Understanding the behavior of linear transformations is central to linear algebra, and few concepts illuminate this behavior as clearly as the algebraic and geometric multiplicity of eigenvalues. While they sound similar, they measure fundamentally different properties: one counts roots in a polynomial, the other counts dimensions in a vector space. These two numbers, attached to every eigenvalue of a matrix, dictate whether a matrix can be diagonalized, how a dynamical system evolves over time, and the structure of the solution space for differential equations. Mastering the distinction between them is the gateway to advanced topics like the Jordan Canonical Form and spectral theory.
The Foundation: Eigenvalues and the Characteristic Polynomial
Before diving into multiplicities, we must establish the source of eigenvalues. For a square matrix $A$ of size $n \times n$, an eigenvalue $\lambda$ is a scalar satisfying the equation $A\mathbf{v} = \lambda\mathbf{v}$ for some non-zero vector $\mathbf{v}$ (the eigenvector). Rearranging this gives $(A - \lambda I)\mathbf{v} = \mathbf{0}$. For a non-trivial solution $\mathbf{v}$ to exist, the matrix $(A - \lambda I)$ must be singular, meaning its determinant is zero.
This leads to the characteristic polynomial: $p(\lambda) = \det(A - \lambda I)$ The roots of this polynomial are precisely the eigenvalues of $A$. Since $p(\lambda)$ is a polynomial of degree $n$, the Fundamental Theorem of Algebra guarantees exactly $n$ roots in the complex numbers, counting repetitions. This counting of repetitions is where algebraic multiplicity begins.
Not obvious, but once you see it — you'll see it everywhere.
Algebraic Multiplicity: Counting Roots
The algebraic multiplicity of an eigenvalue $\lambda$, denoted as $m_a(\lambda)$ or simply $\text{mult}_a(\lambda)$, is defined as the number of times $\lambda$ appears as a root of the characteristic polynomial.
If we factor the characteristic polynomial completely over the complex numbers, it takes the form: $p(\lambda) = (\lambda_1 - \lambda)^{m_a(\lambda_1)} (\lambda_2 - \lambda)^{m_a(\lambda_2)} \dots (\lambda_k - \lambda)^{m_a(\lambda_k)}$ Here, $\lambda_1, \dots, \lambda_k$ are the distinct eigenvalues. The exponent $m_a(\lambda_i)$ is the algebraic multiplicity.
Key Properties of Algebraic Multiplicity:
- Sum Constraint: The sum of the algebraic multiplicities of all distinct eigenvalues always equals $n$, the size of the matrix. $\sum m_a(\lambda_i) = n$.
- Polynomial Nature: It is purely an algebraic property derived from the coefficients of the characteristic polynomial. It does not require finding eigenvectors.
- Upper Bound: It sets a hard upper limit on the geometric multiplicity (discussed next).
Example: Consider the matrix $A = \begin{pmatrix} 2 & 1 \ 0 & 2 \end{pmatrix}$. The characteristic polynomial is $\det\begin{pmatrix} 2-\lambda & 1 \ 0 & 2-\lambda \end{pmatrix} = (2-\lambda)^2$. The eigenvalue is $\lambda = 2$. It appears as a root twice. So, the algebraic multiplicity is $m_a(2) = 2$ Turns out it matters..
Geometric Multiplicity: Counting Dimensions
While algebraic multiplicity counts roots in a polynomial, geometric multiplicity counts vectors in a subspace. It is defined as the dimension of the eigenspace associated with $\lambda$.
The eigenspace $E_\lambda$ is the set of all eigenvectors corresponding to $\lambda$, together with the zero vector. Mathematically, it is the null space (kernel) of the matrix $(A - \lambda I)$: $E_\lambda = \text{Null}(A - \lambda I) = {\mathbf{v} \in \mathbb{C}^n \mid (A - \lambda I)\mathbf{v} = \mathbf{0}}$ The geometric multiplicity, denoted $m_g(\lambda)$ or $\text{mult}g(\lambda)$, is: $m_g(\lambda) = \dim(E\lambda) = \dim(\text{Null}(A - \lambda I))$ By the Rank-Nullity Theorem, this is equivalent to: $m_g(\lambda) = n - \text{rank}(A - \lambda I)$
Key Properties of Geometric Multiplicity:
- Dimension of Solution Space: It tells you how many linearly independent eigenvectors exist for that specific eigenvalue.
- Lower Bound: It is always at least 1 for any eigenvalue (since an eigenvalue must have at least one eigenvector by definition).
- Calculation: It requires row-reducing the matrix $(A - \lambda I)$ to find the number of free variables (nullity).
Example (Continued): For $A = \begin{pmatrix} 2 & 1 \ 0 & 2 \end{pmatrix}$ with $\lambda = 2$. We compute $A - 2I = \begin{pmatrix} 0 & 1 \ 0 & 0 \end{pmatrix}$. The rank of this matrix is 1 (one pivot row). Using Rank-Nullity: $m_g(2) = 2 - 1 = 1$. The eigenspace is spanned by $\begin{pmatrix} 1 \ 0 \end{pmatrix}$. Only one linearly independent eigenvector exists. The geometric multiplicity is $m_g(2) = 1$.
The Fundamental Inequality: $m_g \le m_a$
The most critical relationship in this topic is the inequality linking the two multiplicities: $1 \le m_g(\lambda) \le m_a(\lambda) \le n$
Why is geometric multiplicity always less than or equal to algebraic multiplicity? There are several ways to prove this, but the most intuitive involves the concept of basis extension and block triangularization Not complicated — just consistent..
- Start with a basis for the eigenspace $E_\lambda$ (size $m_g$).
- Extend this to a basis for the whole space $\mathbb{C}^n$.
- Represent the matrix $A$ in this new basis. Because the first $m_g$ basis vectors are eigenvectors, the matrix takes a block upper-triangular form: $A \sim \begin{pmatrix} \lambda I_{m_g} & * \ 0 & B \end{pmatrix}$
- The characteristic polynomial of this block matrix is $(\lambda - \lambda)^{m_g} \cdot \det(B - \lambda I)$.
- This shows $(\lambda - \lambda)$ is a factor at least $m_g$ times. Hence, the algebraic multiplicity (the total exponent) must be at least $m_g$.
This inequality is the "litmus test" for diagonalizability.
Diagonalizability: The Ultimate Application
The primary reason we care about these two numbers is diagonalization. A matrix $A$ is diagonalizable (similar to a diagonal matrix) if and only if there exists a basis of $\mathbb{C}^n$ consisting entirely of eigenvectors of $A$ That alone is useful..
Since eigenvectors corresponding to distinct eigenvalues are always linearly independent, the only obstacle to forming a full basis of $n$ eigenvectors is whether we have enough eigenvectors for each repeated eigenvalue Most people skip this — try not to. Worth knowing..
The Diagonalization Theorem: An $n \times n$ matrix $A$ is diagonalizable if and only if for every eigenvalue $\lambda$: $m_g(\lambda) = m_a(\lambda)$ And equivalently, the sum of the geometric multiplicities equals $n$: $\sum m_g(\lambda_i) = n$ That's the part that actually makes a difference..
Case 1: $m_g = m_a$ (The "Good" Case)
If the geometric multiplicity matches the algebraic multiplicity for every eigenvalue, the matrix is diagonalizable. The eigenspaces are "large enough" to span the whole space. *
