Factoring Characteristic Polynomials For Multiplicities

From a Characteristic Polynomial to Multiplicities

An eigenvalue calculation does not end when a characteristic polynomial has been found. Its roots identify the eigenvalues, while the powers of the corresponding linear factors record their algebraic multiplicities. Factoring can be straightforward when the polynomial is already in a recognizable form, but integer coefficients also provide a systematic way to narrow down possible rational roots. The essential checks are exact: test a candidate by substitution, then divide by its linear factor and test again if a repeated root is possible.

Let \(\mathbf{A}\in\mathbb{R}^{n\times n}\). Its characteristic polynomial, with the convention used here, is \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_n)\). A scalar \(\lambda\) is an eigenvalue over the chosen scalar field precisely when there is a nonzero vector \(\mathbf{v}\) such that \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{v}=\mathbf{0}\). Equivalently, \(\mathbf{A}-\lambda\mathbf{I}_n\) is singular, so its determinant is zero. Thus eigenvalues are exactly the roots of the characteristic polynomial. For a real matrix, we may consider roots over \(\mathbb{R}\) or, when needed, over \(\mathbb{C}\).

The sign convention matters when writing the characteristic polynomial. The polynomial \(\det(\mathbf{A}-\lambda\mathbf{I}_n)\) has leading coefficient \((-1)^n\), not necessarily \(1\). Multiplying a polynomial by \(-1\) does not change its roots or their multiplicities, but keeping the sign correct helps prevent errors in the determinant calculation.

Why a Linear Factor Records a Root

Proof. Polynomial division by the monic linear polynomial \(\lambda-r\) gives polynomials \(q\) and a constant \(c\) such that \(f(\lambda)=(\lambda-r)q(\lambda)+c\). Substituting \(\lambda=r\) gives \(f(r)=c\). Therefore \(f(r)=0\) exactly when the remainder \(c\) is zero, which is exactly the condition that \(\lambda-r\) divides \(f\). \(\square\)

This result explains a practical method for finding multiplicities. Once substitution confirms \(f(r)=0\), divide by \(\lambda-r\). If the quotient also vanishes at \(r\), divide again. Stop at the first quotient whose value at \(r\) is nonzero. The number of successful divisions is the multiplicity. This is not determined merely by how often a candidate appears in an initial list of possible roots; each repetition must be verified.

Rational-Root Checks for Integer Coefficients

For a polynomial with integer coefficients, the rational-root test narrows down the possible rational roots. It does not assert that any candidate is a root: every candidate still needs substitution. The zero root requires separate attention when the constant coefficient is zero.

Proof. First suppose \(u\ne0\) and \(f(u/v)=0\). Multiplying the equation by \(v^n\) gives \(a_nu^n+a_{n-1}u^{n-1}v+\cdots+a_1uv^{n-1}+a_0v^n=0\). Rearranging shows that \(u\) divides \(a_0v^n\). Since \(u\) and \(v\) have no common factor, \(u\) and \(v^n\) have no common factor, so \(u\) divides \(a_0\). The same equation shows that \(v\) divides \(a_nu^n\). Since \(v\) and \(u^n\) have no common factor, \(v\) divides \(a_n\). If \(u=0\), the proposed root is \(0\), and \(f(0)=a_0\), so it is a root exactly when \(a_0=0\); no divisibility claim about \(u\) is needed. Finally, for a monic polynomial \(a_n=1\), the condition \(v\mid a_n\) forces \(v=1\), so each rational root is an integer, including the separately handled root \(0\). \(\square\)

For an integer matrix, every entry of \(\mathbf{A}-\lambda\mathbf{I}_n\) is a polynomial with integer coefficients. The determinant is a sum of products of these entries, so \(p_{\mathbf{A}}\) has integer coefficients. Its leading coefficient is \(1\) or \(-1\). Consequently, if an eigenvalue is rational, it must be an integer. When the constant coefficient is nonzero, the test gives a finite list of candidates: the possible integer roots divide that constant coefficient. When the constant coefficient is zero, \(0\) is already a root; factor out \(\lambda\) before using the test on the remaining polynomial.

Worked Factoring with a Repeated Root

When the Rational-Root Candidates Do Not Work

A failed rational-root check does not mean there are no roots: it only rules out the candidates considered. A polynomial can have irrational real roots or complex roots. The field matters when we state the eigenvalues, and the multiplicities are read from factors over that field.

Handling a Zero Constant Term

Common Pitfalls and a Reliable Procedure

Do not use zero as a divisor in the rational-root test. For example, \(f(\lambda)=\lambda\) has the rational root \(0/1\). A mistaken application of the nonzero-numerator divisibility conclusion would claim \(0\mid0\), which is not a valid divisibility statement: divisors are required to be nonzero. The correction is to handle \(u=0\) separately. In this example \(f(0)=0\), so the constant coefficient is zero and \(0\) is a root; factor out \(\lambda\) if further roots are needed.

A second common mistake is to count a candidate as a repeated root merely because it passes substitution once. If \(f(r)=0\), that establishes one factor \(\lambda-r\), but not a second. Divide by \(\lambda-r\), substitute into the quotient, and continue only if the quotient also vanishes. This procedure directly checks the definition of algebraic multiplicity.

Algebraic multiplicity concerns the characteristic polynomial, not the number of independent eigenvectors. It tells how many times a root occurs as a polynomial factor. Finding the corresponding eigenvectors and eigenspace requires a separate calculation; the factorization alone does not give that information.