📐 Concept diagram

08-09 — Diagonalisation

Phase: 8 — Linear Algebra (Rigorous) Subject: 08-09 Prerequisites: 08-08 — Eigenvalues and Eigenvectors Next subject: 08-10 — Symmetric Matrices and the Spectral Theorem

Learning Objectives

By the end of this subject, you will be able to:

1. Determine whether a matrix is diagonalizable by checking that geometric multiplicity equals algebraic multiplicity for all eigenvalues
2. Diagonalize a matrix A by constructing P (eigenvectors) and D (eigenvalues) such that A = PDP⁻¹
3. Compute powers of diagonalizable matrices efficiently using A^k = P D^k P⁻¹
4. Apply diagonalization to solve systems of first-order linear differential equations and analyze dynamical systems
5. Understand the role of similarity transformations and relate diagonalization to change-of-basis operations

Core Content

1. When Is a Matrix Diagonalizable?

An n × n matrix A is diagonalizable if it is similar to a diagonal matrix:

$A = P D P⁻¹
$

where D = diag(λ₁, ..., λ_n) and columns of P are corresponding eigenvectors.

CRITICAL — Foundational: Diagonalization $A = PDP⁻¹$ is the 'holy grail' — it reveals a complex transformation is just independent scaling along eigen-directions. Enables fast powers $A^k = PD^kP⁻¹$ and solves differential equations.

Theorem (Diagonalizability Criterion): A is diagonalizable if and only if: - The characteristic polynomial splits over the field (for us, usually R or C) - GM(λ) = AM(λ) for every eigenvalue λ

Equivalently: A has n linearly independent eigenvectors.

Sufficient conditions for diagonalizability: 1. A has n DISTINCT eigenvalues (strongest simple condition — automatically diagonalizable)

Common Pitfall: Distinct eigenvalues is SUFFICIENT but NOT necessary. Identity matrix has λ=1 repeated but is diagonal. Conversely, repeated eigenvalues may signal defectiveness — always check GM = AM. 2. A is symmetric (guarantees orthogonal diagonalization over R) 3. A is normal (A A^T = A^T A) over C

Common non-diagonalizable (defective) case:

$A = [λ  1]    (Jordan block)
    [0  λ]
$

λ has AM = 2 but GM = 1. Only one eigenvector.

Checking diagonalizability algorithmically: 1. Find all eigenvalues (roots of characteristic polynomial) 2. For each eigenvalue λ, compute GM = n − rank(A − λI) 3. Compare GM with AM (the multiplicity as a root) 4. Diagonalizable iff GM = AM for all λ

2. The Diagonalization Process

Step-by-step diagonalization of A:

Step 1: Find eigenvalues λ₁, ..., λ_n by solving det(A − λI) = 0.

Step 2: For each eigenvalue λ, find a BASIS of the eigenspace E_λ = N(A − λI).

Step 3: Concatenate all eigenvectors as columns of P.

Step 4: Form D with eigenvalues on the diagonal in the SAME ORDER as the eigenvectors in P.

Step 5: Verify P is invertible (columns are independent; if not, A is not diagonalizable).

Step 6: Compute P⁻¹ (or just verify PDP⁻¹ = A).

Example:

$A = [1  2]
    [2  1]
$

Eigenvalues: det[1−λ 2; 2 1−λ] = (1−λ)² − 4 = λ² − 2λ − 3 = (λ−3)(λ+1) → λ = 3, −1.

λ = 3: A−3I = [−2 2; 2 −2] → x₁ = x₂ → eigenvector (1, 1). λ = −1: A+I = [2 2; 2 2] → x₁ = −x₂ → eigenvector (1, −1).

$P = [1   1]    D = [3   0]    P⁻¹ = [1/2   1/2]
    [1  -1]        [0  -1]          [1/2  -1/2]
$

Check: PDP⁻¹ = [1 1; 1 −1][3 0; 0 −1][1/2 1/2; 1/2 −1/2] = [3 −1; 3 1][1/2 1/2; 1/2 −1/2] = [1 2; 2 1] = A. ✓

3. Powers of Diagonalizable Matrices

If A = PDP⁻¹, then:

$A^k = (PDP⁻¹)(PDP⁻¹)...(PDP⁻¹) = P D^k P⁻¹
$

where D^k = diag(λ₁^k, λ₂^k, ..., λ_n^k).

This reduces computing A^k (O(k n³) directly) to O(n³) for diagonalization + O(n log k) for computing powers.

Example: For A = [1 2; 2 1] above:

$A¹⁰⁰ = P [3¹⁰⁰  0; 0  (−1)¹⁰⁰] P⁻¹
      = P [3¹⁰⁰  0; 0   1] P⁻¹
      = [1 1; 1 −1] [3¹⁰⁰ 0; 0 1] [1/2 1/2; 1/2 −1/2]
      = [(3¹⁰⁰+1)/2  (3¹⁰⁰−1)/2; (3¹⁰⁰−1)/2  (3¹⁰⁰+1)/2]
$

Long-term behavior: As k → ∞, the behavior is dominated by the eigenvalue of largest magnitude: - If |λ_max| < 1, A^k → 0 - If |λ_max| > 1, A^k diverges - If |λ_max| = 1, behavior depends on that eigenvalue

4. Diagonalization via Change of Basis

Diagonalization is fundamentally a change-of-basis operation. In the eigenbasis B, the linear transformation T(x) = Ax becomes the diagonal matrix D:

[T]_B = P⁻¹ A P = D

where P = P_{E←B} is the change-of-basis matrix from B to the standard basis E (its columns are the B-basis vectors in E-coordinates).

Geometric interpretation: In the eigenbasis, the transformation acts by simply scaling each coordinate independently. The "complicated" matrix A is just a diagonal scaling viewed from the "wrong" coordinate system.

5. Application: Systems of Differential Equations

Consider the system: dx/dt = A x, where x(t) ∈ R^n.

If A = PDP⁻¹, let y = P⁻¹ x. Then dy/dt = P⁻¹ dx/dt = P⁻¹ A x = P⁻¹ A P y = D y.

Since D is diagonal, the system decouples:

dy_i/dt = λ_i y_i  →  y_i(t) = y_i(0) e^{λ_i t}

Transforming back: x(t) = P y(t) = Σ c_i e^{λ_i t} v_i, where v_i are eigenvectors.

Example — Predator-Prey model: dx/dt = x + 2y, dy/dt = 2x + y. Matrix: A = [1 2; 2 1] (same as above).

Eigenvalues: 3, −1. Eigenvectors: (1,1), (1,−1).

General solution: x(t) = c₁ e^{3t} (1,1) + c₂ e^{−t} (1,−1) = (c₁ e^{3t} + c₂ e^{−t}, c₁ e^{3t} − c₂ e^{−t})

Phase portrait analysis: - λ = 3 > 0: unstable direction (e^{3t} grows) along (1,1) - λ = −1 < 0: stable direction (e^{−t} decays) along (1,−1) - Origin is a saddle point

6. Application: Markov Chains (Discrete Dynamics)

x_{k+1} = A x_k. If A is diagonalizable: x_k = A^k x₀ = P D^k P⁻¹ x₀.

The steady state (if it exists) corresponds to the eigenvector for λ = 1.

Example: A = [0.9 0.2; 0.1 0.8] (transition matrix).

Eigenvalues: λ₁ = 1 (eigenvector (2/3, 1/3) normalized), λ₂ = 0.7. As k → ∞: D^k → diag(1, 0), so x_∞ = P diag(1,0) P⁻¹ x₀ = projection onto the λ=1 eigenspace.

Key Terms

• 08 09 Diagonalisation
• Application: Markov Chains (Discrete Dynamics)
• Application: Systems of Differential Equations
• Correct: A) Always diagonalizable
• Correct: B) P D^100 P⁻¹
• Correct: B) P D⁻¹ P⁻¹
• Correct: B) P must be invertible
• Correct: B) The eigenvectors of A
• Correct: B) The geometric multiplicity of λ
• Correct: B) The same eigenvalues
• Correct: B) λ = 1
• Correct: C) A saddle point

Worked Examples

Example 1: Complete Diagonalization

Problem: Diagonalize A = [5 −3; 2 0] and compute A¹⁰.

Solution:

Step 1: det(A−λI) = det[5−λ −3; 2 −λ] = (5−λ)(−λ) − (−3)(2) = −5λ + λ² + 6 = λ² − 5λ + 6 = (λ−2)(λ−3).

λ = 2, 3 (distinct → automatically diagonalizable).

Step 2: λ = 2: A−2I = [3 −3; 2 −2] → 3x₁−3x₂=0 → x₁=x₂. Eigenvector: (1, 1). λ = 3: A−3I = [2 −3; 2 −3] → 2x₁−3x₂=0 → x₁ = (3/2)x₂. Choose x₂ = 2: eigenvector (3, 2).

Step 3-4:

$P = [1  3]    D = [2  0]
    [1  2]        [0  3]
$

P⁻¹ = [−2 3; 1 −1]. (Check: P P⁻¹ = I. ✓)

Step 5: A¹⁰ = P D¹⁰ P⁻¹ = [1 3; 1 2] [2¹⁰ 0; 0 3¹⁰] [−2 3; 1 −1]. = [1 3; 1 2] [−2·2¹⁰ 3·2¹⁰; 3¹⁰ −3¹⁰] (not fully expanded — but the pattern is clear).

Example 2: Non-Diagonalizable Matrix

Problem: Show A = [1 1; 0 1] is NOT diagonalizable.

Solution: det(A−λI) = (1−λ)² = 0 → λ = 1, AM = 2. A−I = [0 1; 0 0], rank = 1. GM = 2 − 1 = 1 < 2 = AM. Not enough independent eigenvectors → NOT diagonalizable.

If we try to diagonalize: the only eigenvector is (1, 0) (up to scaling). We can't form an invertible P with only one eigenvector direction.

What to do instead: Jordan canonical form: A = [1 1; 0 1] is already in Jordan form.

Example 3: Differential Equation System

Problem: Solve dx/dt = 2x + y, dy/dt = x + 2y with x(0) = 1, y(0) = 0.

Solution: A = [2 1; 1 2].

Eigenvalues: det[2−λ 1; 1 2−λ] = (2−λ)² − 1 = λ² − 4λ + 3 = (λ−1)(λ−3). λ = 1, 3.

λ = 1: A−I = [1 1; 1 1] → x₁ = −x₂. Eigenvector: (−1, 1) or normalized: (1, −1). λ = 3: A−3I = [−1 1; 1 −1] → x₁ = x₂. Eigenvector: (1, 1).

General solution: [x; y] = c₁ e^t [1; −1] + c₂ e^{3t} [1; 1].

Apply initial condition: [1; 0] = c₁[1; −1] + c₂[1; 1] = [c₁+c₂; −c₁+c₂]. → c₁ + c₂ = 1, −c₁ + c₂ = 0 → c₁ = c₂ = 1/2.

x(t) = (1/2)e^t + (1/2)e^{3t}, y(t) = −(1/2)e^t + (1/2)e^{3t}.

Example 4: Power Computation via Diagonalization

Problem: Compute A⁵ for A = [0 1; 1 1] (Fibonacci matrix).

Solution: Eigenvalues: det[−λ 1; 1 1−λ] = −λ(1−λ) − 1 = λ² − λ − 1 = 0. λ = (1±√5)/2 = φ ≈ 1.618, ψ ≈ −0.618 (where φ is the golden ratio, ψ = 1−φ = −1/φ).

Eigenvectors: For φ: (1, φ). For ψ: (1, ψ).

$P = [1   1]    D = [φ  0]   P⁻¹ = [φ/(φ−ψ)  −1/(φ−ψ)]
    [φ   ψ]        [0  ψ]         [−ψ/(φ−ψ)   1/(φ−ψ)]
$

Note φ−ψ = √5.

A⁵ = P D⁵ P⁻¹. The (1,1) entry (related to Fibonacci numbers) = (φ⁵ − ψ⁵)/√5. φ⁵ ≈ 11.090, ψ⁵ ≈ −0.090. So A⁵ ≈ [5 8; 8 13] — indeed these are consecutive Fibonacci numbers: F₅=5, F₆=8, F₇=13.

Practice Problems

(Answers are below. Try each problem before checking.)

Problem 1: Diagonalize A = [3 1; 0 2] if possible.

Problem 2: Determine whether A = [2 1 0; 0 2 1; 0 0 2] is diagonalizable.

Problem 3: For A = [0 −1; 1 0], is A diagonalizable over R? Over C?

Problem 4: Solve dx/dt = 3x, dy/dt = x + 2y, dz/dt = y − z.

Problem 5: If A = PDP⁻¹, show that A^k = P D^k P⁻¹ by induction.

Problem 6: Diagonalize A = [4 2; 2 1] and compute the limit of A^k as k → ∞ (if it exists).

Problem 7: Find a matrix P that diagonalizes A = [1 2 0; 0 3 0; 0 0 2].

Summary

1. A matrix is diagonalizable (A = PDP⁻¹) iff it has n linearly independent eigenvectors; equivalently, GM = AM for every eigenvalue — distinct eigenvalues guarantee diagonalizability
2. The diagonalization process: find eigenvalues → find eigenbases → assemble P from eigenvectors and D from eigenvalues in matching order → verify P is invertible
3. Powers simplify dramatically: A^k = P D^k P⁻¹ where D^k just raises each diagonal entry to the k-th power; the long-term behavior is governed by the dominant eigenvalue
4. Systems of linear ODEs dx/dt = Ax decouple in the eigenbasis: each coordinate independently follows dy_i/dt = λ_i y_i → y_i(t) = c_i e^{λ_i t}
5. Diagonalization is a change-of-basis operation: A is a diagonal scaling viewed from the standard basis; switching to the eigenbasis reveals the simple nature of the transformation

Pitfalls

1. Assuming distinct eigenvalues is necessary for diagonalizability. Distinct eigenvalues is a SUFFICIENT condition, not a necessary one. The identity matrix has a single repeated eigenvalue but is diagonal (trivially diagonalizable). Always check GM = AM for repeated eigenvalues.

2. Putting eigenvalues in the wrong order. In A = PDP⁻¹, the i-th column of P is the eigenvector corresponding to the i-th diagonal entry of D. If eigenvalues and eigenvectors are misaligned, the decomposition fails. Always match orders carefully.

3. Forgetting to verify that P is invertible. If the eigenvectors assembled in P are linearly dependent, P is singular and A is not diagonalizable. This happens exactly when some eigenvalue has GM < AM — a quick check of det(P) ≠ 0 confirms diagnolizability.

4. Attempting to diagonalize over R when eigenvalues are complex. A real matrix with complex eigenvalues (like a rotation) may be diagonalizable over C but not over R. The diagonalization A = PDP⁻¹ with real P requires real eigenvalues. Over R, the best you can do for complex eigenvalues is block-diagonal real canonical form.

5. Confusing similarity (P⁻¹AP) with congruence (P^T AP). Diagonalization uses similarity transformations, which preserve eigenvalues. Congruence transformations preserve inertia (signs of eigenvalues) but NOT the eigenvalues themselves. Do not use P^T when you need P⁻¹.

Quiz

Answer each question, then read the explanation for your choice.

Q1: An n × n matrix with n distinct eigenvalues is:

A) Always diagonalizable B) Sometimes diagonalizable C) Never diagonalizable D) Always singular

Q2: In A = PDP⁻¹, the columns of P are:

A) The eigenvalues of A B) The eigenvectors of A C) The diagonal entries of A D) The rows of D

Q2b: What must be true about P for diagonalization to work?

A) P must be symmetric B) P must be invertible C) P must be orthogonal D) P must be diagonal

Q3: A^100 for a diagonalizable matrix can be computed as:

A) 100·A B) P D^100 P⁻¹ C) D^100 D) (P D P⁻¹)^100

Q4: For dx/dt = Ax, the solution involves:

A) cos and sin of eigenvalues B) Exponentials e^{λt} of eigenvalues C) Polynomials in t D) All of the above, depending on A

Q5: A shear matrix [1 1; 0 1] is NOT diagonalizable because:

A) It has complex eigenvalues B) It has only one eigenvalue C) GM < AM for its eigenvalue D) It is singular

Q6: If A = PDP⁻¹, then A⁻¹ (when it exists) equals:

A) P⁻¹ D⁻¹ P B) P D⁻¹ P⁻¹ C) D⁻¹ D) P D P⁻¹

Q7: For the system dx/dt = 2x, dy/dt = −3y, the equilibrium at (0,0) is:

A) A stable node B) An unstable node C) A saddle point D) A center

Q8: In the context of Markov chains with transition matrix A, the steady-state distribution corresponds to the eigenvector for:

A) λ = 0 B) λ = 1 C) The largest eigenvalue in magnitude D) The smallest eigenvalue

Q9: The dimension of the solution space for (A − λI)x = 0 is called:

A) The algebraic multiplicity of λ B) The geometric multiplicity of λ C) The rank of λ D) The nullity of A

Q10: Similar matrices have:

A) The same eigenvectors B) The same eigenvalues C) The same entries D) The same rank but different eigenvalues

Next Steps

Move on to 08-10 — Symmetric Matrices and the Spectral Theorem to learn the crowning result of linear algebra: every real symmetric matrix is orthogonally diagonalizable with real eigenvalues, leading to the spectral theorem, quadratic forms, and positive definiteness.