. The candidate-scalar kernel Kλ={v:Mv=λv} includes zero even when λ is not an eigenvalue. Distinct real eigenvalues are the distinct real roots of λ²−tr(M)λ+det(M)=0." } }, { "id": "mko_det_inv", "kind": "cited", "source": { "mko_id": "mko-determinant", "sha256": "d49050d70b58e0d9d2cab2f5ddc64133813f5a99418865b5941af4f1ec240093", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "對實矩陣 M=[[a,b],[c,d]]，本輪 D(M)=det(M)=ad−bc。M 在實數域可逆當且僅當 det(M)≠0；對同為2×2的實矩陣 A、B，det(AB)=det(A)det(B)，且 det(tM)=t²det(M)。", "en": "For M=[[a,b],[c,d]] over R, define D(M)=det(M)=ad−bc. M is invertible over R exactly when det(M)≠0. For 2×2 real matrices A,B, det(AB)=det(A)det(B), and det(tM)=t²det(M)." } }, { "id": "vec_ops", "kind": "prerequisite", "statement": { "zh-Hant": "ℝ² 的列向量逐分量相加與數乘；兩向量相等 ⇔ 兩個分量分別相等；零向量 0=(0,0)ᵀ，v≠0 ⇔ 至少一個分量非零。", "en": "Column vectors in ℝ² add and scale componentwise; two vectors are equal ⇔ both components are equal; the zero vector is 0=(0,0)ᵀ, and v≠0 ⇔ at least one component is nonzero." } }, { "id": "L_kernel_det", "kind": "lemma", "proved_in": [ "S03.1", "S03.2", "S03.3" ], "statement": { "zh-Hant": "對 2×2 實矩陣 N：存在非零 v 使 Nv=0 ⇔ det(N)=0。", "en": "For a 2×2 real matrix N: some nonzero v satisfies Nv=0 ⇔ det(N)=0." } }, { "id": "L_disc", "kind": "lemma", "proved_in": [ "S05.1", "S05.2", "S05.3", "S05.4" ], "statement": { "zh-Hant": "對實數 T、D，令 Δ=T²−4D。方程 λ²−Tλ+D=0 的互異實根個數：Δ<0 時為 0；Δ=0 時為 1（λ=T/2）；Δ>0 時為 2（λ=(T±√Δ)/2）。", "en": "For reals T and D let Δ=T²−4D. The number of distinct real roots of λ²−Tλ+D=0 is 0 if Δ<0, 1 if Δ=0 (λ=T/2), and 2 if Δ>0 (λ=(T±√Δ)/2)." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S04", "vars": [ "a", "b", "c", "d", "l" ], "degree": 2, "lhs": "(a-l)*(d-l)-b*c", "rhs": "l**2-(a+d)*l+(a*d-b*c)" }, { "id": "N2", "step": "S04", "lhs": "4+3", "rhs": "7" }, { "id": "N3", "step": "S04", "lhs": "4*3-1*2", "rhs": "10" }, { "id": "N4", "step": "S05.1", "vars": [ "l", "T", "D" ], "degree": 2, "lhs": "l**2-T*l+D", "rhs": "(l-T/2)**2-(T**2-4*D)/4" }, { "id": "N5", "step": "S05", "lhs": "7**2-4*10", "rhs": "9" }, { "id": "N6", "step": "S05", "lhs": "9", "relation": ">", "rhs": "0" }, { "id": "N7", "step": "S06", "lhs": "3**2", "rhs": "9" }, { "id": "N8", "step": "S06", "lhs": "(7+3)/2", "rhs": "5" }, { "id": "N9", "step": "S06", "lhs": "(7-3)/2", "rhs": "2" }, { "id": "N10", "step": "S06", "vars": [ "l" ], "degree": 2, "lhs": "(l-2)*(l-5)", "rhs": "l**2-7*l+10" }, { "id": "N11", "step": "S07", "lhs": "4*1+1*1", "rhs": "5" }, { "id": "N12", "step": "S07", "lhs": "2*1+3*1", "rhs": "5" }, { "id": "N13", "step": "S08", "lhs": "4*1+1*(-2)", "rhs": "2" }, { "id": "N14", "step": "S08", "lhs": "2*1+3*(-2)", "rhs": "-4" }, { "id": "N15", "step": "S08", "lhs": "2*(-2)", "rhs": "-4" }, { "id": "N16", "step": "S09", "lhs": "(4-3)*(3-3)-1*2", "rhs": "-2" }, { "id": "N17", "step": "S09", "lhs": "3**2-7*3+10", "rhs": "-2" }, { "id": "N18", "step": "S11", "lhs": "0**2-4*1", "rhs": "-4" }, { "id": "N19", "step": "S11", "lhs": "0*0-(-1)*1", "rhs": "1" }, { "id": "N20", "step": "S12", "lhs": "6**2-4*9", "rhs": "0" }, { "id": "N21", "step": "S12", "vars": [ "l" ], "degree": 2, "lhs": "(l-3)**2", "rhs": "l**2-6*l+9" }, { "id": "N22", "step": "S03.1", "vars": [ "p", "q", "r", "s" ], "degree": 2, "lhs": "r*q+s*(-p)", "rhs": "-(p*s-q*r)" }, { "id": "N23", "step": "S03.1", "vars": [ "p", "q" ], "degree": 2, "lhs": "p*q+q*(-p)", "rhs": "0" }, { "id": "N24", "step": "S05.4", "vars": [ "u", "k" ], "degree": 2, "lhs": "(u-k)*(u+k)", "rhs": "u**2-k**2" }, { "id": "N25", "step": "S12", "lhs": "3*1+1*0", "rhs": "3" } ] }, "presentation": { "md_scope": { "zh-Hant": "只在 MKO 範圍內：實數域、2×2 方陣、非零特徵向量慣例。完整證明核非零 ⇔ 行列式為零、特徵多項式公式與判別式分類；行列式非零 ⇒ 可逆引用行列式條目。不處理代數重數、特徵基底、對角化、Jordan 結構或複數譜，也不做 Lean 的 HasEigenvector 橋接；不改動 canonical MKO 或英文 edition。", "en": "Stays inside the MKO's scope: the real field, 2×2 matrices, the nonzero-eigenvector convention. It fully proves 'nonzero kernel ⇔ determinant zero', the characteristic-polynomial formula and the discriminant classification; 'nonzero determinant ⇒ invertible' is cited from the determinant entry. It does not treat algebraic multiplicity, eigenbases, diagonalization, Jordan structure or the complex spectrum, and does not build the Lean HasEigenvector bridge; it does not change the canonical MKO or its English edition." }, "md_symbols": [ [ "\\lambda", "候選實數特徵值", "a candidate real eigenvalue" ], [ "K_\\lambda=\\{v\\mid Mv=\\lambda v\\}", "候選 λ 的核，恆含零向量", "the kernel for a candidate λ, always containing the zero vector" ], [ "\\operatorname{tr}(M)", "跡 a+d", "the trace a+d" ], [ "\\Delta", "判別式 tr(M)²−4det(M)", "the discriminant tr(M)²−4det(M)" ], [ "N=\\begin{pmatrix}p&q\\\\r&s\\end{pmatrix}", "引理中的一般 2×2 實矩陣", "a general 2×2 real matrix in the lemma" ] ], "md_error_controls": [ { "title": "零向量不是特徵向量 / The zero vector is not an eigenvector", "zh": "MKO 的第一個誤解是把零向量當特徵向量。M0=λ0 對每個 λ 都成立；若允許 v=0，每個實數都會是「特徵值」。所以 (3,0) 必須拒絕（S10），而 3 確實不是特徵值（S09）。", "en": "The first misconception identified in the MKO is treating the zero vector as an eigenvector. M0=λ0 holds for every λ; if v=0 were allowed, every real number would be an 'eigenvalue'. So (3,0) must be rejected (S10), and 3 is indeed not an eigenvalue (S09)." }, { "title": "核含零不等於特徵值存在 / A kernel containing zero does not make an eigenvalue", "zh": "K₃={0} 非空，但只含零向量；特徵值要求核中有非零向量。核會員題接受零向量，特徵對題拒絕零向量，兩類問題要分開（MKO 假設 a2）。", "en": "K₃={0} is nonempty, but it contains only the zero vector; an eigenvalue needs a nonzero vector in the kernel. Kernel-membership questions accept the zero vector and eigenpair questions reject it, and the two kinds of question must be kept apart (MKO assumption a2)." }, { "title": "二重根只算一個互異值 / A double root counts as one distinct value", "zh": "J 的特徵多項式 (λ−3)² 有二重根 3，但互異實特徵值只有一個（S12）。把它按重數算成兩個互異值是 MKO 列出的誤解。", "en": "The characteristic polynomial (λ−3)² of J has the double root 3, but there is only one distinct real eigenvalue (S12). Counting it as two distinct values by multiplicity is a misconception the MKO lists." } ], "md_open_items": [ { "zh-Hant": "S03.3 引用行列式條目的「det≠0 ⇒ 在 ℝ 上可逆」與矩陣乘法結合律；它們在本稿中未重新證明。", "en": "S03.3 cites 'det≠0 ⇒ invertible over ℝ' from the determinant entry and uses associativity of matrix multiplication; these are not re-proved here." }, { "zh-Hant": "代數重數、特徵基底、一般對角化與複數譜不在本稿範圍；MKO 的 formal next（HasEigenvector／eigenspace 綁定）未做。", "en": "Algebraic multiplicity, eigenbases, general diagonalization and the complex spectrum are outside this guide; the MKO's formal next step (binding HasEigenvector and eigenspace) is not done." } ] }, "revision": { "id": "production-20261019-adopted-96", "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "adopted_source_sha256": "27f8549732baa2c4dbec3325cb1a0a6e4dda2523f6dd6b29511686514a143988", "original_source_sha256": "27f8549732baa2c4dbec3325cb1a0a6e4dda2523f6dd6b29511686514a143988", "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." } }