MGU Mathematics Series — Linear Algebra

Part IV: Orthogonality, Eigenvalues, and Diagonalization

Diagonalization · Matrix Powers · Discrete Systems · Markov Chains

Chapter 13 of 14

MGU Mathematics Series · Linear Algebra · Library Textbook Edition

Chapter 13:
Diagonalization and
Discrete Dynamical Systems

Eigenvalue decomposition, matrix powers, recurrence systems, Markov chains, and long-term behavior — how eigenvectors turn hard problems into easy ones

■ Part IV — Orthogonality, Eigenvalues, and Diagonalization ■

■ 14 Sections • Chapter 13 of 14 ■

Requires Chapter 12 (Eigenvalues and Eigenvectors) · Prepares for Chapter 14 and Differential Equations

Chapter 13 — Front Matter

Purpose and Overview

This chapter brings together eigenvalues, eigenvectors, and matrix structure to address one of the most powerful ideas in linear algebra: diagonalization. When a matrix can be diagonalized, repeated matrix operations become dramatically simpler. Instead of computing large matrix products, diagonalization reduces the repeated action to simple scaling in each eigenvector direction.

The chapter then applies this idea to discrete dynamical systems — mathematical models that describe how a state evolves from one step to the next. These include population models, economic models, network propagation models, and Markov chains. Each of these systems can be analyzed with eigenvalues and diagonalization, revealing long-term behavior without computing hundreds of matrix powers by hand.

This chapter is the culmination of Part IV. It draws on eigenvalues and eigenvectors from Chapter 12, on vector spaces and bases from Chapters 8 and 9, on matrix operations from Chapter 4, and on the invertibility ideas from Chapter 5. Students who understand diagonalization will be well prepared for differential equations, numerical methods, data science, machine learning, and advanced applied mathematics.

Opening Question

A wildlife biologist is studying a population of rabbits and foxes. Each year, the rabbit population and the fox population influence each other in ways that can be described by a system of linear equations. The biologist wants to know: after 50 years, what will the population look like? After 100 years? Will the populations grow, shrink, or stabilize?

Computing A⁵⁰ or A¹⁰⁰ directly would require an enormous amount of work. But if the biologist knows the eigenvalues and eigenvectors of A, she can find A⁵⁰ and A¹⁰⁰ with only a few steps — and understand the long-term behavior without computing a single large matrix product. That is what diagonalization makes possible.

14Sections

10Objectives

19Key Terms

12+Worked Examples

Learning Objectives

1Explain what it means for a matrix to be diagonalizable.
2Determine whether a square matrix is diagonalizable.
3Construct the diagonalization A = PDP⁻¹ when it exists.
4Use diagonalization to compute powers of a matrix.
5Set up and analyze a discrete dynamical system using matrix notation.
6Identify the role of eigenvalues in determining long-term behavior of a system.
7Analyze recurrence sequences using matrix methods.
8Describe the structure and behavior of a Markov chain.
9Interpret long-term behavior of step-by-step models in applied contexts.
10Recognize common errors in diagonalization and dynamical system analysis.

Chapter 13 — Contents

Sections at a Glance

13.1Why Diagonalization MattersThe problem of repeated multiplication; eigenvectors as a basis; the core insight A^kv = λ^kv
13.2Similar Matrices as an IntroductionDefinition of similar matrices; change-of-basis matrix; shared properties; diagonalization as a special case
13.3Diagonal MatricesProperties and advantages; powers of diagonal matrices; the formula D^k; preview of long-term behavior
13.4Diagonalizing a MatrixThe Diagonalization Theorem; constructing P and D; verification by AP = PD; step-by-step process
13.5When a Matrix Is DiagonalizableSufficient conditions; algebraic and geometric multiplicity; defective matrices; the diagonalizability test
13.6Powers of Diagonal MatricesThe formula D^k = diag(d₁^k, …, d_n^k); negative powers; long-term behavior from eigenvalue magnitudes
13.7Powers of Diagonalizable MatricesThe main formula A^k = PD^kP⁻¹; derivation; worked computation; general eigenvector expansion
13.8Discrete Dynamical SystemsDefinition; matrix form x_{k+1} = Ax_k; solution x_k = A^kx_0; geometric description; dominant eigenvalue
13.9Recurrence SystemsScalar recurrences; second-order recurrences as matrix systems; companion matrix; Fibonacci example; Binet's formula
13.10Markov Chains as an IntroductionState vectors; stochastic matrices; steady-state vectors; connection to eigenvalue 1; weather and population examples
13.11Long-Term Behavior in Step-by-Step ModelsDominant eigenvalue theorem; three categories of behavior; convergence, growth, decay; examples
13.12Applications in Population, Networks, and FinanceLeslie matrices; PageRank; compound growth; two-sector economic model
13.13Common Diagonalization MistakesEight recurring errors; explanations and corrections; prevention strategies
13.14Preparing for Calculus, Differential Equations, and Advanced Linear AlgebraDiscrete to continuous bridge; PCA and spectral theorem preview; matrix exponentials; readiness notes

Section 13.1

Why Diagonalization Matters

The Problem of Repeated Multiplication

Suppose you have a matrix A and you want to compute A¹⁰, meaning the result of multiplying A by itself ten times. You could compute A² = A · A, then A³ = A² · A, and continue. For a 2 × 2 matrix this is tedious but manageable. For a 10 × 10 matrix, computing A¹⁰⁰ this way is essentially impossible by hand.

The situation becomes even more interesting when the repeated multiplication represents something real. If A is the matrix that describes how a population changes from one year to the next, then A⁵⁰ describes the population after 50 years. If A is a transition matrix for a network, then A^k describes how information propagates after k steps. Understanding A^k for large k is often the central question.

The Key Insight: Eigenvectors as a Basis

In Chapter 12 you learned that some vectors are special: eigenvectors. If Av = λv, then applying A to v just scales it by λ. The vector keeps its direction; only its length changes. This means that applying A twice to v gives:

A²v = A(Av) = A(λv) = λ(Av) = λ(λv) = λ²v

And in general, for any positive integer k:

Core Eigenvector-Power Formula

A^kv = λ^kv

On an eigenvector, repeated application of A is just repeated scaling by λ. Powers of a number are easy; powers of a matrix are hard.

If you can express any starting vector as a combination of eigenvectors, you can reduce A^k to a set of scalar power computations. Diagonalization is the formal version of this idea.

Diagonalization: The Big Picture

Diagonalization rewrites a matrix A as A = PDP⁻¹, where D is a diagonal matrix of eigenvalues and P is a matrix whose columns are the corresponding eigenvectors. This turns the hard problem of computing A^k into the easy problem:

A^k = PD^kP⁻¹ Since D is diagonal, D^k is trivial: raise each diagonal entry to the kth power.

Section 13.2

Similar Matrices as an Introduction

Two square matrices A and B are called similar if there exists an invertible matrix P such that B = P⁻¹AP. The matrix P is called the change-of-basis matrix. Similar matrices represent the same linear transformation expressed in different coordinate systems.

Property	Preserved Under Similarity?	Reason
Eigenvalues	Yes	Same characteristic polynomial
Determinant	Yes	det(P⁻¹AP) = det(A)
Rank	Yes	Same null space dimension
Trace	Yes	Sum of eigenvalues is the same
Specific entries	No	Coordinates change with basis
Eigenvectors	Not in general	Expressed in different bases

Diagonalization as Special Similarity

Diagonalization is the special case where B = D is a diagonal matrix. The question becomes: can we find an invertible P such that P⁻¹AP = D is diagonal? The diagonal entries of D will be the eigenvalues of A.

Section 13.3

Diagonal Matrices

A diagonal matrix is a square matrix in which every entry off the main diagonal is zero. Diagonal matrices are the simplest matrices to work with because multiplication, powers, inverses, and determinants all reduce to operations on individual diagonal entries.

Operation	For Diagonal Matrix D = diag(d₁, d₂, …, d_n)
Power D^k	diag(d₁^k, d₂^k, …, d_n^k)
Inverse D⁻¹	diag(1/d₁, 1/d₂, …, 1/d_n) — requires all d_i ≠ 0
Determinant	d₁ · d₂ · ⋯ · d_n
Product D₁D₂	diag(d₁e₁, d₂e₂, …, d_ne_n) — entry-by-entry

Example 13.3.1

Power of a diagonal matrix and long-term behavior

ProblemLet D = diag(3, −1). Compute D⁴ and D⁵. Then describe what happens as k → ∞.
Calculate D⁴ = diag(3⁴, (−1)⁴) = diag(81, 1) D⁵ = diag(3⁵, (−1)⁵) = diag(243, −1)
InterpretThe first component scales by 3^k, growing rapidly since |3| > 1. The second component alternates between +1 and −1 since |(−1)| = 1. For large k, the behavior is dominated entirely by the first component. The eigenvalue with the largest absolute value controls the long-term picture.

Answer: D⁴ = diag(81, 1), D⁵ = diag(243, −1). First component grows; second oscillates.

Section 13.4

Diagonalizing a Matrix

The Diagonalization Theorem

An n × n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors.

If A has n linearly independent eigenvectors v₁, v₂, …, vₙ with corresponding eigenvalues λ₁, λ₂, …, λₙ, form:

P = [v₁ | v₂ | ⋯ | vₙ] (columns are the eigenvectors)

D = diag(λ₁, λ₂, …, λₙ) (eigenvalues on the diagonal, same order)

Then A = PDP⁻¹. The columns of P must be in the same order as the diagonal entries of D.

Why the Formula Works

To verify A = PDP⁻¹, multiply both sides on the right by P to get AP = PD. The jth column of AP is Avⱼ. The jth column of PD is λⱼvⱼ (scaling the jth column of P by λⱼ). Since Avⱼ = λⱼvⱼ by definition of eigenvector, the equation AP = PD holds. Since P is invertible (its columns are linearly independent), we write A = PDP⁻¹.

The Diagonalization Process — Five Steps

Step	Action
1	Find all eigenvalues by solving det(A − λI) = 0
2	For each eigenvalue, find a basis for the eigenspace by solving (A − λI)x = 0
3	Count the total eigenvectors. If that number equals n, the matrix is diagonalizable
4	Form P with eigenvectors as columns; form D with eigenvalues on the diagonal (matching order)
5	Verify: check that AP = PD column by column

Example 13.4.1

Diagonalizing a 2 × 2 matrix

ProblemDiagonalize A = [[4, 1], [2, 3]].
ThinkFind eigenvalues, then eigenvectors, then assemble P and D.
Step 1 det(A − λI) = (4−λ)(3−λ) − 2 = λ² − 7λ + 10 = (λ−5)(λ−2) Eigenvalues: λ₁ = 5, λ₂ = 2
Step 2 For λ₁ = 5: A − 5I = [[ −1, 1], [ 2, −2]] Row reduce → [[1, −1], [0, 0]] x₂ = t, x₁ = t → v₁ = (1, 1)ᵀ For λ₂ = 2: A − 2I = [[ 2, 1], [ 2, 1]] Row reduce → [[2, 1], [0, 0]] x₂ = t, x₁ = −t/2. Choose t = −2 → v₂ = (1, −2)ᵀ
Step 4 P = [[1, 1], D = [[5, 0], [1, −2]] [0, 2]]
Verify AP = [[4,1],[2,3]] · [[1,1],[1,−2]] = [[5, 2],[5, −4]] PD = [[1,1],[1,−2]] · [[5,0],[0,2]] = [[5, 2],[5, −4]] ✓
InterpretThe matrix A acts like simple scaling by 5 and 2 when the state is expressed in the eigenvector basis. The column v₁ = (1,1)ᵀ is stretched by factor 5; the column v₂ = (1,−2)ᵀ is stretched by factor 2.

Answer: P = [[1,1],[1,−2]], D = [[5,0],[0,2]], and A = PDP⁻¹.

Section 13.5

When a Matrix Is Diagonalizable

The most convenient sufficient condition is: if an n × n matrix A has n distinct eigenvalues, then A is diagonalizable. Eigenvectors from distinct eigenvalues are always linearly independent. This is sufficient but not necessary — a matrix can be diagonalizable even when eigenvalues repeat, as long as each repeated eigenvalue has enough independent eigenvectors.

Algebraic and Geometric Multiplicity

When an eigenvalue λ appears as a repeated root of the characteristic polynomial, its algebraic multiplicity is the number of times it appears as a root. The geometric multiplicity is the dimension of the eigenspace for λ — the number of linearly independent eigenvectors for that eigenvalue.

The geometric multiplicity is always at most the algebraic multiplicity. For A to be diagonalizable, every eigenvalue must have geometric multiplicity equal to its algebraic multiplicity.

Example 13.5.1

A matrix that is NOT diagonalizable

ProblemDetermine whether A = [[2, 1], [0, 2]] is diagonalizable.
Eigenvaluesdet(A − λI) = (2−λ)² = 0 → λ = 2, algebraic multiplicity 2
Eigenspace A − 2I = [[0, 1], Row reduce → [[0, 1], [0, 0]] [0, 0]] Only x₂ = 0, x₁ free. One independent eigenvector: v = (1, 0)ᵀ. Geometric multiplicity = 1. Algebraic multiplicity = 2. MISMATCH.
InterpretSince geometric multiplicity (1) is less than algebraic multiplicity (2), this matrix is defective — not diagonalizable. There is no basis of eigenvectors for ℝ².

Answer: Not diagonalizable. The eigenspace for λ = 2 has dimension 1, but algebraic multiplicity is 2.

Student Warning — Not Every Matrix Is Diagonalizable

Never assume a matrix is diagonalizable without checking. A matrix with a repeated eigenvalue may or may not be diagonalizable. Always check whether the geometric multiplicity of each eigenvalue equals its algebraic multiplicity. A single defective eigenvalue makes the entire matrix non-diagonalizable.

Section 13.6

Powers of Diagonal Matrices

For any diagonal matrix D = diag(d₁, d₂, …, d_n) and any non-negative integer k:

Power of a Diagonal Matrix

D^k = diag(d₁^k, d₂^k, …, d_n^k)

Raise each diagonal entry to the k^th power independently. The long-term behavior depends entirely on which entries have |d_i| > 1, = 1, or < 1.

Example 13.6.1

Long-term behavior from diagonal entries

ProblemDescribe the long-term behavior of D = diag(0.9, 1.5) as k → ∞.
Analyze (0.9)ᵏ → 0 because |0.9| < 1 → this component SHRINKS (1.5)ᵏ → ∞ because |1.5| > 1 → this component GROWS without bound
InterpretFor large k, Dᵏ ≈ diag(0, (1.5)ᵏ). The long-term behavior is completely governed by the component with the eigenvalue larger than 1 in absolute value — the dominant eigenvalue.

Answer: First component decays to 0; second component grows without bound. Dominant eigenvalue is 1.5.

Section 13.7

Powers of Diagonalizable Matrices

If A = PDP⁻¹, then A² = (PDP⁻¹)(PDP⁻¹) = PD(P⁻¹P)DP⁻¹ = PD²P⁻¹. By induction the pattern extends to all positive integer powers.

Main Formula — Matrix Powers via Diagonalization

A^k = PD^kP⁻¹

Steps: (1) Find P and D from the eigenvalue decomposition. (2) Compute D^k by raising each eigenvalue to the k^th power. (3) Multiply PD^kP⁻¹.

Example 13.7.1

Computing A¹⁰ using diagonalization

ProblemUse the diagonalization from Example 13.4.1 to compute A¹⁰ where A = [[4,1],[2,3]].
SetupFrom Example 13.4.1: P = [[1,1],[1,−2]], D = [[5,0],[0,2]]. Find P⁻¹ first.
P⁻¹ det(P) = (1)(−2) − (1)(1) = −3 P⁻¹ = (1/−3) · [[−2, −1], = [[ 2/3, 1/3], [−1, 1]] [ 1/3, −1/3]]
D¹⁰ D¹⁰ = [[5¹⁰, 0 ], = [[9 765 625, 0], [ 0, 2¹⁰]] [ 0, 1 024]]
Multiply A¹⁰ = P · D¹⁰ · P⁻¹ PD¹⁰ = [[9 765 625, 1 024], [9 765 625, −2 048]] A¹⁰ entries involve (2·5¹⁰ ± 2¹⁰)/3 and (5¹⁰ ∓ 2¹⁰)/3 — all integers ✓
InterpretComputing A¹⁰ directly would require 9 separate matrix multiplications. Diagonalization reduces this to raising two numbers to the 10th power and multiplying three fixed matrices — a dramatic simplification for large k.

Answer: A¹⁰ = PD¹⁰P⁻¹, computed by raising eigenvalues 5 and 2 to the 10th power.

General Solution Formula for a Diagonalizable System

If x₀ = c₁v₁ + c₂v₂ + ⋯ + c_n v_n (initial vector expanded in the eigenvector basis), then:

A^kx₀ = c₁λ₁^kv₁ + c₂λ₂^kv₂ + ⋯ + c_n λ_n^kv_n

Each component evolves independently in its eigenvector direction, scaled by the corresponding eigenvalue raised to the kth power. The coefficients c₁, …, c_n come from solving Pc = x₀.

Section 13.8

Discrete Dynamical Systems

A discrete dynamical system is a model in which a state changes from one step to the next according to a fixed rule. The state might represent a population, an amount of money, proportions in a network, or a physical configuration. The rule is applied repeatedly, and we want to understand how the state evolves over time.

Discrete means the time steps are separate and countable — year 0, year 1, year 2, etc. — rather than continuous. This contrasts with differential equations, which model continuous change. Since this book does not require calculus, we focus entirely on the discrete case.

Matrix Form of a Discrete Dynamical System

The system is described by x_{k+1} = A x_k, where:

x_k — the state vector at step k (all quantities in the system)

A — the transition matrix (how the state changes each step)

Given initial state x₀, the state at step k is: x_k = A^k x₀

If A is diagonalizable: x_k = c₁λ₁^kv₁ + c₂λ₂^kv₂ + ⋯ where c = P⁻¹x₀.

Example 13.8.1

Setting up a discrete dynamical system — city neighborhoods

ProblemA city has two neighborhoods: Downtown and Uptown. Each year, 10% of Downtown residents move to Uptown and 20% of Uptown residents move to Downtown. The rest stay. Set up the dynamical system if Downtown starts at 8000 residents and Uptown at 6000.
RepresentLet x_k = (d_k, u_k)ᵀ be the population vector at year k. The transition rules: Next Downtown = 0.90 · d_k + 0.20 · u_k Next Uptown = 0.10 · d_k + 0.80 · u_k
Matrix form A = [[0.90, 0.20], x₀ = [[8000], [0.10, 0.80]] [6000]] x_{k+1} = A x_k
CheckEach column of A sums to 1 (90% + 10% = 1; 20% + 80% = 1). Total population is preserved. ✓
InterpretWe will find the eigenvalues of A in Section 13.11 to determine the long-term population split. The column sum = 1 property will appear again in Markov chains.

Answer: Transition matrix A = [[0.90, 0.20], [0.10, 0.80]], initial state x₀ = (8000, 6000)ᵀ.

Geometric Description — Eigenvector Decomposition of a State

Draw a timeline of vectors in ℝ². The initial vector x₀ is shown as a combination of v₁ and v₂. At each step, the v₁ component is multiplied by λ₁ and the v₂ component is multiplied by λ₂. If |λ₁| > |λ₂|, then for large k the vector x_k points increasingly in the v₁ direction — the direction of the dominant eigenvector.

Section 13.9

Recurrence Systems

A recurrence relation defines each term in a sequence in terms of previous terms. The second-order linear recurrence a_{k+2} = p · a_{k+1} + q · a_k can be converted to a matrix system by introducing the state vector x_k = (a_{k+1}, a_k)ᵀ. Then x_{k+1} = A x_k with the companion matrix:

A = [[p, q], (companion matrix of the recurrence) [1, 0]]

Diagonalizing the companion matrix yields a closed-form solution for a_k.

Example 13.9.1

The Fibonacci sequence as a matrix system

ProblemThe Fibonacci recurrence: F₀ = 0, F₁ = 1, F_{k+2} = F_{k+1} + F_k. Find the companion matrix and describe the long-term growth rate.
Represent x_k = (F_{k+1}, F_k)ᵀ, A = [[1, 1], [1, 0]]
Eigenvalues det(A − λI) = (1−λ)(−λ) − 1 = λ² − λ − 1 = 0 λ = (1 ± √5) / 2 φ = (1 + √5)/2 ≈ 1.618 (the golden ratio) φ̂ = (1 − √5)/2 ≈ −0.618
ResultTwo distinct eigenvalues → A is diagonalizable. Diagonalizing yields Binet's formula: F_k = (φᵏ − φ̂ᵏ) / √5
InterpretSince |φ̂| < 1, the term φ̂ᵏ → 0, so for large k, F_k ≈ φᵏ/√5. Each Fibonacci number is approximately φ ≈ 1.618 times the previous one. The golden ratio is the dominant eigenvalue of the Fibonacci companion matrix.

Answer: Companion matrix A = [[1,1],[1,0]] has eigenvalues φ and φ̂. Dominant eigenvalue φ = (1+√5)/2 controls long-term growth.

Tip — Every Linear Recurrence Has a Matrix Form

Any second-order linear recurrence a_{k+2} = p · a_{k+1} + q · a_k becomes the matrix system x_{k+1} = [[p,q],[1,0]] · x_k. The characteristic polynomial of the recurrence and the characteristic polynomial of the companion matrix are the same. The eigenvalues of the companion matrix are exactly the roots of the recurrence's characteristic equation.

Section 13.10

Markov Chains as an Introduction

A Markov chain is a special discrete dynamical system in which the state vector represents probabilities. Each entry x_i of the state vector gives the probability of being in state i. The transition matrix describes the probability of moving from one state to another at each step.

Component	Definition	Required Property
State vector x_k	Column vector of probabilities at time k	All entries ≥ 0; entries sum to 1
Stochastic matrix A	Transition matrix encoding movement probabilities	All entries ≥ 0; each column sums to 1
Steady-state vector q	Long-term probability distribution	Aq = q; entries ≥ 0, sum to 1

Example 13.10.1

A weather Markov chain

ProblemWeather is either Sunny (S) or Cloudy (C). If today is Sunny, tomorrow is Sunny with probability 0.7 and Cloudy with probability 0.3. If today is Cloudy, tomorrow is Sunny with probability 0.4 and Cloudy with probability 0.6. Set up the Markov chain and compute the first few state vectors.
Matrix A = [[ 0.7, 0.4], Columns: current state (S or C) [ 0.3, 0.6]] Rows: next state (S or C) Each column sums to 1. ✓
Iterations x₀ = (1, 0)ᵀ (start: Sunny) x₁ = A x₀ = (0.7, 0.3)ᵀ x₂ = A x₁ = [[0.7,0.4],[0.3,0.6]] · (0.7, 0.3)ᵀ = (0.61, 0.39)ᵀ
InterpretAfter two days starting from Sunny, there is a 61% chance of Sunny and 39% chance of Cloudy. The distribution is already shifting toward a long-term balance.

Answer: Transition matrix A = [[0.7, 0.4], [0.3, 0.6]]. After two days: (0.61, 0.39)ᵀ.

Finding the Steady-State Vector

The steady-state vector q satisfies Aq = q, which means (A − I)q = 0. This makes q an eigenvector of A with eigenvalue λ = 1. For any regular stochastic matrix, one eigenvalue is always exactly 1, and all other eigenvalues satisfy |λ| < 1. The steady state is the eigenvector for λ = 1, normalized so its entries sum to 1.

Example 13.10.2

Finding the steady-state distribution

ProblemFind the steady-state vector for the weather Markov chain in Example 13.10.1.
Solve A − I = [[ −0.3, 0.4], [ 0.3, −0.4]] Row reduce: → [[ 0.3, −0.4], [ 0, 0]] 0.3 q₁ = 0.4 q₂ → q₁ = (4/3) q₂
Normalize q₁ + q₂ = 1 (4/3)q₂ + q₂ = 1 → (7/3)q₂ = 1 → q₂ = 3/7, q₁ = 4/7 q = (4/7, 3/7)ᵀ ≈ (0.571, 0.429)ᵀ
InterpretIn the long run, about 57.1% of days will be Sunny and 42.9% will be Cloudy — regardless of today's weather. This equilibrium is the eigenvector for λ = 1, scaled to sum to 1.

Answer: Steady-state vector q = (4/7, 3/7)ᵀ ≈ (0.571, 0.429)ᵀ.

Student Warning — Normalize Markov Steady-State Vectors

The eigenvector for λ = 1 gives the correct direction but must be scaled so all entries are non-negative and sum to 1 to be a valid probability distribution. Simply reporting the eigenvector without normalizing gives the wrong answer in any probability context.

Section 13.11

Long-Term Behavior in Step-by-Step Models

When A is diagonalizable and x₀ = c₁v₁ + c₂v₂ + ⋯ + c_n v_n, the state at step k is:

x_k = c₁λ₁ᵏv₁ + c₂λ₂ᵏv₂ + \dots + c_n λ_nᵏ v_n

If |λ₁| > |λ₂| ≥ ⋯ ≥ |λ_n|, then for large k the term c₁λ₁ᵏv₁ dominates all others. The eigenvalue λ₁ with the largest absolute value is the dominant eigenvalue, and its eigenvector v₁ gives the long-term direction of the state vector.

Dominant Eigenvalue \|λ₁\|	Long-Term Behavior	Example
\|λ₁\| > 1	State grows without bound (explosive growth)	Growing population, runaway system
\|λ₁\| = 1 and λ₁ = +1	State converges to a stable equilibrium	Markov chain steady state, balanced population
\|λ₁\| = 1 and λ₁ = −1	State oscillates with constant magnitude	Alternating-sign behavior
\|λ₁\| < 1	State decays to 0 (stable decay)	Declining population, dying-out process

Example 13.11.1

Long-term behavior of the city neighborhood model

ProblemFind the eigenvalues of A = [[0.90, 0.20], [0.10, 0.80]] from Example 13.8.1 and determine the long-term population split.
Eigenvalues det(A − λI) = λ² − 1.7λ + 0.70 = 0 λ = (1.7 ± √(2.89 − 2.80))/2 = (1.7 ± 0.3)/2 λ₁ = 1, λ₂ = 0.7
Analyzeλ₁ = 1 is the dominant eigenvalue (|λ₂| = 0.7 < 1, so that component decays). The total population is preserved (columns sum to 1). Find the eigenvector for λ₁ = 1: A − I = [[ −0.10, 0.20], [ 0.10, −0.20]] → x₁ = 2x₂ Eigenvector: v₁ = (2, 1)ᵀ
InterpretIn the long run, the population splits in ratio 2:1, Downtown to Uptown. With total population 14 000, the eventual split is Downtown ≈ 9333, Uptown ≈ 4667. This equilibrium is reached from any starting distribution, because |λ₂| < 1 causes all deviations to decay away.

Answer: Eigenvalues 1 and 0.7. Long-term distribution: Downtown/Uptown ratio = 2:1.

Section 13.12

Applications in Population, Networks, and Finance

Population Models — Leslie Matrices

In ecology, a Leslie matrix models a population divided into age groups. The dominant eigenvalue of A determines the long-term fate of the population: if |λ₁| > 1, the population grows; if |λ₁| = 1, it stabilizes; if |λ₁| < 1, the population declines toward extinction. This is why conservation biologists care about eigenvalues — they directly reveal whether a managed species will survive.

Network Models — PageRank

Search engines use a form of Markov chain to rank web pages. A user randomly following links is modeled as a random walk through a network. The steady-state distribution of this Markov chain gives each page's importance score — called PageRank. The transition matrix encodes the link structure: entry a_{ij} is the probability of moving from page j to page i. The steady-state eigenvector (for λ = 1) gives the ranking. This is the same steady-state analysis from Section 13.10, applied at enormous scale.

Example 13.12.1

A two-sector economic model

ProblemAn economy has manufacturing (M) and services (S) sectors with transition matrix A = [[1.2, 0.3], [0.1, 1.1]]. Analyze long-term behavior.
Eigenvalues det(A − λI) = λ² − 2.3λ + 1.29 = 0 λ = (2.3 ± √(5.29 − 5.16))/2 = (2.3 ± √0.13)/2 λ₁ ≈ (2.3 + 0.361)/2 ≈ 1.330 λ₂ ≈ (2.3 − 0.361)/2 ≈ 0.970
Interpretλ₁ ≈ 1.330 > 1: the economy grows at approximately 33% per step in the dominant direction. Since λ₂ < 1, the other component decays. In the long run, the ratio of manufacturing to services output approaches the eigenvector for λ₁.

Answer: Dominant eigenvalue ≈ 1.330 → economy grows. Long-term direction given by eigenvector for λ₁.

MGU Library Connection

Data Science Foundations — PageRank and graph-based ranking algorithms
Machine Learning Mathematics — dimensionality reduction and principal component analysis
Differential Equations — systems of ODEs and matrix exponentials (continuous analog of this chapter)
Appendix O: Diagonalization Reference — formula summaries and worked method guide
Appendix R: Calculus Readiness Bridge Guide — connections to continuous systems

Section 13.13

Common Diagonalization Mistakes

Mistake 1 — Inverting P in the Wrong Position

A = P⁻¹DP

Correct form: A = PDP⁻¹, where columns of P are eigenvectors. Verify by checking AP = PD: the jth column of AP is Avⱼ = λⱼvⱼ, which equals the jth column of PD.

Mistake 2 — Misaligning Eigenvalues and Eigenvectors

P columns and D diagonal entries in different orders

The eigenvalue in position (j,j) of D must correspond to the eigenvector in column j of P. Mixing this correspondence produces an incorrect decomposition. Always check AP = PD column by column.

Mistake 3 — Assuming Every Matrix Is Diagonalizable

"I found the eigenvalues, so I can diagonalize A."

Finding eigenvalues is only the first step. You must check that the sum of all eigenspace dimensions equals n. A single defective eigenvalue (geometric multiplicity < algebraic multiplicity) makes the whole matrix non-diagonalizable.

Mistake 4 — Forgetting to Normalize Markov Steady-State Vectors

Reporting q = (2, 3)ᵀ as a steady-state probability distribution

The eigenvector (2, 3)ᵀ gives the correct direction, but it is not a valid probability vector. Normalize: q = (2/5, 3/5)ᵀ so entries sum to 1 and each entry represents an actual probability.

Mistake 5 — Using A^k = PD^kP⁻¹ When A Is Not Diagonalizable

Computing A^k = PD^kP⁻¹ for a defective matrix

This formula requires A to be diagonalizable. If A is defective, there is no valid decomposition A = PDP⁻¹, and the formula does not apply. For non-diagonalizable matrices, Jordan normal form is needed (beyond this course).

Mistake 6 — Ignoring the Initial Condition

"The dominant eigenvalue governs everything" — regardless of x₀

The formula x_k = c₁λ₁ᵏv₁ + ⋯ depends on the coefficients c₁, c₂, … from expressing x₀ in the eigenvector basis. If c₁ = 0 (the initial state has no component in the dominant direction), then the dominant eigenvalue does not control the long-term behavior.

Mistake 7 — Treating λ = −1 as a Steady State

"The eigenvalue has |λ| = 1, so the system is at steady state."

An eigenvalue λ = −1 has magnitude 1 but produces alternating signs: (−1)ᵏ = +1, −1, +1, −1, … Only λ = +1 produces a true steady state in a real system. Always check the sign, not just the magnitude.

Mistake 8 — Confusing Similar Matrices with Equal Matrices

Substituting D in place of A in a formula that depends on A's actual entries

Similar matrices A and D = P⁻¹AP have the same eigenvalues, determinant, and rank, but different entries. They cannot be substituted for each other in calculations that use the actual matrix values — only in those that depend on these shared structural invariants.

Section 13.14

Preparing for Calculus, Differential Equations, and Advanced Linear Algebra

The Bridge from Discrete to Continuous

This chapter focused on discrete dynamical systems, where the state changes in countable steps. The formula x_k = A^kx₀ and the role of eigenvalues in long-term behavior are the discrete analogs of deeper ideas in calculus and differential equations.

In Differential Equations, the discrete system x_{k+1} = Ax_k becomes the continuous system dx/dt = Ax. The solution involves matrix exponentials, and the eigenvalues of A still control long-term behavior — now through expressions of the form e^{λt} instead of λ^k. The conceptual structure is identical; only the tools differ. By mastering the discrete version, you have built the intuition you will need for the continuous case.

Discrete (This Chapter)	Continuous (Differential Equations)
x_{k+1} = Ax_k	dx/dt = Ax
Solution: x_k = A^kx₀	Solution: x(t) = e^{At}x₀
Grows if \|λ\| > 1	Grows if Re(λ) > 0
Decays if \|λ\| < 1	Decays if Re(λ) < 0
A^k = PD^kP⁻¹	e^{At} = Pe^{Dt}P⁻¹

Principal Component Analysis — A Preview

In statistics and machine learning, Principal Component Analysis (PCA) finds the eigenvectors of a covariance matrix to identify the most important directions of variation in a data set. These eigenvectors — called principal components — form a new coordinate system aligned with the data. The eigenvalues measure how much variation occurs in each direction.

Diagonalization underlies PCA entirely. Students who understand eigenvalues and diagonalization are ready to understand one of the most widely used tools in data science and machine learning.

The Spectral Theorem — A Preview

For symmetric matrices — those satisfying A = Aᵀ — the diagonalization has a special form. The eigenvectors can be chosen to be orthogonal and normalized, and the diagonalization becomes A = QDQᵀ where Q is an orthogonal matrix. This is the Spectral Theorem, the foundation for PCA, the Singular Value Decomposition, and many other matrix factorizations. Students who complete this chapter are prepared to study the Spectral Theorem in a future course.

Calculus Readiness — What You Have Built

This chapter completes Part IV. You now understand how eigenvalues and eigenvectors reveal the structure of repeated transformations, how to compute matrix powers efficiently, how to analyze systems that evolve step by step, and how to determine whether those systems grow, decay, or stabilize in the long run.

These ideas — dominant eigenvalue, decomposition of initial conditions, long-term direction — reappear in every course that follows: Calculus (tangent approximation, directional derivatives), Multivariable Calculus (Jacobians, Hessians), Differential Equations (solution of systems), Statistics (regression, PCA), Data Science (dimensionality reduction), and Machine Learning (gradient descent, neural networks, SVD).

Chapter 13 — Back Matter

Chapter Summary

Sec 13.1The core insight: on an eigenvector, Aᵏv = λᵏv. This makes repeated matrix multiplication tractable when starting vectors are expressed in terms of eigenvectors.
Sec 13.2Similar matrices A and B = P⁻¹AP represent the same transformation in different coordinate systems. They share eigenvalues, determinant, rank, and trace. Diagonalization is the case where B = D is diagonal.
Sec 13.3For diagonal D = diag(d₁, …, d_n), the power Dᵏ = diag(d₁ᵏ, …, d_nᵏ). Entries with |d_i| > 1 grow; |d_i| < 1 shrink; |d_i| = 1 with d_i = 1 stabilize.
Sec 13.4To diagonalize A: find eigenvalues, find eigenvectors, assemble P (eigenvectors as columns) and D (eigenvalues on diagonal, matching order). Verify with AP = PD.
Sec 13.5A matrix is diagonalizable iff it has n independent eigenvectors, equivalently iff geometric multiplicity = algebraic multiplicity for every eigenvalue. n distinct eigenvalues always guarantee diagonalizability.
Sec 13.6–7If A = PDP⁻¹, then Aᵏ = PDᵏP⁻¹. Computing Aᵏ reduces to raising eigenvalues to the kth power. The general solution is x_k = c₁λ₁ᵏv₁ + ⋯ + c_nλ_nᵏv_n.
Sec 13.8A discrete dynamical system: x_{k+1} = Ax_k, giving x_k = Aᵏx₀. The dominant eigenvalue (largest |λ|) controls long-term direction. The coefficients come from expressing x₀ in the eigenvector basis.
Sec 13.9Second-order recurrences become matrix systems via the companion matrix. Diagonalizing the companion matrix yields closed-form solutions. The Fibonacci sequence has growth rate φ = (1+√5)/2, the dominant eigenvalue of its companion matrix.
Sec 13.10Markov chains: state vectors are probability distributions; transition matrices are stochastic (columns sum to 1). The steady-state vector is the eigenvector for λ = 1, normalized to sum to 1.
Sec 13.11Long-term behavior: if |λ₁| > 1, state grows; if λ₁ = +1 with all others |λ| < 1, state converges to steady state; if all |λ| < 1, state decays to 0.
Sec 13.12Applications include Leslie population matrices (dominant eigenvalue determines survival), PageRank (steady-state of a web-link Markov chain), and economic growth models (dominant eigenvalue gives long-term growth rate).
Sec 13.13Eight common errors: wrong inverse position, misaligned P and D, assuming all matrices are diagonalizable, unnormalized Markov vectors, using the formula on non-diagonalizable matrices, ignoring initial conditions, treating λ=−1 as a steady state, and confusing similar with equal.
Sec 13.14Discrete systems are the foundation for continuous differential equations. Diagonalization underlies PCA, the Spectral Theorem, and the SVD. These ideas are central to calculus, differential equations, statistics, data science, and machine learning.

Key Terms Review

algebraic multiplicity

The number of times an eigenvalue appears as a root of the characteristic polynomial.

companion matrix

A matrix encoding a linear recurrence relation as a matrix dynamical system; has the same characteristic polynomial as the recurrence.

defective matrix

A matrix that is not diagonalizable because the geometric multiplicity of some eigenvalue is strictly less than its algebraic multiplicity.

diagonal matrix

A square matrix with all zero entries off the main diagonal. Powers, inverses, and products are trivial to compute.

diagonalizable matrix

A matrix that can be written as A = PDP⁻¹ with D diagonal. Requires n linearly independent eigenvectors.

diagonalization

The process of writing A = PDP⁻¹ where columns of P are eigenvectors and diagonal entries of D are corresponding eigenvalues.

discrete dynamical system

A model in which a state vector evolves step by step via x_{k+1} = Ax_k, giving x_k = Aᵏx₀.

dominant eigenvalue

The eigenvalue with the largest absolute value. Determines the long-term direction and growth rate of a dynamical system.

eigenvalue decomposition

Another name for diagonalization: A = PDP⁻¹. Reveals how A acts as simple scaling in each eigenvector direction.

geometric multiplicity

The dimension of the eigenspace for an eigenvalue; the number of linearly independent eigenvectors for that eigenvalue.

Markov chain

A discrete dynamical system with a stochastic transition matrix, modeling how a probability distribution evolves over time.

recurrence relation

A formula defining each term of a sequence in terms of earlier terms. Can be converted to matrix form via the companion matrix.

similar matrices

Matrices A and B related by B = P⁻¹AP for some invertible P. They share eigenvalues, determinant, rank, and trace.

state vector

A vector whose entries describe the current state of a dynamical system at a given step.

steady-state vector

The long-term stable distribution in a Markov chain; the eigenvector for eigenvalue 1, normalized to sum to 1.

stochastic matrix

A matrix with non-negative entries whose columns all sum to 1. Used as the transition matrix in a Markov chain.

transition matrix

The matrix A in x_{k+1} = Ax_k, describing how the state evolves from one step to the next.

long-term behavior

The eventual pattern of x_k as k → ∞. Determined by the dominant eigenvalue and its corresponding eigenvector.

golden ratio φ

φ = (1+√5)/2 ≈ 1.618. The dominant eigenvalue of the Fibonacci companion matrix; governs the long-term growth rate of Fibonacci numbers.

Concept Review Questions

What does it mean for a matrix to be diagonalizable? State the condition in terms of eigenvectors.
Explain why the formula Aᵏ = PDᵏP⁻¹ simplifies the computation of matrix powers.
What is the difference between algebraic multiplicity and geometric multiplicity? Which must be at least as large as the other?
Give an example of a 2 × 2 matrix that is not diagonalizable and explain why.
What is a discrete dynamical system? How is the state at step k expressed in terms of the transition matrix and initial state?
What is the dominant eigenvalue, and how does it determine long-term behavior?
What makes a matrix stochastic? What property ensures that probabilities are preserved at each step?
What is a steady-state vector in a Markov chain, and how is it related to eigenvalues?
How can a second-order linear recurrence be converted to a matrix dynamical system?
Why does the Fibonacci sequence have a long-term growth rate approximately equal to the golden ratio?

Diagonalization Practice

Part AFinding P and D

For each matrix, find eigenvalues and eigenvectors and construct P and D such that A = PDP⁻¹, or explain why the matrix is not diagonalizable.

A = [[3, 0], [0, −2]]
A = [[5, 2], [0, 1]]
A = [[1, 1], [−2, 4]]
A = [[2, 1], [0, 2]] — investigate carefully before proceeding
A = [[0, 1], [−6, 5]]

Part BVerifying and Checking

Verify your diagonalization in Problem 3 above by confirming that AP = PD column by column.
For A = [[6, −1], [2, 3]]: find eigenvalues, find a basis for each eigenspace, construct P and D, compute P⁻¹, and verify A = PDP⁻¹.
Determine, without finding all eigenvectors, whether the following matrices are diagonalizable. Justify each answer.
1. A 3 × 3 matrix with eigenvalues 2, 3, and 7.
2. A 3 × 3 matrix with eigenvalues 2, 2, and 7 where the eigenspace for λ = 2 has dimension 1.
3. A 3 × 3 matrix with eigenvalues 2, 2, and 7 where the eigenspace for λ = 2 has dimension 2.
True or false? Explain each.
1. Every invertible matrix is diagonalizable.
2. Every diagonal matrix is diagonalizable.
3. If A and B are similar, they have the same eigenvalues.
4. If A has n distinct eigenvalues, it is diagonalizable.
5. Two matrices with the same eigenvalues must be similar.

Matrix Power Practice

Part CComputing Powers via Diagonalization

Let A = diag(2, 3). Compute A⁸ directly from the diagonal formula.
Let A = [[4, 2], [1, 3]]. Find eigenvalues, diagonalize, and use Aᵏ = PDᵏP⁻¹ to compute A⁶.
Let A = [[−1, 0], [0, 2]]. Compute A¹⁰ and A¹¹. What do you notice about the sign pattern in A¹¹?
For the Fibonacci companion matrix A = [[1,1],[1,0]]: diagonalize A, then use Aᵏx₀ with x₀ = (1,0)ᵀ to verify that the first entry gives F₁, F₂, F₃, F₄, F₅ correctly.
A matrix satisfies A = PDP⁻¹ where D = [[3,0],[0,−1]]. As k → ∞, does Aᵏ converge to the zero matrix? Explain why or why not.

Discrete Dynamical System Practice

Part DSetting Up and Analyzing Systems

A bird population is divided into juveniles (J) and adults (A). Each year: each adult produces 1.8 juveniles; 60% of juveniles survive to become adults; 50% of adults survive.
1. Write the transition matrix.
2. Find the eigenvalues.
3. What does the dominant eigenvalue tell you about the long-term population trend?
4. Find the eigenvector for the dominant eigenvalue. What does it represent?
Let x_{k+1} = A x_k with A = [[0.8, 0.1],[0.2, 0.9]] and x₀ = (300, 700)ᵀ.
1. Compute x₁, x₂, x₃ by direct multiplication.
2. Find the eigenvalues of A.
3. Find the steady-state vector (eigenvector for λ = 1, normalized to sum to 1000).
4. Compare the trajectory from (a) to the steady state in (c). Is the system converging?
A recurrence satisfies a_{k+2} = 5a_{k+1} − 6a_k with a₀ = 1, a₁ = 3.
1. Write this as a matrix system x_{k+1} = Ax_k.
2. Find the eigenvalues of A.
3. Diagonalize A.
4. Find a closed-form expression for a_k.

Application Problems

(Population Ecology) A managed fish population has transition matrix A = [[0, 4],[0.3, 0.8]] (rows: young, mature; columns: young, mature). Find the eigenvalues. Is the population growing, shrinking, or stable in the long run? Find the long-term ratio of young to mature fish.
(Markov Chain — Messaging) A person's messaging behavior follows the Markov chain with transition matrix A = [[0.6, 0.3],[0.4, 0.7]] (state H: on phone a lot; state L: barely on phone). If today is state H, find the state distribution after 1, 2, and 3 days. Find the steady-state distribution. What fraction of days is this person in state H in the long run?
(Finance) An investment fund evolves by A = [[1.10, 0.05],[0, 1.04]] (states: stock value, bond value). The eigenvalues of this upper triangular matrix are on the diagonal. Write the general form of x_k. If x₀ = (10000, 5000)ᵀ, find the stock value after 10 years.
(Genetics) Two alleles in a population migrate between two regions with transition matrix M = [[0.95, 0.10],[0.05, 0.90]]. Find the eigenvalues, identify the steady-state allele distribution, and determine how quickly the population approaches steady state from the magnitude of the non-unit eigenvalue.

Error Analysis

Each problem contains an error. Identify it and write the corrected version.

Problem 1

A student claims A = [[1,1],[0,1]] is diagonalizable because it's 2×2 and has eigenvalue λ = 1 with algebraic multiplicity 2.

What is the error? Is this matrix diagonalizable? Show the check.

Problem 2

To diagonalize A = [[3,1],[0,5]], a student writes P = [[1,0],[0,1]] and D = [[3,0],[0,5]] and declares A = PDP⁻¹.

Check whether AP = PD. If not, find the correct P.

Problem 3

A student finds the steady-state eigenvector q = (2, 3)ᵀ for a Markov chain and reports this as the long-run probability distribution.

What is wrong? What should the student report?

Problem 4

A student says: "Since A has eigenvalues 2 and 0.5, the system x_{k+1} = Ax_k will decay to zero."

Is this correct? What actually happens, and what determines it?

Problem 5

One student writes Aᵏ = P⁻¹DᵏP; another writes Aᵏ = PDᵏP⁻¹. Both claim to be correct.

Which is correct? Why? What goes wrong with the other form?

Chapter 13 Checkpoint

Complete without assistance — assesses all major skills and concepts from Chapter 13

Part A — Diagonalization

Let A = [[7, 2], [−4, 1]].

Find the characteristic polynomial of A.
Find all eigenvalues of A.
Find a basis for each eigenspace.
Write P and D such that A = PDP⁻¹.
Verify your answer by checking AP = PD.
Use diagonalization to compute A⁵.

Part B — Discrete Dynamical System

Two competing species are modeled by x_{k+1} = [[1.2, −0.1],[0.2, 0.8]] x_k with x₀ = (100, 80)ᵀ.

Find the eigenvalues of the transition matrix.
Determine whether each species grows, decays, or stabilizes in the long run.
Find the eigenvector for the dominant eigenvalue.
Describe the long-term ratio of the two species.

Part C — Markov Chain

A campus shuttle system has three stops: Library (L), Science Building (S), and Student Center (C). Transition matrix:

A = [[ 0.5, 0.2, 0.3], [ 0.3, 0.6, 0.2], [ 0.2, 0.2, 0.5]]

Verify that A is a stochastic matrix.
If a student starts at the Library, find the probability distribution after 2 trips.
Set up the equation for the steady-state vector.
Interpret: in the long run, roughly what fraction of students will be at each stop? (Technology may be used for computation.)

Part D — Reflection

A classmate says: "Eigenvalues only matter for rotating or stretching vectors — they have nothing to do with predicting the future." Write two paragraphs responding to this claim, explaining what eigenvalues actually tell us about long-term behavior in discrete dynamical systems.

Looking Ahead

Chapter 14: Linear Algebra Applications and Capstone Modeling

Chapter 13 closes Part IV and brings the major theoretical machinery of this course to completion. Chapter 14 brings it all together. You will revisit systems, vectors, matrices, transformations, vector spaces, projections, eigenvalues, and diagonalization through an applied lens — building linear models of real systems, analyzing computer graphics transformations, fitting data with least squares, studying Markov models, and completing capstone projects. Chapter 14 also includes an explicit readiness bridge showing how every major topic in this book connects to Calculus, Multivariable Calculus, Differential Equations, Physics, Data Science, and Machine Learning. It is both a capstone and a launch point.