Eigenvalues, Eigenvectors and the Characteristic Polynomial

An eigenvector of a square matrix A is a nonzero vector v with Av = λv for some number λ, called the eigenvalue; the eigenvalues are the roots of the characteristic polynomial det(A − λI) = 0. In brief: for A = [[4, 1], [2, 3]] the characteristic polynomial is λ² − 7λ + 10 = (λ − 5)(λ − 2), so the eigenvalues are 5 and 2, with the eigenvectors (1, 1) and (1, −2). The eigenvalues add up to the trace (7) and multiply to the determinant (10).

An eigenvector is a direction that the matrix only stretches (or flips), without turning it. Eigenvalues describe the long-term behavior of dynamical systems, vibration frequencies and the principal axes of data. This guide explains how to find eigenvalues and eigenvectors, the role of multiplicities, the basic properties, and some applications, with worked examples and a table.

What are eigenvalues and eigenvectors?

The equation Av = λv with v ≠ 0 can be rewritten as (A − λI)v = 0. A nonzero solution exists exactly when A − λI is singular, that is det(A − λI) = 0. The left side is a polynomial of degree n in λ, the characteristic polynomial p(λ). For a 2 × 2 matrix, p(λ) = λ² − (tr A)λ + det A. For each eigenvalue λ, the set of all solutions of (A − λI)v = 0 (with the zero vector included) is a subspace, the eigenspace.

How do you compute them?

First solve the characteristic equation for λ, then, for each root, solve the homogeneous system (A − λI)v = 0 using row reduction.

Example 1: a 2 × 2 matrix

A = [[4, 1], [2, 3]]. p(λ) = (4 − λ)(3 − λ) − 2 = λ² − 7λ + 10, so λ = 5 or λ = 2. For λ = 5: A − 5I = [[−1, 1], [2, −2]], giving x = y, and v = (1, 1). For λ = 2: A − 2I = [[2, 1], [2, 1]], giving y = −2x, and v = (1, −2). Check: A(1, −2) = (4 − 2, 2 − 6) = (2, −4) = 2(1, −2) ✓.

Example 2: complex eigenvalues

The rotation R = [[0, −1], [1, 0]] has p(λ) = λ² + 1, so λ = ±i. A real matrix may have no real eigenvalue: a rotation by 90° leaves no direction of the plane unchanged. The eigenvector for λ = i: (R − iI)v = 0 gives −ix − y = 0, so v = (1, −i).

Example 3: a 3 × 3 matrix

A = [[2, 0, 0], [0, 3, 4], [0, 4, 9]]. p(λ) = (2 − λ)((3 − λ)(9 − λ) − 16) = (2 − λ)(λ² − 12λ + 11) = (2 − λ)(λ − 1)(λ − 11). The eigenvalues are 2, 1, 11. Eigenvectors: for 2, (1, 0, 0). For 1, the block [[2, 4], [4, 8]] gives (0, 2, −1); check: A(0, 2, −1) = (0, 6 − 4, 8 − 9) = (0, 2, −1) ✓. For 11, the block [[−8, 4], [4, −2]] gives (0, 1, 2); check: A(0, 1, 2) = (0, 3 + 8, 4 + 18) = (0, 11, 22) = 11(0, 1, 2) ✓.

Example 4: a triangular matrix

For [[1, 2, 3], [0, 4, 5], [0, 0, 6]], the characteristic polynomial is (1 − λ)(4 − λ)(6 − λ), so the eigenvalues are the diagonal entries 1, 4, 6.

What are algebraic and geometric multiplicity?

The algebraic multiplicity of an eigenvalue is its multiplicity as a root of p(λ). The geometric multiplicity is the dimension of its eigenspace. Always 1 ≤ geometric ≤ algebraic. When they differ, the matrix is defective.

Example 5: repeated eigenvalues

I₂ has p(λ) = (λ − 1)², and every nonzero vector is an eigenvector, so the geometric multiplicity is 2. But N = [[2, 1], [0, 2]] has p(λ) = (λ − 2)², and (N − 2I)v = 0 means y = 0, so the eigenspace is spanned by (1, 0): algebraic multiplicity 2, geometric multiplicity 1. This matrix is defective and cannot be diagonalized.

What are the main properties?

• The sum of the eigenvalues (with multiplicity) is tr A, and their product is det A.
• A is invertible if and only if 0 is not an eigenvalue.
• The eigenvalues of A^k are λ^k, those of A⁻¹ are 1/λ, and those of A + cI are λ + c; the eigenvectors stay the same.
• A and A^T have the same characteristic polynomial; similar matrices too.
• Eigenvectors belonging to distinct eigenvalues are linearly independent.
• A real symmetric matrix has only real eigenvalues, and eigenvectors for distinct eigenvalues are orthogonal.

Example 6: using the properties

For A of Example 1, tr A = 7 = 5 + 2 and det A = 10 = 5 · 2 ✓. The matrix A² = [[18, 7], [14, 11]] has the eigenvalues 25 and 4 (trace 29 = 25 + 4, determinant 198 − 98 = 100 = 25 · 4 ✓), and A⁻¹ has the eigenvalues 1/5 and 1/2.

Example 7: a symmetric matrix

S = [[2, 1], [1, 2]]: p(λ) = λ² − 4λ + 3, so λ = 3 and λ = 1, with the eigenvectors (1, 1) and (1, −1), which are orthogonal: 1 − 1 = 0, as the dot product confirms.

Where do eigenvalues appear in applications?

In a Markov chain, the steady state is the eigenvector for the eigenvalue 1. In the Fibonacci recurrence, the matrix [[1, 1], [1, 0]] has the characteristic polynomial λ² − λ − 1, with the eigenvalues (1 ± √5)/2, the golden ratio and its conjugate, which give the growth rate of the sequence. In mechanics, eigenvalues of the stiffness matrix give vibration frequencies, and in statistics, the eigenvectors of the covariance matrix give the principal components.

Example 8: a steady state

A weather model: from sunny, the next day is sunny with probability 0.9, and from rainy, sunny with probability 0.5. The transition matrix (columns sum to 1) is T = [[0.9, 0.5], [0.1, 0.5]]. p(λ) = λ² − 1.4λ + 0.4 = (λ − 1)(λ − 0.4). For λ = 1: (T − I)v = 0 gives −0.1x + 0.5y = 0, so v = (5, 1). Normalizing to probabilities: (5/6, 1/6): in the long run, 5/6 of the days are sunny.

Table of examples

Where does the concept come from?

The roots of the theory lie in the study of principal axes of quadratic forms and of systems of differential equations by Euler, Lagrange and Cauchy in the eighteenth and early nineteenth centuries; Cauchy treated the characteristic equation of symmetric matrices in 1829. The German prefix “eigen” (own, characteristic) became standard after David Hilbert’s work on integral equations in the early 1900s. The name “characteristic” polynomial remains common.

Which mistakes are common?

Accepting the zero vector

An eigenvector must be nonzero. The zero vector satisfies Av = λv for every λ, but it is excluded by definition.

Confusing A − λI with A − λ

Subtract λ only from the diagonal: the scalar λ stands for the matrix λI.

Forgetting to check the eigenvectors

After solving (A − λI)v = 0, verify Av = λv. Rows of A − λI must be dependent; if row reduction gives the identity, the eigenvalue was computed incorrectly.

Assuming enough eigenvectors exist

A matrix with a repeated eigenvalue can lack a full set of eigenvectors (Example 5).