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Eigen Decomposition — Definition, Formula & Examples

Eigen decomposition (also called eigendecomposition) is a way of breaking a square matrix into a product of three matrices built from its eigenvalues and eigenvectors. It reveals the fundamental stretching directions and magnitudes encoded in the matrix.

A square matrix A∈Rn×nA \in \mathbb{R}^{n \times n} that has nn linearly independent eigenvectors admits the factorization A=PDP−1A = PDP^{-1}, where PP is the matrix whose columns are the eigenvectors of AA and DD is the diagonal matrix whose diagonal entries are the corresponding eigenvalues.

Key Formula

A=PDP−1A = PDP^{-1}
Where:
  • AA = The original n×n square matrix
  • PP = Matrix whose columns are the linearly independent eigenvectors of A
  • DD = Diagonal matrix with the eigenvalues of A on the diagonal
  • P−1P^{-1} = The inverse of the eigenvector matrix P

How It Works

First, find the eigenvalues of AA by solving the characteristic equation det⁡(A−λI)=0\det(A - \lambda I) = 0. Next, for each eigenvalue λi\lambda_i, solve (A−λiI)x=0(A - \lambda_i I)\mathbf{x} = \mathbf{0} to get the eigenvector vi\mathbf{v}_i. Form PP by placing these eigenvectors as columns, and form DD by placing the eigenvalues along the diagonal. Verify that A=PDP−1A = PDP^{-1}. This decomposition exists only when AA has nn linearly independent eigenvectors (i.e., AA is diagonalizable).

Worked Example

Problem: Find the eigen decomposition of A = [[4, 1], [2, 3]].
Find eigenvalues: Solve det(A − λI) = 0.
det⁡(4−λ123−λ)=(4−λ)(3−λ)−2=λ2−7λ+10=0\det\begin{pmatrix}4-\lambda & 1\\2 & 3-\lambda\end{pmatrix} = (4-\lambda)(3-\lambda)-2 = \lambda^2 - 7\lambda + 10 = 0
Solve for λ: Factor the characteristic polynomial.
(λ−5)(λ−2)=0  ⇒  λ1=5,  λ2=2(\lambda - 5)(\lambda - 2) = 0 \;\Rightarrow\; \lambda_1 = 5,\; \lambda_2 = 2
Find eigenvectors: For λ₁ = 5: (A − 5I)v = 0 gives v₁ = (1, 1). For λ₂ = 2: (A − 2I)v = 0 gives v₂ = (1, −2).
P=(111−2),D=(5002)P = \begin{pmatrix}1 & 1\\1 & -2\end{pmatrix},\quad D = \begin{pmatrix}5 & 0\\0 & 2\end{pmatrix}
Write decomposition: Compute P⁻¹ and express A = PDP⁻¹.
P−1=1−3(−2−1−11)=(231313−13)P^{-1} = \frac{1}{-3}\begin{pmatrix}-2 & -1\\-1 & 1\end{pmatrix} = \begin{pmatrix}\frac{2}{3} & \frac{1}{3}\\\frac{1}{3} & -\frac{1}{3}\end{pmatrix}
Answer: A = PDP⁻¹ with P = [[1, 1], [1, −2]], D = diag(5, 2), confirming the eigen decomposition.

Why It Matters

Eigen decomposition makes computing matrix powers trivial: Ak=PDkP−1A^k = PD^kP^{-1}, since raising a diagonal matrix to a power is just raising each diagonal entry. This is essential in solving systems of differential equations, principal component analysis in data science, and stability analysis in engineering.

Common Mistakes

Mistake: Assuming every square matrix can be eigen-decomposed.
Correction: A matrix must have n linearly independent eigenvectors (be diagonalizable) for the decomposition to exist. Defective matrices—those with repeated eigenvalues that lack a full set of eigenvectors—cannot be eigen-decomposed.