Lesson 3: Eigenspaces and Diagonalization
Estimated time: 45-55 minutes
Learning Objectives
- Define and compute eigenspaces for each eigenvalue
- State the conditions for a matrix to be diagonalizable
- Carry out the diagonalization procedure: find P and D such that A = PDP^{-1}
- Recognize when diagonalization fails and why
Eigenspaces
Eigenspace: For eigenvalue lambda, the eigenspace is E_lambda = null(A - lambda I) = {v : Av = lambda v}. It is a subspace of R^n containing all eigenvectors for lambda (plus the zero vector).
Worked Example 1
A = [4 -1 6; 2 1 6; 2 -1 8]. Suppose eigenvalues are lambda = 2 (mult 2) and lambda = 9 (mult 1).
E_2: Row reduce A - 2I = [2 -1 6; 2 -1 6; 2 -1 6] → [2 -1 6; 0 0 0; 0 0 0].
Two free variables (x2, x3). x1 = (x2 - 6x3)/2. Basis: {(1/2, 1, 0), (-3, 0, 1)} or equivalently {(1, 2, 0), (-3, 0, 1)}.
Geometric multiplicity of lambda = 2 is 2 (matches algebraic multiplicity).
What Is Diagonalization?
Diagonalizable: An n x n matrix A is diagonalizable if there exists an invertible matrix P and diagonal matrix D such that A = PDP^{-1} (equivalently, P^{-1}AP = D).
The columns of P are the eigenvectors of A, and the diagonal entries of D are the corresponding eigenvalues.
Why Diagonalize?
Diagonal matrices are trivial to work with. If A = PDP^{-1}, then A^n = PD^nP^{-1}, and D^n just raises each diagonal entry to the nth power. This makes computing matrix powers fast and elegant.
The Diagonalization Theorem
Diagonalization Theorem: An n x n matrix A is diagonalizable if and only if it has n linearly independent eigenvectors. Equivalently:
- A is diagonalizable if and only if the sum of the geometric multiplicities equals n
- A is diagonalizable if the geometric multiplicity equals the algebraic multiplicity for every eigenvalue
Sufficient condition: If an n x n matrix has n distinct eigenvalues, it is automatically diagonalizable (eigenvectors from distinct eigenvalues are always linearly independent).
The Diagonalization Procedure
Step 1: Find all eigenvalues by solving det(A - lambda I) = 0.
Step 2: For each eigenvalue, find a basis for its eigenspace by solving (A - lambda I)v = 0.
Step 3: Check: do the total number of basis vectors equal n? If yes, A is diagonalizable.
Step 4: Form P = [v1 | v2 | ... | vn] with eigenvectors as columns. Form D with eigenvalues on the diagonal (in the same order as the corresponding eigenvectors in P).
Worked Example 2: Full Diagonalization
Diagonalize A = [1 2; 0 3].
Step 1: A is upper triangular. Eigenvalues: lambda_1 = 1, lambda_2 = 3.
Step 2 (lambda = 1): A - I = [0 2; 0 2] → [0 1; 0 0]. x2 = 0, x1 free. v1 = (1, 0).
Step 2 (lambda = 3): A - 3I = [-2 2; 0 0] → [1 -1; 0 0]. x1 = x2. v2 = (1, 1).
Step 3: Two eigenvectors for a 2x2 matrix. Diagonalizable!
Step 4: P = [1 1; 0 1], D = [1 0; 0 3]. Then A = PDP^{-1}.
Check: P^{-1} = [1 -1; 0 1]. PDP^{-1} = [1 1; 0 1][1 0; 0 3][1 -1; 0 1] = [1 2; 0 3] = A.
Worked Example 3: 3x3 Diagonalization
Diagonalize A = [2 0 0; 1 2 1; -1 0 1].
Step 1: det(A - lambda I) = (2-lambda)[(2-lambda)(1-lambda) - 0] = (2-lambda)^2(1-lambda). Eigenvalues: lambda = 2 (mult 2), lambda = 1 (mult 1).
Step 2 (lambda = 2): A - 2I = [0 0 0; 1 0 1; -1 0 -1] → [1 0 1; 0 0 0; 0 0 0]. x2, x3 free. Basis: {(0,1,0), (-1,0,1)}. Geo mult = 2.
Step 2 (lambda = 1): A - I = [1 0 0; 1 1 1; -1 0 0] → [1 0 0; 0 1 1; 0 0 0]. x3 free. Basis: {(0,-1,1)}. Geo mult = 1.
Step 3: Total eigenvectors: 2 + 1 = 3 = n. Diagonalizable!
Step 4: P = [0 -1 0; 1 0 -1; 0 1 1], D = [2 0 0; 0 2 0; 0 0 1].
When Diagonalization Fails
Worked Example 4: Not Diagonalizable
A = [3 1; 0 3]. Eigenvalue lambda = 3 with algebraic multiplicity 2.
A - 3I = [0 1; 0 0]. Only one free variable, so geometric multiplicity = 1.
We can only find ONE linearly independent eigenvector, but we need 2. A is NOT diagonalizable.
A matrix fails to be diagonalizable precisely when some eigenvalue has geometric multiplicity strictly less than its algebraic multiplicity -- there are not enough eigenvectors.
Check Your Understanding
1. A 3x3 matrix has eigenvalues 2, 2, 5 and the eigenspace for lambda=2 is one-dimensional. Is A diagonalizable?
2. A 4x4 matrix has four distinct eigenvalues. Is it diagonalizable?
3. If A = PDP^{-1} with D = [2 0; 0 7], what is A^3?
Key Takeaways
- Eigenspace E_lambda = null(A - lambda I) is a subspace
- A is diagonalizable iff it has n linearly independent eigenvectors
- n distinct eigenvalues guarantees diagonalizability
- Procedure: find eigenvalues → eigenspaces → P = [eigenvectors], D = diag(eigenvalues)
- A = PDP^{-1} implies A^n = PD^nP^{-1}
- Fails when geometric multiplicity < algebraic multiplicity for some eigenvalue