Understanding the behavior of linear transformations is central to linear algebra, and few concepts illuminate the structure of a matrix quite like the algebraic multiplicity and geometric multiplicity of an eigenvalue. These two numbers, attached to every eigenvalue of a square matrix, dictate whether a matrix can be diagonalized, how solutions to differential equations behave, and the very geometry of the transformation itself. While they sound similar, they measure fundamentally different properties: one counts roots of a polynomial, while the other measures the dimension of a subspace Practical, not theoretical..
Not obvious, but once you see it — you'll see it everywhere.
The Foundation: Eigenvalues and the Characteristic Polynomial
Before distinguishing between the two multiplicities, we must establish the groundwork. For a square matrix $A$ of size $n \times n$, an eigenvalue $\lambda$ is a scalar such that there exists a non-zero vector $\mathbf{v}$ (the eigenvector) satisfying $A\mathbf{v} = \lambda\mathbf{v}$. This leads to this equation rewrites as $(A - \lambda I)\mathbf{v} = \mathbf{0}$. For a non-trivial solution $\mathbf{v} \neq \mathbf{0}$ to exist, the matrix $(A - \lambda I)$ must be singular, meaning its determinant is zero Easy to understand, harder to ignore..
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 plane, counting repetitions. This repetition count is where algebraic multiplicity enters the picture Small thing, real impact..
Algebraic Multiplicity: Counting Roots
The algebraic multiplicity of an eigenvalue $\lambda$, denoted as $m_a(\lambda)$ or simply $alg(\lambda)$, is defined as the number of times $\lambda$ appears as a root of the characteristic polynomial $p(\lambda)$ Most people skip this — try not to..
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 serves as an upper bound for the geometric multiplicity (proven below).
Example: Consider a matrix with characteristic polynomial $p(\lambda) = (\lambda - 2)^3(\lambda + 1)^2$. The eigenvalue $\lambda = 2$ has algebraic multiplicity 3. The eigenvalue $\lambda = -1$ has algebraic multiplicity 2. The matrix is $5 \times 5$ Turns out it matters..
Geometric Multiplicity: Dimension of the Eigenspace
While algebraic multiplicity counts roots, geometric multiplicity counts dimensions. The geometric multiplicity of an eigenvalue $\lambda$, denoted $m_g(\lambda)$ or $geo(\lambda)$, is defined as the dimension of the eigenspace associated with $\lambda$.
The eigenspace $E_\lambda$ is the set of all eigenvectors corresponding to $\lambda$, plus the zero vector: $E_\lambda = {\mathbf{v} \in \mathbb{C}^n \mid (A - \lambda I)\mathbf{v} = \mathbf{0}} = \text{Null}(A - \lambda I)$ This is precisely the null space (kernel) of the matrix $(A - \lambda I)$. That's why, the geometric multiplicity is the nullity of $(A - \lambda I)$: $m_g(\lambda) = \dim(\text{Null}(A - \lambda I)) = n - \text{rank}(A - \lambda I)$
Key Properties of Geometric Multiplicity:
- Minimum Value: Since $\lambda$ is an eigenvalue, the null space is non-trivial, so $m_g(\lambda) \ge 1$.
- Linear Independence: It represents the maximum number of linearly independent eigenvectors associated with $\lambda$.
- Computational Nature: Finding it requires row reduction (Gaussian elimination) on $(A - \lambda I)$ to determine the number of free variables.
Example: If for $\lambda=2$, the matrix $(A - 2I)$ row-reduces to a form with 2 pivot columns and 3 free variables (in a $5 \times 5$ matrix), the nullity is 3. Thus, $m_g(2) = 3$ It's one of those things that adds up..
The Fundamental Inequality: Geometric $\le$ Algebraic
The most critical relationship between these two concepts is the inequality: $1 \le m_g(\lambda) \le m_a(\lambda) \le n$
Why does this hold? Suppose $m_g(\lambda) = k$. This means there are $k$ linearly independent eigenvectors $\mathbf{v}_1, \dots, \mathbf{v}_k$ for $\lambda$. We can extend this set to a basis for $\mathbb{C}^n$. In this basis, the matrix representation of the linear transformation takes a block upper-triangular form where the top-left block is $\lambda I_k$ (a $k \times k$ diagonal matrix with $\lambda$ on the diagonal). The characteristic polynomial of this block is $(\lambda - \lambda)^k = (\lambda - \lambda)^k$. Since similar matrices have the same characteristic polynomial, the factor $(\lambda - \lambda)^k$ must divide the characteristic polynomial of $A$. So, the exponent $m_a(\lambda)$ must be at least $k$.
This inequality is not just a theoretical curiosity; it is the litmus test for diagonalizability Most people skip this — try not to..
Diagonalizability: The Perfect Balance
A matrix $A$ is diagonalizable if and only if there exists a basis of $\mathbb{C}^n$ consisting entirely of eigenvectors of $A$. This happens precisely when, for every eigenvalue $\lambda$, the geometric multiplicity equals the algebraic multiplicity: $m_g(\lambda) = m_a(\lambda) \quad \text{for all eigenvalues } \lambda$
If this condition holds for all eigenvalues, the sum of the geometric multiplicities equals the sum of the algebraic multiplicities, which is $n$. We have $n$ linearly independent eigenvectors, forming the columns of the invertible matrix $P$ such that $P^{-1}AP = D$ (a diagonal matrix).
The Defective Case: When Geometry Falls Short
If for at least one eigenvalue $m_g(\lambda) < m_a(\lambda)$, the matrix is defective (non-diagonalizable). There are "missing" eigenvectors. The total number of linearly independent eigenvectors is strictly less than $n$ But it adds up..
In this scenario, we cannot diagonalize the matrix. Instead, we must use the Jordan Normal Form. The "defect" $d = m_a(\lambda) - m_g(\lambda)$ determines the number of Jordan blocks associated with $\lambda$ and the sizes of the off-diagonal 1s (superdiagonal entries) in those blocks. The geometric multiplicity equals the number of Jordan blocks for $\lambda$, while the algebraic multiplicity equals the sum of the sizes of those blocks.
This is where a lot of people lose the thread Not complicated — just consistent..
Illustrative Examples
Example 1: The Identity Matrix (Perfect Symmetry)
Let $A = I_3$ (the $3 \times 3$ identity matrix).
- Characteristic polynomial: $(\lambda - 1)^3$.
- Algebraic multiplicity of $\lambda=1$: $m_a =
3.
- Eigenspace: $E_1 = \text{null}(I - I) = \text{null}(0) = \mathbb{C}^3$.
- Geometric multiplicity: $m_g = \dim(\mathbb{C}^3) = 3$.
Since $m_g = m_a = 3$, the matrix is diagonalizable (it is already diagonal). Every vector is an eigenvector; the transformation is pure scaling And it works..
Example 2: A Diagonalizable Matrix with Distinct Eigenvalues
Let $A = \begin{pmatrix} 2 & 0 \ 0 & 3 \end{pmatrix}$.
- Characteristic polynomial: $(\lambda - 2)(\lambda - 3)$.
- Eigenvalues: $\lambda_1 = 2, \lambda_2 = 3$.
- Algebraic multiplicities: $m_a(2) = 1, m_a(3) = 1$.
- Eigenspaces: $E_2 = \text{span}{\mathbf{e}_1}, E_3 = \text{span}{\mathbf{e}_2}$.
- Geometric multiplicities: $m_g(2) = 1, m_g(3) = 1$.
Here $m_g = m_a = 1$ for each eigenvalue. The sum of geometric multiplicities is $2 = n$. Distinct eigenvalues always yield $m_g = m_a = 1$, guaranteeing diagonalizability.
Example 3: The Defective Matrix (The Jordan Block)
Let $J = \begin{pmatrix} 2 & 1 \ 0 & 2 \end{pmatrix}$.
- Characteristic polynomial: $(\lambda - 2)^2$.
- Algebraic multiplicity of $\lambda=2$: $m_a = 2$.
- Eigenspace: $E_2 = \text{null}(J - 2I) = \text{null}\begin{pmatrix} 0 & 1 \ 0 & 0 \end{pmatrix} = \text{span}\left{ \begin{pmatrix} 1 \ 0 \end{pmatrix} \right}$.
- Geometric multiplicity: $m_g = 1$.
Here $m_g(2) = 1 < 2 = m_a(2)$. That's why the matrix is defective. So there is only one linearly independent eigenvector, insufficient to form a basis for $\mathbb{C}^2$. The "missing" dimension manifests as the superdiagonal $1$ in the Jordan form—the matrix is already in its Jordan Normal Form.
Example 4: Higher Defect
Consider $A = \begin{pmatrix} 5 & 1 & 0 \ 0 & 5 & 0 \ 0 & 0 & 5 \end{pmatrix}$.
- Characteristic polynomial: $(\lambda - 5)^3 \implies m_a(5) = 3$.
- $A - 5I = \begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{pmatrix}$.
- Nullspace basis: $\mathbf{e}_1, \mathbf{e}_3 \implies m_g(5) = 2$.
- Defect $d = 3 - 2 = 1$.
- Jordan Form: Two blocks for $\lambda=5$: one $2 \times 2$ block (with a superdiagonal 1) and one $1 \times 1$ block. The geometric multiplicity (2) counts the blocks; the algebraic multiplicity (3) sums their sizes.
Generalized Eigenvectors: Bridging the Gap
When $m_g < m_a$, standard eigenvectors fail to span the space. A generalized eigenvector of rank $k$ for eigenvalue $\lambda$ satisfies: $(A - \lambda I)^k \mathbf{v} = \mathbf{0} \quad \text{but} \quad (A - \lambda I)^{k-1} \mathbf{v} \neq \mathbf{0}$ Standard eigenvectors are rank 1. Also, the vectors $\mathbf{v}_1, \dots, \mathbf{v}_k$ forming a Jordan chain satisfy: $(A - \lambda I)\mathbf{v}_1 = \mathbf{0}, \quad (A - \lambda I)\mathbf{v}_2 = \mathbf{v}1, \quad \dots, \quad (A - \lambda I)\mathbf{v}k = \mathbf{v}{k-1}$ These chains span the generalized eigenspace $G\lambda = \text{null}((A - \lambda I)^{m_a})$, whose dimension is exactly $m_a(\lambda)$. We recover a full basis by relaxing the eigenvector definition. The direct sum of all generalized eigenspaces is $\mathbb{C}^n$, guaranteeing the existence of the Jordan Normal Form for every square matrix Most people skip this — try not to..
Conclusion
The relationship $m_g(\lambda) \le m_a(\lambda)$ is the structural backbone of linear algebra. It quantifies the tension between a matrix's algebraic signature—encoded in the roots of its characteristic polynomial—and its geometric action—the invariant subspaces it preserves.
When equality
