Algebraic and Geometric Multiplicity¶

Definition¶

If the characteristic polynomial factors as $c(\lambda-\lambda_1)^{k_1}\cdots(\lambda-\lambda_l)^{k_l}$ with distinct $\lambda_1,\dots,\lambda_l$ , then $k_i$ is the algebraic multiplicity of the eigenvalue $\lambda_i$ , written $\operatorname{am}(\lambda_i)$ .

The geometric multiplicity of an eigenvalue $\lambda$ is the maximal number of linearly independent eigenvectors associated with $\lambda$ , written $\operatorname{gm}(\lambda)$ .

Interpretation¶

Each eigenvalue comes with two counts. Algebraic multiplicity counts how often it appears as a root of the characteristic polynomial; geometric multiplicity counts how many independent eigendirections it actually has. The first is read from a formula, the second from the eigenvectors.

The characteristic polynomial (lambda - 1) squared times (lambda - 3) over 2, touching the axis at the double root 1 and crossing it at the simple root 3.

For $\boldsymbol{A}=\begin{pmatrix}1&1&0\\0&1&0\\0&0&3\end{pmatrix}$ $100110003$ , $p(\lambda)=\det(\boldsymbol{A}-\lambda\boldsymbol{I})=(1-\lambda)^2(3-\lambda)$ $λ)$ . The plotted $(\lambda-1)^2(\lambda-3)/2$ differs from it by the constant factor $-\tfrac{1}{2}$ and has the same roots. It touches the axis at the double root $1$ , where the factor $(\lambda-1)^2$ does not change sign, and crosses it at the simple root $3$ , so $\operatorname{am}(1)=2$ and $\operatorname{am}(3)=1$ $1$ . Yet $\boldsymbol{A}-\boldsymbol{I}$ has rank $2$ , so its kernel is one-dimensional and $\operatorname{gm}(1)=1$ $1$ .

The geometric multiplicity is the dimension of the eigenvectors for $\lambda$ together with $\boldsymbol{0}$ : $\operatorname{gm}(\lambda)=\dim\ker(\boldsymbol{A}-\lambda\boldsymbol{I})=n-\operatorname{rank}(\boldsymbol{A}-\lambda\boldsymbol{I})$ $λ I)$ , by the fundamental theorem of linear maps. It is at least $1$ , because every eigenvalue has an eigenvector, and never exceeds the algebraic multiplicity. Over $\mathbb{C}$ the algebraic multiplicities add up to $n$ , by the fundamental theorem of algebra.

Comparing the two counts decides diagonalizability: every eigenvalue needs as many independent eigenvectors as its algebraic multiplicity. When $\operatorname{gm}(\lambda)<\operatorname{am}(\lambda)$ $am (λ)$ , the missing directions are supplied by generalized eigenvectors, and $\operatorname{gm}(\lambda)$ is the number of Jordan blocks for $\lambda$ in the Jordan normal form.

A repeated eigenvalue does not by itself prevent diagonalization. The identity $\boldsymbol{I}_2$ has the double eigenvalue $1$ with $\operatorname{gm}(1)=\operatorname{am}(1)=2$ $2$ , because every nonzero vector is an eigenvector; only $\operatorname{gm}<\operatorname{am}$ is a problem.

Key Results¶

For every eigenvalue of an $n\times n$ matrix, $1\leq \operatorname{gm}(\lambda)\leq \operatorname{am}(\lambda)\leq n$ $n$ .
A matrix is diagonalizable if and only if $\operatorname{gm}(\lambda)=\operatorname{am}(\lambda)$ for every eigenvalue, the eigenvalues being taken in $\mathbb{C}$ so that the characteristic polynomial factors completely. A real matrix is diagonalizable over $\mathbb{R}$ only if, in addition, all its eigenvalues are real.

Examples¶

For $(\lambda-2)^3(\lambda+1)$ $1)$ , eigenvalue $2$ has algebraic multiplicity $3$ .
For $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ , eigenvalue $1$ has algebraic multiplicity $2$ but geometric multiplicity $1$ .

Prerequisites¶

Sources¶

Math II: Linear Algebra · Week 6

Keywords: algebraic multiplicity, geometric multiplicity, multiplicity of an eigenvalue, eigenvalue multiplicity Symbols: complex numbers, lambda, real numbers

Last updated: 20 October 2026 03:02