Eigenvalues of 2×2 real matrices: characteristic polynomial, discriminant and nonzero eigenvectors

On this page
  1. Goal and scope
  2. Givens and symbols
  3. Rules and sources
  4. Main derivation
  5. Substeps and lemmas
  6. Reverse audits
  7. Error controls and negative controls
  8. Result and limitations
  9. Verification and source binding

← Guide index · Original MKO

derivation-mko-eigenvalue-01 · production-20261019-adopted-96

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.

Givens and symbols

  • matrix_product_def
  • distributive
  • square_nonneg
  • sqrt_def
  • zero_product
  • numeric_evaluation
  • G1 — 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).
  • G2 — Example: M=[[4,1],[2,3]]; candidate value λ=3; candidate eigenpair (3,0).
  • G3 — Contrasts: R=[[0,−1],[1,0]] and J=[[3,1],[0,3]].
  • \lambda — a candidate real eigenvalue
  • K_\lambda=\{v\mid Mv=\lambda v\} — the kernel for a candidate λ, always containing the zero vector
  • \operatorname{tr}(M) — the trace a+d
  • \Delta — the discriminant tr(M)²−4det(M)
  • N=\begin{pmatrix}p&q\\r&s\end{pmatrix} — a general 2×2 real matrix in the lemma

Rules and sources

  • matrix_product_def · prerequisite — 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.
  • distributive · prerequisite — Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu.
  • square_nonneg · prerequisite — Squares are nonnegative: for every real u, u²≥0, and u²=0 ⇔ u=0.
  • sqrt_def · prerequisite — 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.
  • zero_product · prerequisite — Zero-product property: in ℤ, ℚ, ℝ, ℂ (or any integral domain), uv=0 ⇔ u=0 or v=0.
  • numeric_evaluation · prerequisite — Exact evaluation of an explicit integer or rational expression; independently recomputable.
  • def_unfold · prerequisite — Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true.
  • add_inverse · prerequisite — Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v).
  • logic_chain · prerequisite — Chaining: from P⇒Q and Q⇒R infer P⇒R; likewise for ⇔.
  • neg_mul · prerequisite — Signs and multiplication: (−u)v=−(uv) and (−u)(−v)=uv.
  • mul_comm · prerequisite — Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ).
  • case_analysis · prerequisite — 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.
  • mul_zero · prerequisite — 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.
  • contrapositive · prerequisite — Contrapositive: P⇒Q is equivalent to ¬Q⇒¬P. The converse Q⇒P is in general not equivalent to P⇒Q.
  • add_comm · prerequisite — Commutativity of addition: u+v=v+u.
  • frac_rules · prerequisite — 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.
  • two_nonzero · prerequisite — In the reals 0<1, hence 0<1+1=2, so 2≠0.
  • order_basics · prerequisite — Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw.
  • forall_intro · prerequisite — ∀-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.
  • eq_congruence · prerequisite — 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.
  • exists_intro · prerequisite — ∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x).
  • mul_inverse_nonzero · prerequisite — 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.
  • and_intro_elim · prerequisite — Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q.
  • mko_eig_def · cited — 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. · mko-eigenvalue
  • mko_det_inv · cited — 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). · mko-determinant
  • vec_ops · prerequisite — 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.
  • L_kernel_det · lemma — For a 2×2 real matrix N: some nonzero v satisfies Nv=0 ⇔ det(N)=0.
  • L_disc · lemma — 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).

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)=49-40

After

\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}=0

After

K_5=\{(t,t)^{\mathsf T}\mid t\in\mathbb{R}\},\ v=(1,1)^{\mathsf T}:\ Mv=(5,5)^{\mathsf T}=5v

Conditions: 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}=0

After

K_2=\{(t,-2t)^{\mathsf T}\mid t\in\mathbb{R}\},\ v=(1,-2)^{\mathsf T}:\ Mv=(2,-4)^{\mathsf T}=2v

Conditions: 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}=0

After

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}=0

Conditions: \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=0

Conditions: 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=0

Conditions: \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^2-4D

Conditions: 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.

Reverse audits

  • S13 · sufficient_goal_reduction · S03, S05, S06, S07, S08, S09, S10, S11, S12 — 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.
  • S12 · premise_audit · S03, L_disc, numeric_evaluation — Only the number of distinct real eigenvalues, 1, is claimed; no algebraic multiplicity or eigenbasis is established.
  • S11 · premise_audit · S03, L_disc, square_nonneg — This guide works over ℝ only and does not discuss complex eigenvalues.
  • S10 · premise_audit · S01, mko_eig_def — The only reason for rejection is v=0; kernel membership and being an eigenpair are different questions.
  • S09 · premise_audit · S02, S03, eq_congruence, mul_inverse_nonzero — Solving the equations directly and the determinant criterion agree (N16, N17). The zero vector in the kernel does not create an eigenvalue.
  • S08 · premise_audit · S02, exists_intro, numeric_evaluation — Direct computation (N13 to N15) confirms it; the eigenvector is not unique, and any nonzero multiple also works.
  • S07 · premise_audit · S02, exists_intro, numeric_evaluation — Being an eigenvalue needs only one nonzero vector; computing Mv directly (N11, N12) confirms it independently.
  • S06 · inverse_equivalence · sqrt_def, zero_product, numeric_evaluation — The factorization and the zero-product property give the same two roots in both directions (N7 to N10).
  • S05.4 · premise_audit · S05.1, sqrt_def, zero_product, distributive — 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).
  • S05.3 · premise_audit · S05.1, square_nonneg — u²=0 ⇔ u=0 holds both ways, so the root exists and is unique.
  • S05.2 · premise_audit · S05.1, square_nonneg, order_basics — The conclusion is for every real λ; the argument makes no assumption about λ.
  • S05.1 · inverse_equivalence · distributive, frac_rules, two_nonzero — An identity for all reals λ, T, D (N4), two-way.
  • S05 · premise_audit · S03, S04, L_disc — It needs the equivalence of S03 to turn the eigenvalue question into a root question; the lemma counts distinct roots, not multiplicities.
  • S04 · inverse_equivalence · distributive, neg_mul, add_comm — A polynomial identity (N1), two-way.
  • S03.3 · premise_audit · mko_det_inv, contrapositive, vec_ops — 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.
  • S03.2 · premise_audit · S03.1, case_analysis, matrix_product_def — The three cases (first row nonzero; first row zero but second row nonzero; all zero) exhaust every possibility.
  • S03.1 · premise_audit · matrix_product_def, vec_ops, neg_mul — The second component vanishes exactly because det N=0; the first row being nonzero guarantees v≠0.
  • S03 · inverse_equivalence · L_kernel_det, S02 — Both directions of the lemma are proved (S03.1 and S03.2 for ⇐, S03.3 for ⇒).
  • S02 · inverse_equivalence · distributive, vec_ops, add_inverse — Subtracting λv from both sides and adding it back undo each other, so it is two-way.
  • S01 · premise_audit · G1, mko_eig_def — This step only states the definition; it derives no new fact.

Error controls and negative controls

The zero vector is not an eigenvector

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).

A kernel containing zero does not make an eigenvalue

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).

A double root counts as one distinct value

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.

  • M·0=3·0, so 3 is an eigenvalue of M → reject — A nonzero vector is needed; K₃={0} and det(M−3I)=−2≠0.
  • K₃ is nonempty, so 3 is an eigenvalue → reject — K₃ contains only the zero vector; every kernel K_λ contains the zero vector.
  • J=[[3,1],[0,3]] has two distinct real eigenvalues because its characteristic polynomial is quadratic → reject — Δ=0 and the only root is 3; there is one distinct value.
  • Every 2×2 real matrix has at least one real eigenvalue → reject — The characteristic polynomial λ²+1 of R=[[0,−1],[1,0]] has no real root.

Result and limitations

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).

  • S03.3 cites 'det≠0 ⇒ invertible over ℝ' from the determinant entry and uses associativity of matrix multiplication; these are not re-proved here.
  • 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.

Verification and source binding

25 declared checks for this guide; only the listed expressions are covered.

  • source commit: b28a5edbd46279b1db2a3f3f48bb3f447e18899f
  • adopted source SHA-256: 27f8549732baa2c4dbec3325cb1a0a6e4dda2523f6dd6b29511686514a143988
  • original source SHA-256: 27f8549732baa2c4dbec3325cb1a0a6e4dda2523f6dd6b29511686514a143988
  • Lean: NOT_RUN · independent AI: NOT_MEASURED

Same-source full JSON (formulas, ASTs, controls and revision log)