Linear Algebra
7.512 min read

Eigenspaces and Multiplicity

A single eigenvalue can appear multiple times as a root of the characteristic polynomial. The number of times λ\lambda appears as a root is its algebraic multiplicity. For example, if p(λ)=(λ−3)2(λ−1)p(\lambda) = (\lambda - 3)^2(\lambda - 1), then λ=3\lambda = 3 has algebraic multiplicity 2 and λ=1\lambda = 1 has algebraic multiplicity 1.

The geometric multiplicity of λ\lambda is the dimension of the eigenspace Eλ=Null(A−λI)E_\lambda = \text{Null}(A - \lambda I). It tells you how many linearly independent eigenvectors correspond to λ\lambda. Geometric multiplicity is always at least 1 (since λ\lambda is an eigenvalue) and at most the algebraic multiplicity.

The relationship between these two multiplicities determines whether a matrix is diagonalizable. When geometric multiplicity equals algebraic multiplicity for every eigenvalue, you have enough linearly independent eigenvectors to form a basis — and the matrix is diagonalizable. When geometric multiplicity is strictly less, the matrix is said to be defective and cannot be diagonalized.

Formal View

Definition 7.9 — Algebraic and Geometric Multiplicity
For an eigenvalue λ\lambda of AA: the algebraic multiplicity is its multiplicity as a root of the characteristic polynomial p(λ)p(\lambda). The geometric multiplicity is dim⁡(Eλ)=dim⁡(Null(A−λI))\dim(E_\lambda) = \dim(\text{Null}(A - \lambda I)).
Theorem 7.10 — Multiplicity Inequality
For every eigenvalue λ\lambda:
1≤geometric multiplicity(λ)≤algebraic multiplicity(λ).1 \leq \text{geometric multiplicity}(\lambda) \leq \text{algebraic multiplicity}(\lambda).

The lower bound holds because λ\lambda is an eigenvalue (so Eλ≠{0}E_\lambda \neq \{\mathbf{0}\}). The upper bound is a deeper result from the Rank-Nullity theorem applied to powers of A−λIA - \lambda I.

Example 7.11
Compare A=(3003)A = \begin{pmatrix}3 & 0 \\ 0 & 3\end{pmatrix} and B=(3103)B = \begin{pmatrix}3 & 1 \\ 0 & 3\end{pmatrix}. Both have characteristic polynomial (λ−3)2(\lambda - 3)^2, so λ=3\lambda = 3 has algebraic multiplicity 2 in both. For AA: E3=Null(A−3I)=Null(0)=R2E_3 = \text{Null}(A - 3I) = \text{Null}(\mathbf{0}) = \mathbb{R}^2, so geometric multiplicity is 2 — AA is already diagonal. For BB: B−3I=(0100)B - 3I = \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix} has nullity 1, so geometric multiplicity is 1 — BB is defective.

Why This Matters

Multiplicity is the key quantity that separates well-behaved diagonalizable matrices from defective ones.

  • Control theory: repeated eigenvalues at the stability boundary (eigenvalue = 0) require checking geometric multiplicity to determine system behavior
  • Numerical linear algebra: defective matrices are ill-conditioned for eigenvalue computation and require special treatment
  • Differential equations: repeated eigenvalues with defective matrices generate polynomial-times-exponential solutions
  • Google PageRank requires the dominant eigenvalue (1) to have geometric multiplicity 1 for uniqueness of the ranking vector

Quiz

Question 1

The characteristic polynomial is (λ−4)2(λ−2)(\lambda - 4)^2(\lambda - 2). What is the algebraic multiplicity of λ=4\lambda = 4?

Question 2

The geometric multiplicity of an eigenvalue can exceed its algebraic multiplicity.

Question 3

An n×nn \times n matrix has nn distinct eigenvalues. What can you conclude about the geometric multiplicity of each?

Common Mistakes

  • Confusing algebraic multiplicity (a property of the polynomial) with geometric multiplicity (a property of the matrix).
  • Assuming repeated eigenvalues always mean defective matrices — A=3IA = 3I has λ=3\lambda = 3 repeated nn times, yet every vector is an eigenvector.
  • Forgetting that geometric multiplicity is always at least 1 for an actual eigenvalue.