A real polynomial can have complex roots. Likewise, a real matrix can have complex eigenvalues and eigenvectors. For example, the matrix $A= \begin{pmatrix} 2 & -1 \cr 1 & 2 \end{pmatrix}$ represents a rotation in ${\bf R}^2$, followed by a stretch, so it doesn't have any real eigenvectors. ($A {\bf x}$ is always $\tan^{-1}(1/2)$ counterclockwise of ${\bf x}$, so we can't have $A{\bf x} = \lambda {\bf x}$.) However, $$ \begin{pmatrix} 2& -1 \cr 1 & 2 \end{pmatrix} \begin{pmatrix}i \cr 1 \end{pmatrix} = \begin{pmatrix} -1+2i \cr 2+i \end{pmatrix} = (2+i) \begin{pmatrix}i \cr 1 \end{pmatrix}.$$ We have an eigenvector, just not a real eigenvector.

As explained in the following video, we find complex eigenvalues and eigenvectors the same way that we find real eigenvalues and eigenvectors.

• Compute the characteristic $p_A(\lambda) = \det(\lambda I - A)$.
• The eigenvalues of $A$ are the roots of $p_A(\lambda)$.
• If some of these roots are complex, so be it.
• For each eigenvalue $\lambda$, whether real or complex, the eigenspace $E_\lambda$ is the null space of $A-\lambda I$, which you can compute by row-reducing $A-\lambda I$.



If $A$ is a real matrix, then complex eigenvalues and eigenvectors come in conjugate pairs. If ${\bf \xi} = {\bf x} + i {\bf y}$ is an eigenvector with eigenvalue $a+bi$, then $\bar{\bf \xi} = {\bf x} -i {\bf y}$ is an eigenvector with eigenvalue $a-bi$. In the plane spanned by ${\bf y}$ and ${\bf x}$, $A$ acts just like multiplication by $\begin{pmatrix} a&-b \cr b & a \end{pmatrix}$, namely rotation by $\tan^{-1}(b/a)$ and stretching by $\sqrt{a^2+b^2}$.