Algebraic Multiplicity Of An Eigenvalue

What Algebraic Multiplicity Counts

An eigenvalue is a scalar for which a matrix has a nonzero vector that it scales. The characteristic polynomial records which scalars have this property, but it can also record that a root occurs more than once. Algebraic multiplicity is the precise count of that repetition. It comes from the polynomial alone: it is not a count of eigenvectors or of independent directions.

For a matrix \(\mathbf{A}\in\mathbb{R}^{n\times n}\), use the characteristic polynomial convention \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_n)\). Its roots are the eigenvalues, by the eigenvalue equation and singularity criterion established in the earlier material on eigenvalues. The polynomial has degree \(n\) and leading coefficient \((-1)^n\). Its overall sign under this convention does not change its roots or their multiplicities. When discussing roots of a real matrix’s characteristic polynomial, we allow \(\lambda\in\mathbb{C}\); a real polynomial can have complex roots.

The largest integer exists: a nonzero polynomial has finite degree, so no power of \((\lambda-\alpha)\) with exponent greater than its degree can divide it. A scalar that is not a root has multiplicity \(0\); an eigenvalue has positive algebraic multiplicity. Factoring out the largest possible power leaves a quotient that does not vanish at the root. That final nonzero value is what confirms that the exponent has not been overstated.

Reading Multiplicity from Factors

Suppose a characteristic polynomial is displayed as a constant times factors of the form \((\lambda-\alpha)^k\), with distinct values of \(\alpha\). The exponent on each factor gives the algebraic multiplicity of that root. If a root occurs in several factors, first rewrite them as a single power: \((\lambda-\alpha)^r(\lambda-\alpha)^s=(\lambda-\alpha)^{r+s}\). A root written once as a value is not necessarily simple; the exponent, not the number of times the value is listed, determines its multiplicity.

Proof. Fix an index \(j\). Factor \(p\) as \((\lambda-\alpha_j)^{k_j}q_j(\lambda)\), where \(q_j(\lambda)=c\prod_{\ell\ne j}(\lambda-\alpha_\ell)^{k_\ell}\). Because the roots are distinct, \(\alpha_j-\alpha_\ell\ne 0\) for every \(\ell\ne j\). Thus \(q_j(\alpha_j)=c\prod_{\ell\ne j}(\alpha_j-\alpha_\ell)^{k_\ell}\ne 0\); if there are no other factors, this product is \(1\), so \(q_j(\alpha_j)=c\ne0\). The factor \((\lambda-\alpha_j)\) therefore occurs exactly \(k_j\) times, as required by the definition. \(\square\)

This theorem also explains how to handle a polynomial written in stages rather than in a single collected product. If \(p(\lambda)=(\lambda-\alpha)^r u(\lambda)\) and \(u(\alpha)\ne0\), then \(\alpha\) has multiplicity exactly \(r\). If instead \(u(\alpha)=0\), the displayed exponent is not the full multiplicity: the remaining factor supplies at least one more copy.

Worked Examples

Two Useful Multiplicity Rules

The preceding examples use a general rule: exponents add when factors with the same root are multiplied. This rule does not require the whole polynomial to be factored into distinct roots.

Proof. Multiplying the given expressions yields \(p(\lambda)q(\lambda)=(\lambda-\alpha)^{r+s}u(\lambda)v(\lambda)\). The remaining factor evaluates at \(\alpha\) to \(u(\alpha)v(\alpha)\), which is nonzero because both factors are nonzero. By the definition of root multiplicity, the exponent \(r+s\) is exact. \(\square\)

For example, \((\lambda-4)(\lambda-1)^2(\lambda-4)^2\) has the root \(4\) with multiplicity \(1+2=3\), not two separate multiplicities to report. The root \(1\) has multiplicity \(2\). The theorem just proved supplies the reason that the two factors for \(4\) combine by adding exponents.

When a characteristic polynomial is completely factored over \(\mathbb{C}\), the sum of the listed exponents is its degree. This follows directly by multiplying the factors: each \((\lambda-\alpha)^k\) has degree \(k\), and the degree of a product of nonzero polynomials is the sum of their degrees. For an \(n\times n\) matrix, the characteristic polynomial has degree \(n\), so a complete factorization lists \(n\) roots when each root is counted according to its algebraic multiplicity. This count concerns polynomial roots only; by itself it says nothing about the dimensions of eigenspaces.

Common Mistakes and Checks

Mistake: counting distinct roots instead of repeated factors. From \(p(\lambda)=(\lambda-1)^2(\lambda+2)\), a wrong calculation might say “the root \(1\) appears once in the list of roots, so its algebraic multiplicity is \(1\).” The correction is to inspect the power: \((\lambda-1)^2\) divides the polynomial, and the remaining factor at \(1\) is \(1+2=3\ne0\). Thus the exact multiplicity is \(2\).

Mistake: stopping before checking the quotient. If a polynomial is written as \((\lambda-\alpha)^r u(\lambda)\), the exponent \(r\) is only a candidate until \(u(\alpha)\ne0\) is confirmed. For instance, \((\lambda-1)^2(\lambda-1)\) is visibly not a factorization with distinct roots: the second factor vanishes at \(1\), too. Collecting gives \((\lambda-1)^3\), so the multiplicity is \(3\), not \(2\).

Mistake: treating the sign convention as a change in multiplicity. The polynomials \(\det(\mathbf{A}-\lambda\mathbf{I}_n)\) and \(\det(\lambda\mathbf{I}_n-\mathbf{A})\) differ by the nonzero constant \((-1)^n\). Multiplying by a nonzero constant does not change which powers of \((\lambda-\alpha)\) divide the polynomial. The sign may change the polynomial’s values, but it does not change eigenvalues or algebraic multiplicities.

For a reliable calculation, identify each root, collect every factor that corresponds to that same root, and evaluate what remains at the root. If the remaining value is nonzero, the collected exponent is the algebraic multiplicity. If it is zero, factor further before reporting the answer.