Diagonalizable Matrix — Definition, Formula & Examples

A diagonalizable matrix is a square matrix that can be written in the form

PDP^{-1}

, where

D

is a diagonal matrix and

P

is an invertible matrix whose columns are the eigenvectors of the original matrix.

An $n \times n$ matrix $A$ is diagonalizable if and only if there exists an invertible matrix $P$ and a diagonal matrix $D$ such that $A = PDP^{-1}$ . Equivalently, $A$ is diagonalizable if and only if it has $n$ linearly independent eigenvectors.

Key Formula

A = PDP^{-1}

Where:

$A$ = The original n × n matrix
$P$ = Invertible matrix whose columns are linearly independent eigenvectors of A
$D$ = Diagonal matrix with eigenvalues of A on the diagonal

How It Works

To diagonalize a matrix

A

, first find its eigenvalues by solving

\det(A - \lambda I) = 0

. Then find the eigenvectors for each eigenvalue. If you obtain

n

linearly independent eigenvectors for an

n \times n

matrix, form

P

by placing those eigenvectors as columns and

D

by placing the corresponding eigenvalues along the diagonal. The factorization

A = PDP^{-1}

makes computing powers of

A

especially efficient, since

A^k = PD^kP^{-1}

and raising a diagonal matrix to a power simply raises each diagonal entry to that power.

Worked Example

Problem: Determine whether the matrix A is diagonalizable, and if so, find P and D.

Given: Let A be the following 2×2 matrix.

A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}

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

\det\begin{pmatrix} 4-\lambda & 1 \\ 2 & 3-\lambda \end{pmatrix} = (4-\lambda)(3-\lambda) - 2 = \lambda^2 - 7\lambda + 10 = (\lambda - 5)(\lambda - 2) = 0

Step 2: The eigenvalues are λ₁ = 5 and λ₂ = 2. Find an eigenvector for each. For λ₁ = 5, solve (A − 5I)v = 0.

\begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}\mathbf{v} = \mathbf{0} \implies \mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}

Step 3: For λ₂ = 2, solve (A − 2I)v = 0.

\begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}\mathbf{v} = \mathbf{0} \implies \mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}

Step 4: Since there are 2 linearly independent eigenvectors for a 2×2 matrix, A is diagonalizable. Form P and D.

P = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}, \quad D = \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}

Answer: A is diagonalizable with P = [[1,1],[1,−2]] and D = diag(5, 2), so A = PDP⁻¹.

Why It Matters

Diagonalization simplifies computing matrix powers, which appears in solving systems of linear differential equations, analyzing Markov chains, and modeling population dynamics. Many numerical algorithms in data science and engineering rely on diagonalizing matrices (or the closely related spectral decomposition) to reduce computational cost.

Common Mistakes

Mistake: Assuming every square matrix is diagonalizable.

Correction: A matrix is diagonalizable only if it has n linearly independent eigenvectors. For example, some matrices with repeated eigenvalues lack enough independent eigenvectors and are therefore not diagonalizable (these are called defective matrices).