Eigenvalues and Eigenvectors — Complete Guide

§01What Is an Eigenvector?

Most vectors change direction when multiplied by a matrix. Eigenvectors are the rare exceptions, they only scale.

When you multiply a matrix A by most vectors v, the output Av points in a completely different direction. But certain special vectors satisfy a remarkable property: the matrix only stretches or compresses them, never tilts them.

Eigenvalue Equation

Av = λv (v ≠ 0)

v is the eigenvector; λ (a scalar) is the eigenvalue.

Geometrically: Av lands on the same line through the origin as v. The eigenvalue λ records the scale factor: λ = 3 triples the vector, λ = −1 flips it, λ = 0 collapses it to zero.

Why "eigen"? From the German for "own" or "characteristic". Eigenvectors are the matrix's own special directions.

§02Finding Eigenvalues — The Characteristic Polynomial

Rewrite Av = λv as a homogeneous system and demand a non-trivial solution.

Rearranging Av = λv:

Key derivation

(A − λI)v = 0

For a non-zero solution v to exist, the matrix (A − λI) must be singular:

Characteristic Equation

det(A − λI) = 0

Expanding det(A − λI) produces a polynomial in λ, the characteristic polynomial p(λ). Its roots are the eigenvalues.

🔑 Algorithm — Finding Eigenvalues

Form A − λI
Subtract λ from every diagonal entry of A.
Compute det(A − λI)
Expand to get a polynomial p(λ) of degree n.
Solve p(λ) = 0
The roots λ₁, λ₂, … are the eigenvalues.

The 2×2 Shortcut

2×2 characteristic polynomial

p(λ) = λ² − tr(A)·λ + det(A)

tr(A) = sum of diagonal entries.

§03Finding Eigenvectors — Eigenspaces

Once you know λ, substitute back and solve (A − λI)v = 0.

For each eigenvalue λ, the set of all solutions to (A − λI)v = 0 is the eigenspace E(λ):

Eigenspace

E(λ) = Null(A − λI) = { v : (A − λI)v = 0 }

Non-zero vectors in E(λ) are the eigenvectors. Note: scalar multiples of eigenvectors are also eigenvectors: they are never unique.

Algebraic vs Geometric Multiplicity

Multiplicity	Definition	Symbol
Algebraic (AM)	Power of (λ − λ₀) in p(λ) — how many times λ₀ is a root.	a(λ₀)
Geometric (GM)	dim E(λ₀) = dim Null(A − λ₀I).	g(λ₀)

Fundamental Inequality

For every eigenvalue λ₀: 1 ≤ g(λ₀) ≤ a(λ₀).
When GM < AM for any eigenvalue, the matrix is defective (not diagonalisable).

§04Key Properties & Special Cases

Trace, Determinant, and Eigenvalues

Two golden identities

tr(A) = λ₁ + λ₂ + ··· + λₙ det(A) = λ₁ · λ₂ · ··· · λₙ

These are indispensable verification tools: after computing eigenvalues, check that their sum equals the trace and their product equals the determinant.

Special Matrix Types

Matrix Type	Eigenvalue Guarantee
Symmetric (Aᵀ = A)	All eigenvalues real; eigenvectors for distinct values are orthogonal.
Upper/Lower Triangular	Eigenvalues are the diagonal entries.
Orthogonal (AᵀA = I)	All eigenvalues satisfy \|λ\| = 1.
Positive Definite	All eigenvalues strictly positive.
Nilpotent (Aᵏ = 0)	All eigenvalues are 0.

λ = 0 as an eigenvalue Zero is an eigenvalue if and only if A is singular, since det(A − 0·I) = det(A) = 0.

§05Ten Worked Examples

From simple 2×2 cases to repeated roots and complex eigenvalues, full procedures throughout.

Example 012×2 matrix with distinct eigenvalues

Find the eigenvalues and eigenvectors of A = [[4,1],[2,3]].

Step 1 — Characteristic polynomial

tr(A) = 7, det(A) = 10 p(λ) = λ² − 7λ + 10 = (λ−2)(λ−5)

Step 2 — Eigenvalues

λ₁ = 2, λ₂ = 5

Step 3 — Eigenvector for λ = 2

(A−2I)v=0: 2v₁+v₂=0 ⟹ v = t(1, −2)

Step 4 — Eigenvector for λ = 5

(A−5I)v=0: −v₁+v₂=0 ⟹ v = t(1, 1)

λ=2: v=(1,−2); λ=5: v=(1,1)

Example 02Repeated eigenvalue — defective matrix

Find eigenvalues and eigenvectors of A = [[3,1],[0,3]].

Characteristic polynomial

p(λ) = (3−λ)² = 0 ⟹ λ = 3, AM = 2

Eigenspace for λ = 3

(A−3I)v=0: [0,1;0,0]v=0 ⟹ v₂=0 v = t(1,0) — only one free variable, GM = 1

GM = 1 < AM = 2, so the matrix is defective: not diagonalisable. This is a Jordan block.

λ = 3 (AM=2, GM=1) — defective

Example 03Repeated eigenvalue — NOT defective

Find eigenvectors of A = 5I₂ (2×2 scalar matrix).

Characteristic polynomial

p(λ) = (5−λ)² = 0 ⟹ λ = 5, AM = 2

Eigenspace

(A−5I)v = 0v = 0 ⟹ every v ∈ ℝ² is an eigenvector GM = 2 = AM ✓ fully diagonalisable

Every non-zero vector is an eigenvector with eigenvalue 5.

Example 04Triangular matrix — eigenvalues by inspection

Find eigenvalues of the upper triangular matrix A = [[2,5,−3],[0,−1,4],[0,0,7]].

Theorem

For triangular matrices: eigenvalues = diagonal entries.

Verification: tr(A) = 2+(−1)+7 = 8; sum of eigenvalues = 2−1+7 = 8 ✓

λ₁ = 2, λ₂ = −1, λ₃ = 7

Example 05Rotation matrix — complex eigenvalues

Find eigenvalues of the rotation matrix R = [[0,−1],[1,0]].

Characteristic polynomial

p(λ) = λ² + 1 = 0 ⟹ λ = ±i

Over ℝ there are no eigenvectors, rotation moves every non-zero vector. Over ℂ the eigenvalues form the conjugate pair ±i, reflecting pure 90° rotation.

λ = i and λ = −i (complex conjugate pair)

Example 06Full 3×3 procedure

Find all eigenvalues and eigenvectors of A = [[1,2,0],[0,3,0],[2,−4,2]].

Characteristic polynomial

det(A−λI) = (1−λ)(3−λ)(2−λ) = 0

Eigenvalues

λ₁=1, λ₂=2, λ₃=3

Eigenvectors (row reduce (A−λI))

λ=1: v=(1,0,−1) | λ=2: v=(0,0,1) | λ=3: v=(1,1,−1)

Three distinct eigenvalues → three independent eigenvectors → diagonalisable

Example 07Eigenvalues from trace and determinant

A 2×2 matrix has tr(A) = 5 and det(A) = 6. Find the eigenvalues.

Set up system

λ₁+λ₂ = 5 and λ₁λ₂ = 6 t² − 5t + 6 = (t−2)(t−3) = 0

λ₁ = 2, λ₂ = 3

Example 08Eigenvalues of A², A⁻¹, and Aᵏ

If Av = λv, derive the eigenvalues of A², A⁻¹, and Aᵏ.

For A²

A²v = A(Av) = A(λv) = λ(Av) = λ²v

For A⁻¹ (λ ≠ 0)

Av = λv ⟹ v = λA⁻¹v ⟹ A⁻¹v = (1/λ)v

Aᵏ has eigenvalue λᵏ; A⁻¹ has eigenvalue λ⁻¹

Example 09Projection matrix — geometric argument

Let P project onto a 1-dimensional subspace L ⊂ ℝ². Determine its eigenvalues without computing.

Vectors along L

Pv = v ⟹ λ = 1

Vectors perpendicular to L

Pv = 0 ⟹ λ = 0

Check: det(P) = 1·0 = 0 (P is singular) ✓

λ = 1 (along L) and λ = 0 (perpendicular to L)

Example 10Symmetric matrix — orthogonal eigenvectors

Find eigenvectors of the symmetric matrix A = [[2,1],[1,2]] and verify orthogonality.

Characteristic polynomial

p(λ) = λ²−4λ+3 = (λ−1)(λ−3) λ₁=1, λ₂=3

Eigenvectors

λ=1: v₁=(1,−1) | λ=3: v₂=(1,1)

Orthogonality check

v₁ · v₂ = 1·1 + (−1)·1 = 0 ✓

Spectral theorem: symmetric matrices always yield orthogonal eigenvectors for distinct eigenvalues.

§06Four Common Mistakes

Mistake 1: Forgetting v ≠ 0The zero vector satisfies Av=λv for any λ. Always state v ≠ 0 — eigenvectors must be non-zero.

Mistake 2: Confusing AM and GMAM is about the characteristic polynomial; GM is about the null space dimension. They can differ.

Mistake 3: Reporting only one eigenvector when GM > 1Row reduce fully and give an entire basis for the eigenspace, not just a single vector.

Mistake 4: Thinking A and Aᵀ have different eigenvaluesThey share the same characteristic polynomial and thus the same eigenvalues — only their eigenvectors differ (left vs right).

§07Practice Quiz — 10 Questions

Test yourself before moving on to Diagonalisation.

Score: 0 / 10

Question 01 — Multiple Choice

To find eigenvalues, which equation must hold?

det(A + λI) = 0
det(A − λI) = 0
tr(A − λI) = 0
Av = 0

Question 02 — Fill In

For A = [[5,0],[0,3]], the larger eigenvalue is .

Diagonal matrices have eigenvalues equal to their diagonal entries.

Question 03 — Multiple Choice

A 3×3 matrix has characteristic polynomial (λ−2)²(λ+1). The algebraic multiplicity of λ = 2 is:

Question 04 — Fill In

A 2×2 matrix has tr(A) = 7, det(A) = 12. The smaller eigenvalue is .

Solve λ² − 7λ + 12 = 0 by factoring.

Question 05 — Multiple Choice

If A has eigenvalue λ = 4, what is the eigenvalue of A³ for the same eigenvector?

Question 06 — Multiple Choice

A matrix has eigenvalue λ = 0. This means the matrix is:

Invertible
Singular (non-invertible)
The identity
Positive definite

Question 07 — Fill In

A 3×3 matrix has eigenvalues 1, 2, 3. Its determinant is .

det(A) = product of all eigenvalues.

Question 08 — Multiple Choice

A matrix is defective when, for some eigenvalue:

GM = AM
GM < AM
GM > AM
The eigenvalue is negative

Question 09 — Fill In

A 3×3 matrix has eigenvalues 2, 2, 5. Its trace is .

tr(A) = sum of all eigenvalues = 2+2+5.

Question 10 — Multiple Choice

For a symmetric matrix, eigenvectors for distinct eigenvalues are always:

Parallel
Orthogonal
Equal in length
Complex