Solving Two-By-Two Eigenpair Problems

From Eigenvalues to Eigenpairs

For a two-by-two matrix, finding eigenpairs has two distinct stages. First, the characteristic polynomial gives the possible eigenvalues. Then, for each eigenvalue, a homogeneous system gives the eigenvectors. Keeping these stages separate matters: a root of the characteristic polynomial is a candidate eigenvalue, while the corresponding eigenvectors must be found by solving a new system.

Let \(\mathbf{A}\in\mathbb{R}^{2\times2}\). We may regard \(\mathbf{A}\) as acting on \(\mathbb{R}^2\) when working with real eigenvalues, or on \(\mathbb{C}^2\) when its characteristic polynomial has nonreal roots. An eigenpair is a scalar \(\lambda\) and a nonzero vector \(\mathbf{v}\) satisfying \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). For a real eigenvalue, the eigenvectors can be taken in \(\mathbb{R}^2\); for a nonreal eigenvalue of a real matrix, they are in \(\mathbb{C}^2\).

The equation defining an eigenvector can be rearranged as \((\mathbf{A}-\lambda\mathbf{I}_2)\mathbf{v}=\mathbf{0}\). Thus, after finding a candidate \(\lambda\), row reduce \(\mathbf{A}-\lambda\mathbf{I}_2\) and solve its homogeneous system. The zero vector is always a solution of that system, but it is not an eigenvector; at least one nonzero solution is required.

Proof. By definition, \((\lambda,\mathbf{v})\) is an eigenpair exactly when \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) and \(\mathbf{v}\ne\mathbf{0}\). Subtracting \(\lambda\mathbf{v}\) from both sides gives \((\mathbf{A}-\lambda\mathbf{I}_2)\mathbf{v}=\mathbf{0}\). This says precisely that \(\mathbf{v}\in\operatorname{Nul}(\mathbf{A}-\lambda\mathbf{I}_2)=E_{\lambda}\). Reversing the subtraction proves the converse. \(\square\)

Why the Characteristic Roots Give the Eigenvalues

The characteristic polynomial step is justified by the two-by-two determinant criterion: a two-by-two matrix has determinant zero exactly when its homogeneous system has a nonzero solution. Here is a direct verification of that criterion, so the connection between roots and eigenvalues is explicit.

Proof. Write \(\mathbf{B}=\mathbf{A}-\lambda\mathbf{I}_2=\begin{bmatrix}a&b\\c&d\end{bmatrix}\), with entries now regarded as complex numbers. If \(\mathbf{B}\mathbf{x}=\mathbf{0}\) for a nonzero \(\mathbf{x}=(x,y)^{\top}\), then \(ax+by=0\) and \(cx+dy=0\). If \(x\ne0\), multiply the first equation by \(d\), the second by \(b\), and subtract; the result is \((ad-bc)x=0\), so \(ad-bc=0\). If \(x=0\), then \(y\ne0\), and the equations give \(b=d=0\), again yielding \(ad-bc=0\). Conversely, suppose \(ad-bc=0\). If the first row is nonzero and \(a\ne0\), the determinant equation gives \(d=bc/a\), so the second row is \((c/a)\) times the first. The nonzero vector \((-b,a)^{\top}\) is annihilated by both rows. If the first row is nonzero with \(a=0\), then \(b\ne0\); the determinant equation gives \(c=0\), and \((1,0)^{\top}\) is a nonzero solution. If the first row is zero, then either the second row is nonzero and \((-d,c)^{\top}\) is a nonzero solution, or both rows are zero and \((1,0)^{\top}\) is a nonzero solution. Therefore \(\det\mathbf{B}=0\) exactly when \(\mathbf{B}\mathbf{x}=\mathbf{0}\) has a nonzero solution. Applying this to \(\mathbf{B}=\mathbf{A}-\lambda\mathbf{I}_2\), and using the definition of an eigenpair, proves the claim. \(\square\)

For a two-by-two matrix \(\mathbf{A}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\), the determinant to factor is \(\det(\mathbf{A}-\lambda\mathbf{I}_2)=(a-\lambda)(d-\lambda)-bc\). The minus sign between the diagonal product and \(bc\) is part of the determinant formula. Once the roots are found, substitute each one into \(\mathbf{A}-\lambda\mathbf{I}_2\), solve the resulting homogeneous system, and exclude the zero vector.

Real Eigenvalues: Two Null-Space Calculations

When the Eigenvalues Are Complex

A real matrix need not have real eigenvalues. If its characteristic polynomial has nonreal roots, the eigenpair calculation can be carried out over \(\mathbb{C}\). The null-space procedure is unchanged: substitute each root and solve the resulting system, now allowing complex entries and components.

A Sign Error to Avoid

Common mistake: using a plus sign for the off-diagonal product. For \(\mathbf{B}=\begin{bmatrix}2&1\\1&2\end{bmatrix}\), an incorrect calculation is \(\det(\mathbf{B}-\lambda\mathbf{I}_2)=(2-\lambda)^2+1\). This uses the wrong sign: for \(\begin{bmatrix}a&b\\c&d\end{bmatrix}\), the determinant is \(ad-bc\), not \(ad+bc\). The corrected calculation is

So the eigenvalues are \(1\) and \(3\), not the nonreal roots suggested by the mistaken polynomial. For \(\lambda=1\), \(\mathbf{B}-\mathbf{I}_2=\begin{bmatrix}1&1\\1&1\end{bmatrix}\), whose null space is spanned by \((1,-1)^{\top}\); indeed \(\mathbf{B}(1,-1)^{\top}=(1,-1)^{\top}\). For \(\lambda=3\), \(\mathbf{B}-3\mathbf{I}_2=\begin{bmatrix}-1&1\\1&-1\end{bmatrix}\), whose null space is spanned by \((1,1)^{\top}\); indeed \(\mathbf{B}(1,1)^{\top}=(3,3)^{\top}=3(1,1)^{\top}\). Factoring the characteristic polynomial and checking the eigenvectors by multiplication catches errors from either stage.

A Reliable Calculation Pattern

For a two-by-two eigenpair problem, use the same sequence every time:

The characteristic polynomial determines which scalars to test, but it does not by itself give the eigenvectors. The null-space calculation supplies those vectors, and direct multiplication confirms that each proposed pair satisfies the defining equation. This separation is useful in any eigenpair computation: a correct eigenvalue with an incorrectly solved system is not yet a correct eigenpair.