Diagonalization — Mathematics Knowledge System

Overview

A square matrix A is diagonalizable if it can be written as A = PDP⁻¹, where D is diagonal and P's columns are eigenvectors of A. Diagonalization exists exactly when A has n linearly independent eigenvectors (for an n×n matrix), which is guaranteed, for instance, whenever A has n distinct eigenvalues. Once diagonalized, computing powers of A becomes trivial — Aᵏ = PDᵏP⁻¹, and raising a diagonal matrix to a power just means raising each diagonal entry to that power — which makes diagonalization a central tool for solving recurrences, systems of differential equations, and Markov chain long-run behavior.

Intuition

Diagonalization finds a special coordinate system — the eigenvector directions — in which the transformation A acts by simple independent scaling along each axis, with no mixing between directions. In the original coordinates, applying A repeatedly can look complicated, twisting and stretching in a tangled way; but change to eigenvector coordinates via P, and A becomes just D, a diagonal matrix that scales each axis independently. That's why Aᵏ is easy once diagonalized: scaling each independent axis k times is just raising each scale factor to the k-th power.

Formal Definition

Definition

An n×n matrix A is diagonalizable if there exists an invertible matrix P and diagonal matrix D such that:

A = PDP^{-1}

Diagonalization

P = [v_1\ v_2\ \cdots\ v_n], \qquad D = \operatorname{diag}(\lambda_1,\ldots,\lambda_n)

P's columns are eigenvectors, D holds the eigenvalues

A^k = P D^k P^{-1}, \qquad D^k = \operatorname{diag}(\lambda_1^k,\ldots,\lambda_n^k)

Fast matrix powers

Worked Examples

Since A is upper triangular, its eigenvalues are the diagonal entries.
$\lambda_1 = 3, \quad \lambda_2 = 2$
For λ=3: solve (A-3I)v=0, i.e. [[0,1],[0,-1]]v=0, giving v_2=0, v_1 free.
$v_1 = (1, 0)$
For λ=2: solve (A-2I)v=0, i.e. [[1,1],[0,0]]v=0, giving v_1+v_2=0.
$v_2 = (1, -1)$

Answer: A = PDP⁻¹ with P = [[1,1],[0,-1]], D = [[3,0],[0,2]].

Practice Problems

Difficulty 6/10

Diagonalize A = [[4, 0], [0, -1]] (already diagonal). What are P and D?

Difficulty 7/10

Find the eigenvalues of A = [[2,1],[1,2]] and state whether A is diagonalizable.

Difficulty 7/10

A population model has state vector x_{t+1} = A x_t where A = [[3,1],[0,2]] (from the worked example). Using A = PDP⁻¹ with D = diag(3,2), describe qualitatively what happens to the population as t → ∞, and why.

Common Mistakes

Common Mistake

Assuming every square matrix is diagonalizable.

A matrix is diagonalizable only if it has n linearly independent eigenvectors. Repeated eigenvalues can fail to produce enough independent eigenvectors (a 'defective' matrix), e.g. [[1,1],[0,1]] has only one independent eigenvector for its repeated eigenvalue 1.

Quiz

Summary

A is diagonalizable if A = PDP⁻¹ with D diagonal and P's columns independent eigenvectors of A.
Distinct eigenvalues guarantee diagonalizability; repeated eigenvalues may or may not, depending on eigenspace dimension.
Diagonalization makes computing Aᵏ = PDᵏP⁻¹ easy, powering long-run analysis of dynamical systems and Markov chains.

References

WebsiteWikipedia — Diagonalizable matrix