Complex Eigenvalues In Real 3x3 Examples

From a Real Cubic to a Complex Eigenpair

A real \(3\times3\) matrix can have one real eigenvalue and two nonreal eigenvalues. The nonreal eigenvalues occur as a conjugate pair, so their two linear factors combine into a quadratic with real coefficients. Factoring the characteristic polynomial identifies the eigenvalues; solving a complex homogeneous system then gives an eigenvector for each one.

We work over \(\mathbb{C}\) when solving for eigenvectors, even though the entries of the matrix are real. For a real matrix \(\mathbf{A}\in\mathbb{R}^{3\times3}\), its characteristic polynomial is \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_3)\). A complex number \(\lambda\) is an eigenvalue if there is a nonzero vector \(\mathbf{v}\in\mathbb{C}^3\) such that \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). Equivalently, \((\mathbf{A}-\lambda\mathbf{I}_3)\mathbf{v}=\mathbf{0}\) has a nonzero solution. The determinant criterion says this occurs exactly when \(p_{\mathbf{A}}(\lambda)=0\).

Proof. The entries of \(\mathbf{A}\) are real, so the determinant defining \(p_{\mathbf{A}}\) has real coefficients. Consequently, for every complex number \(z\), \(p_{\mathbf{A}}(\overline{z})=\overline{p_{\mathbf{A}}(z)}\). Since \(p_{\mathbf{A}}(a+bi)=0\), this identity gives \(p_{\mathbf{A}}(a-bi)=0\) as well. The two distinct roots \(a+bi\) and \(a-bi\) give the real quadratic factor \((\lambda-(a+bi))(\lambda-(a-bi))=(\lambda-a)^2+b^2\). To see that this quadratic divides \(p_{\mathbf{A}}\) over \(\mathbb{R}\), divide \(p_{\mathbf{A}}\) by it. The remainder has degree at most one and vanishes at both distinct roots; a polynomial of degree at most one with two distinct roots is zero. The quotient is a real linear polynomial. Since \(p_{\mathbf{A}}(\lambda)\) has leading term \(-\lambda^3\), the quotient has leading term \(-\lambda\), so it is \(r-\lambda\) for some real \(r\). \(\square\)

This factor form follows the course convention \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_3)\); its leading coefficient is negative. The quadratic factor has no real roots because \((\lambda-a)^2+b^2>0\) for every real \(\lambda\) when \(b\ne0\). In each example, first factor the polynomial, then solve the system for one member of the nonreal pair. The other eigenvector can be obtained by conjugating the first.

Finding the Eigenvectors

The conjugation rule for eigenvectors is worth proving before using it. It explains why solving for one eigenvalue in the pair is enough.

Proof. Take the complex conjugate of each component of \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). Because every entry of \(\mathbf{A}\) is real, conjugation leaves the entries of \(\mathbf{A}\) unchanged and gives \(\mathbf{A}\overline{\mathbf{v}}=\overline{\lambda}\,\overline{\mathbf{v}}\). Since \(\mathbf{v}\ne\mathbf{0}\), its conjugate is also nonzero. Thus \(\overline{\mathbf{v}}\) is an eigenvector for \(\overline{\lambda}\). \(\square\)

When the matrix is not block diagonal, solving for a complex eigenvector can involve all three coordinates. The characteristic polynomial still reduces the eigenvalue calculation to a real linear factor and a real quadratic factor.

A Companion-Matrix Example

A matrix need not visibly contain a \(2\times2\) rotation block for its characteristic polynomial to have a conjugate pair. The direct method remains the same: compute the determinant, factor it, and solve the system at one complex root.

Common Mistake: Conjugating Only the Eigenvalue

For a real matrix, conjugating an eigenpair requires conjugating both the eigenvalue and the eigenvector. For the second example, a tempting but incorrect claim is that \((2,-2i,-1+i)^{\top}\) is an eigenvector for \(-i\), because that vector was found while studying the pair \(i,-i\). But its first component after matrix multiplication is \(2i\), whereas the first component of \((-i)(2,-2i,-1+i)^{\top}\) is \(-2i\). They differ, so the claimed eigenpair fails.

The correction is to conjugate the whole equation: the eigenvector for \(-i\) is \((2,2i,-1-i)^{\top}\), not the vector for \(i\). Directly, the first component of its matrix product is \(-2i\), which agrees with the first component of \((-i)(2,2i,-1-i)^{\top}\). The conjugation theorem guarantees the other components agree too.

A Reliable Calculation Order

Keeping the polynomial calculation separate from the eigenvector calculation helps prevent sign errors. In particular, do not read roots from a familiar factorization without checking the convention used in the determinant: here the real factor is \(r-\lambda\), not \(\lambda-r\). The roots are the same either way if the whole polynomial is adjusted consistently, but a mistaken overall sign can make a claimed factorization incorrect.

The check is not optional bookkeeping: it tests both the eigenvalue and the computed vector. A zero vector is never an eigenvector, even though it solves every homogeneous system, so the chosen vector must have at least one nonzero coordinate. Different nonzero scalar multiples of a valid eigenvector are also valid, but they must still satisfy the same matrix equation.