Characteristic Polynomials And Their Roots

From Eigenvalues to a Polynomial

An eigenvalue is a scalar \(\lambda\) for which a matrix acts on some nonzero vector by multiplication by \(\lambda\). Finding such a scalar directly means asking whether \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\) has a nonzero solution. Rearranging gives \((\lambda\mathbf{I}-\mathbf{A})\mathbf{x}=\mathbf{0}\). The determinant turns this question into a polynomial equation in \(\lambda\): the values that make that determinant zero are precisely the values for which the homogeneous system has a nonzero solution.

Let \(n\geq 1\), let \(\mathbb{F}\) be \(\mathbb{R}\) or \(\mathbb{C}\), and let \(\mathbf{A}\in\mathbb{F}^{n\times n}\). Here \(\mathbf{I}\) denotes the \(n\times n\) identity matrix. An eigenvalue of \(\mathbf{A}\) over \(\mathbb{F}\) is a scalar \(\lambda\in\mathbb{F}\) for which there is a vector \(\mathbf{x}\in\mathbb{F}^n\), \(\mathbf{x}\ne\mathbf{0}\), satisfying \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\).

The notation distinguishes two determinant expressions rather than two different tests for eigenvalues. In fact, \(\mathbf{A}-\lambda\mathbf{I}=-(\lambda\mathbf{I}-\mathbf{A})\), and multiplying all \(n\) rows by \(-1\) multiplies the determinant by \((-1)^n\). Thus \(p_{\mathbf{A}}(\lambda)=(-1)^n\chi_{\mathbf{A}}(\lambda)\). This relation lets us use the requested \(\lambda\mathbf{I}-\mathbf{A}\) expression while retaining the course notation for \(p_{\mathbf{A}}\).

Why the Determinant Is a Polynomial

Each entry of \(\lambda\mathbf{I}-\mathbf{A}\) is either \(\lambda-a_{ii}\) on the diagonal or a constant off the diagonal. The determinant is a sum of products of entries, so it is a polynomial in \(\lambda\). More precisely, the determinant expansion over permutations is

where \(S_n\) is the set of permutations of \(\{1,\ldots,n\}\), \(\operatorname{sgn}(\pi)\) is the sign of a permutation, and \(\delta_{ij}=1\) when \(i=j\) and \(0\) otherwise. The identity permutation contributes \(\prod_{i=1}^{n}(\lambda-a_{ii})\), whose highest-degree term is \(\lambda^n\). Every nonidentity permutation moves at least one index; for a moved index, its factor is an off-diagonal constant. Its product therefore has degree at most \(n-1\). This proves the following degree statement.

Proof. In the determinant expansion, the identity-permutation term has leading term \(\lambda^n\) with coefficient \(1\). Every other permutation moves at least one index, so at least one factor in its product is constant rather than containing \(\lambda\). Such a product has degree at most \(n-1\), and cannot alter the coefficient of \(\lambda^n\). Hence the determinant is a degree-\(n\) polynomial with leading coefficient \(1\). \(\square\)

This proposition says that the determinant expression is a polynomial of a known degree; it does not say that the polynomial has \(n\) distinct roots in the chosen field. A polynomial may repeat a root, and a real polynomial may have roots that are not real. The field matters when deciding which roots are eigenvalues: the eigenvector must have entries in that same field.

The Root–Eigenvalue Theorem

Proof. Fix \(\lambda\in\mathbb{F}\) and set \(\mathbf{B}=\lambda\mathbf{I}-\mathbf{A}\). By the determinant criterion for a square matrix, \(\det\mathbf{B}=0\) if and only if \(\mathbf{B}\) is singular. By the Invertible Matrix Theorem, a square matrix is singular if and only if its homogeneous equation has a nonzero solution. Therefore, \(\chi_{\mathbf{A}}(\lambda)=0\) if and only if there is an \(\mathbf{x}\ne\mathbf{0}\) such that \((\lambda\mathbf{I}-\mathbf{A})\mathbf{x}=\mathbf{0}\). Expanding the left side gives \(\lambda\mathbf{x}-\mathbf{A}\mathbf{x}=\mathbf{0}\), which is equivalent to \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). This last equation, with \(\mathbf{x}\ne\mathbf{0}\), is exactly the definition that \(\lambda\) is an eigenvalue. Each equivalence is reversible, proving both directions. \(\square\)

The key condition is singularity, not merely a determinant calculation. A zero determinant means the columns of \(\lambda\mathbf{I}-\mathbf{A}\) are linearly dependent, so some nonzero coefficient vector \(\mathbf{x}\) satisfies \((\lambda\mathbf{I}-\mathbf{A})\mathbf{x}=\mathbf{0}\). That vector is an eigenvector. Conversely, an eigenvector supplies such a nonzero solution and forces the matrix to be singular. The theorem is exact over the stated field: if \(\lambda\) is not in \(\mathbb{F}\), it is not an eigenvalue over \(\mathbb{F}\), even if it is a root after extending the field.

Worked Examples

Common Sign and Field Pitfalls

Do not confuse a change in sign convention with a change in the roots. For example, take \(\mathbf{A}=2\mathbf{I}_3\). A mistaken calculation might write \(\det(\mathbf{A}-\lambda\mathbf{I})=(\lambda-2)^3\), silently changing the signs inside the determinant. The corrected calculations are

The polynomials are not equal, but their roots are both \(2\). The correction matters because the sign of the polynomial is part of the polynomial itself, even when the eigenvalue test gives the same answer.

Also specify the field. For the rotation matrix above, saying “there are no eigenvalues” without qualification is incomplete: there are no real eigenvalues, but there are two complex eigenvalues. The theorem compares roots and eigenvalues over the same field. If the matrix is real and the question concerns complex roots, regard its entries as complex and solve the eigenvector equation in \(\mathbb{C}^n\).

The characteristic polynomial is useful because it packages the singularity test into one scalar equation. Instead of solving \((\lambda\mathbf{I}-\mathbf{A})\mathbf{x}=\mathbf{0}\) separately for every possible \(\lambda\), one first finds the roots of a polynomial and then, for each root, solves the corresponding homogeneous system to find eigenvectors. The theorem proves that this approach finds exactly the eigenvalues—no more and no fewer. The polynomial’s roots alone do not provide the eigenvectors; those come from the null space of \(\lambda\mathbf{I}-\mathbf{A}\).