Eigenpairs Of Small Integer Matrices

A Complete Eigenpair Calculation

An eigenpair calculation is more than finding numbers that make a characteristic polynomial vanish. For each eigenvalue, we must solve a homogeneous system to describe every corresponding eigenvector, give a basis for the eigenspace, and report the two multiplicities that measure different aspects of the eigenvalue. Small integer matrices are useful practice because their arithmetic can be checked exactly, while still requiring care with signs, repeated roots, and free variables.

Unless specified otherwise, the matrices and eigenvectors in this tutorial are over \(\mathbb{R}\). For a matrix \(\mathbf{A}\in\mathbb{R}^{n\times n}\), a real number \(\lambda\) is an eigenvalue if there is a nonzero vector \(\mathbf{v}\in\mathbb{R}^n\) such that \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). Such a vector \(\mathbf{v}\) is an eigenvector for \(\lambda\). The requirement \(\mathbf{v}\ne\mathbf{0}\) matters: the zero vector satisfies \(\mathbf{A}\mathbf{0}=\lambda\mathbf{0}\) for every \(\lambda\), but is never an eigenvector.

A basis for an eigenspace is an ordered list of independent vectors that spans the whole eigenspace. To find one, rearrange the eigenvector equation as \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{v}=\mathbf{0}\), then solve this homogeneous system. A basis must describe all solutions, not just one eigenvector. In the examples, a displayed basis will be followed by a direct multiplication check.

Proof. The equation \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) is equivalent to \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{v}=\mathbf{0}\). Thus \(\lambda\) is an eigenvalue exactly when this homogeneous system has a nonzero solution. By the standard determinant criterion for invertibility, a square matrix has a nonzero vector in its null space exactly when its determinant is zero. Applying that criterion to \(\mathbf{A}-\lambda\mathbf{I}_n\) gives the equivalence with \(\det(\mathbf{A}-\lambda\mathbf{I}_n)=0\), which is \(p_{\mathbf{A}}(\lambda)=0\). \(\square\)

This criterion identifies the eigenvalues, but does not give eigenspace bases or geometric multiplicities. Those require solving the homogeneous system for each root separately.

Two-by-Two Matrices: Distinct Eigenvalues

For a two-by-two matrix with distinct characteristic roots, each root has algebraic multiplicity \(1\). Its eigenspace is nonzero by the eigenvalue criterion and is contained in \(\mathbb{R}^2\); in this example the row reduction makes its one-dimensional form explicit. Do not infer an eigenspace basis from the root alone: the homogeneous equations show which vectors actually belong to it.

Repeated Roots Require a Separate Eigenspace Calculation

This example illustrates why the algebraic and geometric multiplicities must be reported separately. Algebraic multiplicity counts how often a root occurs in a polynomial. Geometric multiplicity counts the number of independent directions in the eigenspace, measured by its dimension. They answer different questions, so one cannot be substituted for the other.

Proof. By definition, \(\mathbf{v}\in E_{\lambda}\) exactly when \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). Subtracting \(\lambda\mathbf{v}\) from both sides gives \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{v}=\mathbf{0}\), which is exactly the condition \(\mathbf{v}\in\operatorname{Nul}(\mathbf{A}-\lambda\mathbf{I}_n)\). The two sets are therefore equal. The geometric multiplicity is the dimension of \(E_{\lambda}\), and the nullity is the dimension of the null space, so the equality of spaces gives the stated equality of dimensions. \(\square\)

A Three-by-Three Calculation

A Reliable Workflow and a Common Error

For a small integer matrix, keep the calculation in a fixed order. This separates the polynomial calculation from the linear-system calculations and makes it easier to notice an error before it affects every reported result.

Common mistake: using \(\mathbf{A}+\lambda\mathbf{I}\) instead of \(\mathbf{A}-\lambda\mathbf{I}\). For the first matrix above, an incorrect setup would produce \(\det(\mathbf{A}+\lambda\mathbf{I}_2)=(4+\lambda)(3+\lambda)-2=\lambda^2+7\lambda+10\), with roots \(-5\) and \(-2\). Those are not the eigenvalues of \(\mathbf{A}\). The correction is to start from \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\), move the right-hand side to the left, and obtain \((\mathbf{A}-\lambda\mathbf{I}_2)\mathbf{v}=\mathbf{0}\). Its determinant is \(\lambda^2-7\lambda+10=(\lambda-5)(\lambda-2)\), matching the directly checked eigenpairs.

Another useful check is to count basis vectors, not just to display a nonzero solution. In the repeated-root example, \((1,0)^{\top}\) is an eigenvector, but it does not by itself prove that the eigenspace is one-dimensional. The row-reduced system shows that every solution has the form \(x(1,0)^{\top}\), which establishes both spanning and the claimed geometric multiplicity.