Goal and scope
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).
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.
Main derivation
S01 · definition
Before
\lambda\in\mathbb{R}After
\lambda\ \text{eigenvalue}\iff\exists v\neq 0:\ Mv=\lambda v;\quad K_\lambda=\{v\mid Mv=\lambda v\}Conditions: none
Rule: mko_eig_def · Uses: G1
Leaf basis: mko_eig_def + def_unfold
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.
S02 · equivalent
Before
Mv=\lambda v
After
(M-\lambda I)v=0
Conditions: none
Rule: distributive · Uses: S01, matrix_product_def, vec_ops
Leaf basis: distributive + matrix_product_def + vec_ops + add_inverse
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.
S03 · equivalent
Before
\lambda\ \text{eigenvalue}After
\det(M-\lambda I)=0
Conditions: none
Rule: L_kernel_det · Uses: S01, S02, L_kernel_det
Leaf basis: L_kernel_det + logic_chain
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.
S04 · equivalent
Before
\det(M-\lambda I)=(a-\lambda)(d-\lambda)-bc
After
\lambda^2-(a+d)\lambda+(ad-bc)=\lambda^2-\operatorname{tr}(M)\lambda+\det(M)Conditions: none
Rule: distributive · Uses: S03, G1
Leaf basis: distributive + neg_mul + add_comm + mko_det_inv
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.
S05 · theorem_application
Before
\Delta=\operatorname{tr}(M)^2-4\det(M)=49-40After
\Delta=9>0\ \Rightarrow\ \text{two distinct real eigenvalues}Conditions: none
Rule: L_disc · Uses: S04, L_disc
Leaf basis: L_disc + numeric_evaluation
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.
S06 · equivalent
Before
\lambda=\tfrac{7\pm\sqrt{9}}{2}After
\lambda\in\{2,5\},\quad \lambda^2-7\lambda+10=(\lambda-2)(\lambda-5)Conditions: none
Rule: sqrt_def · Uses: S05, sqrt_def
Leaf basis: sqrt_def + zero_product + numeric_evaluation
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}.
S07 · implies
Before
(M-5I)v=\begin{pmatrix}-1&1\\2&-2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=0After
K_5=\{(t,t)^{\mathsf T}\mid t\in\mathbb{R}\},\ v=(1,1)^{\mathsf T}:\ Mv=(5,5)^{\mathsf T}=5vConditions: none
Rule: exists_intro · Uses: S02, S06, G2
Leaf basis: exists_intro + matrix_product_def + numeric_evaluation
(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.
S08 · implies
Before
(M-2I)v=\begin{pmatrix}2&1\\2&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=0After
K_2=\{(t,-2t)^{\mathsf T}\mid t\in\mathbb{R}\},\ v=(1,-2)^{\mathsf T}:\ Mv=(2,-4)^{\mathsf T}=2vConditions: none
Rule: exists_intro · Uses: S02, S06, G2
Leaf basis: exists_intro + matrix_product_def + numeric_evaluation
(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.
S09 · implies
Before
(M-3I)v=\begin{pmatrix}1&1\\2&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=0After
K_3=\{0\},\ 3\ \text{not an eigenvalue}Conditions: none
Rule: eq_congruence · Uses: S02, S03, S04, G2
Leaf basis: eq_congruence + mul_inverse_nonzero + two_nonzero + numeric_evaluation
(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).
S10 · implies
Before
(\lambda,v)=(3,0)
After
M0=0=3\cdot 0\ \text{but}\ v=0\ \Rightarrow\ \text{reject as eigenpair}Conditions: none
Rule: mko_eig_def · Uses: S01, S09, G2
Leaf basis: mko_eig_def + mul_zero
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.
S11 · implies
Before
R=\begin{pmatrix}0&-1\\1&0\end{pmatrix}After
\lambda^2+1,\ \Delta=0^2-4\cdot 1=-4<0\ \Rightarrow\ \text{no real eigenvalue}Conditions: none
Rule: L_disc · Uses: S03, S04, S05, G3
Leaf basis: L_disc + square_nonneg + numeric_evaluation
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).
S12 · implies
Before
J=\begin{pmatrix}3&1\\0&3\end{pmatrix}After
(\lambda-3)^2,\ \Delta=36-36=0\ \Rightarrow\ E(J)=\{3\}Conditions: none
Rule: L_disc · Uses: S03, S04, S05, G3
Leaf basis: L_disc + numeric_evaluation + exists_intro
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.
S13 · implies
Before
\text{tasks of the goal}After
E(M)=\{2,5\},\ K_3=\{0\},\ (3,0)\ \text{rejected},\ E(R)=\varnothing,\ E(J)=\{3\}Conditions: none
Rule: and_intro_elim · Uses: S03, S05, S06, S07, S08, S09, S10, S11, S12
Leaf basis: and_intro_elim
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.
Substeps and lemmas
S03.1 · implies
Before
\det N=ps-qr=0,\ (p,q)\neq(0,0)
After
v=(q,-p)^{\mathsf T}\neq 0,\ Nv=(pq-qp,\ rq-sp)^{\mathsf T}=(0,\ -\det N)^{\mathsf T}=0Conditions: \det N=0; (p,q)\neq(0,0)
Rule: matrix_product_def · Uses: matrix_product_def, vec_ops
Leaf basis: matrix_product_def + vec_ops + neg_mul + mul_comm
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.
S03.2 · implies
Before
p=q=0
After
v=(s,-r)^{\mathsf T}\ \text{if}\ (r,s)\neq(0,0),\ \text{else}\ v=(1,0)^{\mathsf T};\ \ v\neq 0,\ Nv=0Conditions: p=q=0
Rule: case_analysis · Uses: S03.1, matrix_product_def, vec_ops
Leaf basis: case_analysis + matrix_product_def + mul_zero + vec_ops
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.
S03.3 · implies
Before
\det N\neq 0,\ Nv=0
After
v=N^{-1}(Nv)=N^{-1}0=0Conditions: \det N\neq 0
Rule: contrapositive · Uses: mko_det_inv, vec_ops
Leaf basis: mko_det_inv + contrapositive + mul_zero
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.
S05.1 · equivalent
Before
\lambda^2-T\lambda+D
After
\left(\lambda-\tfrac{T}{2}\right)^2-\tfrac{\Delta}{4},\quad \Delta=T^2-4DConditions: none
Rule: distributive · Uses: distributive, frac_rules
Leaf basis: distributive + frac_rules + two_nonzero
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.
S05.2 · implies
Before
\Delta<0
After
\left(\lambda-\tfrac{T}{2}\right)^2-\tfrac{\Delta}{4}>0\ \text{for all }\lambda:\ \text{no real root}Conditions: \Delta<0
Rule: square_nonneg · Uses: S05.1, square_nonneg
Leaf basis: square_nonneg + order_basics + forall_intro
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.
S05.3 · implies
Before
\Delta=0
After
\left(\lambda-\tfrac{T}{2}\right)^2=0\iff\lambda=\tfrac{T}{2}:\ \text{exactly one root}Conditions: \Delta=0
Rule: square_nonneg · Uses: S05.1, square_nonneg
Leaf basis: square_nonneg + eq_congruence
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).
S05.4 · implies
Before
\Delta>0
After
\lambda=\tfrac{T+\sqrt{\Delta}}{2}\ \lor\ \lambda=\tfrac{T-\sqrt{\Delta}}{2},\ \text{two distinct roots}Conditions: \Delta>0
Rule: zero_product · Uses: S05.1, sqrt_def, zero_product
Leaf basis: zero_product + sqrt_def + distributive + order_basics
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.