Diagonalization

Definition and Concept

A square matrix A is diagonalizable if there exists an invertible matrix P and a diagonal matrix D such that:

P⁻¹AP = D

The diagonal entries of D are the eigenvalues of A
The columns of P are the corresponding eigenvectors of A
We say that P diagonalizes A

Criteria for Diagonalizability

A square n×n matrix A is diagonalizable if and only if one of the following equivalent conditions is met:

A has n linearly independent eigenvectors
For each eigenvalue λ of A, the geometric multiplicity equals the algebraic multiplicity
The sum of the dimensions of all eigenspaces equals n

Diagonalization Process

Step 1: Find the Eigenvalues

Compute the characteristic polynomial: det(A - λI) = 0
Solve this polynomial equation to find all eigenvalues λ₁, λ₂, …, λₙ
Note the algebraic multiplicity of each eigenvalue (how many times it appears as a root)

Step 2: Find the Eigenvectors

For each eigenvalue λᵢ, solve the homogeneous system (A - λᵢI)v = 0
Find a basis for the null space of (A - λᵢI), which forms a basis for the eigenspace
Note the geometric multiplicity (dimension of the eigenspace)

Step 3: Check Diagonalizability

Compare the algebraic and geometric multiplicities for each eigenvalue
If they match for all eigenvalues, the matrix is diagonalizable
Additionally, check that the total number of linearly independent eigenvectors equals n

Step 4: Construct the Matrix P

Form P by using the eigenvectors as columns
The order of eigenvectors should match the order of eigenvalues in D

Step 5: Form the Diagonal Matrix D

D = diag(λ₁, λ₂, …, λₙ)
Each eigenvalue appears on the diagonal as many times as its algebraic multiplicity

Step 6: Verify

Compute P⁻¹AP and confirm it equals D

Examples

Example 1: 2×2 Matrix

For A = $\begin{bmatrix} 3 & 8 \ 2 & 3 \end{bmatrix}$:

Step 1: Find eigenvalues

Characteristic equation: det(A - λI) = (3-λ)² - 16 = 0
Solving: λ² - 6λ + 9 - 16 = 0 → λ² - 6λ - 7 = 0
Eigenvalues: λ₁ = 7, λ₂ = -1

Step 2: Find eigenvectors

For λ₁ = 7:
- (A - 7I)v = $\begin{bmatrix} -4 & 8 \ 2 & -4 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}$
- This gives -4v₁ + 8v₂ = 0, so v₁ = 2v₂
- Let v₂ = 1, then v₁ = 2
- Eigenvector: v₁ = $\begin{bmatrix} 2 \ 1 \end{bmatrix}$
For λ₂ = -1:
- (A + I)v = $\begin{bmatrix} 4 & 8 \ 2 & 4 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}$
- This gives 4v₁ + 8v₂ = 0, so v₁ = -2v₂
- Let v₂ = 1, then v₁ = -2
- Eigenvector: v₂ = $\begin{bmatrix} -2 \ 1 \end{bmatrix}$

Step 3: Check diagonalizability

We found 2 linearly independent eigenvectors for a 2×2 matrix
The matrix is diagonalizable

Step 4: Construct P

P = $\begin{bmatrix} 2 & -2 \ 1 & 1 \end{bmatrix}$

Step 5: Form D

D = $\begin{bmatrix} 7 & 0 \ 0 & -1 \end{bmatrix}$

Step 6: Verify

Calculate P⁻¹AP = D

Example 2: 3×3 Matrix with Repeated Eigenvalue

For A = $\begin{bmatrix} 2 & 0 & 0 \ 0 & 1 & -1 \ 0 & -1 & 1 \end{bmatrix}$:

Step 1: Find eigenvalues

Characteristic equation: det(A - λI) = (2-λ)[(1-λ)² - 1] = 0
Eigenvalues: λ₁ = 2, λ₂ = 0, λ₃ = 2

Step 2: Find eigenvectors

For λ₁ = 2:
- (A - 2I)v = $\begin{bmatrix} 0 & 0 & 0 \ 0 & -1 & -1 \ 0 & -1 & -1 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \ v_3 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix}$
- This gives v₂ + v₃ = 0, so v₃ = -v₂
- Let v₁ = 1, v₂ = 0, v₃ = 0
- First eigenvector: v₁ = $\begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix}$
λ₁ = 2 has algebraic multiplicity 2, so we need another linearly independent eigenvector:
- Let v₁ = 0, v₂ = 1, v₃ = -1
- Second eigenvector: v₂ = $\begin{bmatrix} 0 \ 1 \ -1 \end{bmatrix}$
For λ₂ = 0:
- (A)v = $\begin{bmatrix} 2 & 0 & 0 \ 0 & 1 & -1 \ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} v_1 \ v_2 \ v_3 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix}$
- This gives 2v₁ = 0, v₂ - v₃ = 0, -v₂ + v₃ = 0
- So v₁ = 0, v₂ = v₃
- Let v₂ = v₃ = 1
- Eigenvector: v₃ = $\begin{bmatrix} 0 \ 1 \ 1 \end{bmatrix}$

Step 3: Check diagonalizability

We found 3 linearly independent eigenvectors for a 3×3 matrix
The matrix is diagonalizable

Step 4 & 5: Construct P and D

P = $\begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 1 \ 0 & -1 & 1 \end{bmatrix}$
D = $\begin{bmatrix} 2 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 0 \end{bmatrix}$

Special Cases and Properties

1. Symmetric Matrices

If A is symmetric (A = A^T):

A is always diagonalizable
All eigenvalues are real
Eigenvectors corresponding to distinct eigenvalues are orthogonal
A complete set of orthogonal eigenvectors always exists
P can be chosen to be orthogonal (P^T = P^(-1)), resulting in an orthogonal diagonalization

2. Triangular Matrices

For a triangular matrix:

The eigenvalues are the diagonal entries
It is diagonalizable if and only if all eigenvalues have geometric multiplicity equal to their algebraic multiplicity

3. Defective Matrices

A matrix is defective if it is not diagonalizable:

This happens when there are not enough linearly independent eigenvectors
For some eigenvalue, the geometric multiplicity is less than the algebraic multiplicity
This requires more advanced techniques like Jordan canonical form

Applications of Diagonalization

Matrix Powers: If A = PDP^(-1), then A^n = PD^nP^(-1)
Solving Systems of Differential Equations: ẋ = Ax can be solved using diagonalization
Spectral Decomposition: A = ∑λᵢvᵢvᵢ^T (for symmetric matrices)
Computing Matrix Functions: f(A) = Pf(D)P^(-1)

Common Techniques and Tricks

1. Recognizing Diagonalizability

If a matrix has n distinct eigenvalues, it is diagonalizable
If a matrix is symmetric, it is diagonalizable
If the characteristic polynomial splits into linear factors and the geometric multiplicity equals algebraic multiplicity for each eigenvalue, the matrix is diagonalizable

2. Dealing with Repeated Eigenvalues

For an eigenvalue λ with algebraic multiplicity m, check if the null space of (A - λI) has dimension m
If dimension < m, the matrix is not diagonalizable
To find multiple linearly independent eigenvectors for the same eigenvalue, solve (A - λI)v = 0 and find a basis for the null space

3. Verification Short-cuts

To verify diagonalization, it’s often easier to check AP = PD instead of P^(-1)AP = D
This avoids having to compute P^(-1)

Solving Diagonalization Exercises

Strategy 1: For 2×2 Matrices

Calculate the determinant and trace to find eigenvalues quickly:
- λ₁ + λ₂ = tr(A)
- λ₁ × λ₂ = det(A)
- Solve x² - tr(A)x + det(A) = 0
Find eigenvectors directly from (A - λI)v = 0
Form P and D, verify AP = PD

Strategy 2: Exploiting Special Structures

For matrices with special structures:
- Symmetric: use orthogonality properties
- Triangular: easy eigenvalues from diagonal
- Block diagonal: work with each block separately
Look for patterns that make calculations easier

Strategy 3: Checking Diagonalizability Directly

Compute the rank of (A - λI) for each eigenvalue λ
Compare n - rank(A - λI) with the algebraic multiplicity of λ
If they’re equal for all eigenvalues, A is diagonalizable

Exercises

Consider $A = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & -1 & 1 \end{pmatrix} \in \mathbb{R}^{3 \times 3}$ Establish whether $A$ is diagonalizable. If yes, find an invertible matrix $P \in \mathbb{R}^{3 \times 3}$ and a diagonal matrix $D \in \mathbb{R}^{3 \times 3}$ such that $P^{-1}AP = D$.
Consider $A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 3 & 0 \\ 1 & 0 & 1 \end{pmatrix} \in \mathbb{R}^{3 \times 3}$ Establish whether $A$ is diagonalizable. If yes, find an invertible matrix $P \in \mathbb{R}^{3 \times 3}$ and a diagonal matrix $D \in \mathbb{R}^{3 \times 3}$ such that $P^{-1}AP = D$.
Consider $A = \begin{pmatrix} -2 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{pmatrix} \in \mathbb{R}^{3 \times 3}$ Establish whether $A$ is diagonalizable. If yes, find an invertible matrix $P \in \mathbb{R}^{3 \times 3}$ and a diagonal matrix $D \in \mathbb{R}^{3 \times 3}$ such that $P^{-1}AP = D$.
Check if the following matrices $A_i$ are diagonalizable over $\mathbb{R}$ and, if they exist, find for each $A_i$ a diagonal matrix $D$ and an invertible matrix $C$ such that $C^{-1}A_iC = D$:
- $A_1 = \begin{pmatrix} 3 & 8 \ 2 & 3 \end{pmatrix}$
- $A_2 = \begin{pmatrix} 1 & -3 \ 5 & 4 \end{pmatrix}$
- $A_3 = \begin{pmatrix} 1 & 0 \ 5 & 3 \end{pmatrix}$
- $A_4 = \begin{pmatrix} 1 & -3 & 3 \ 3 & -5 & 3 \ 6 & -6 & 4 \end{pmatrix}$
- $A_5 = \begin{pmatrix} 2 & 1 & 0 \ 0 & 1 & -1 \ 0 & 2 & 4 \end{pmatrix}$
Find the eigenvalues of the following matrix: $A = \begin{pmatrix} 2 & 1 & 0 \ 0 & 1 & -1 \ 0 & 2 & 4 \end{pmatrix}$