TITLE: Can a real symmetric matrix have complex eigenvectors?
QUESTION [46 upvotes]: A Hermitian matrix always has real eigenvalues and real or complex orthogonal eigenvectors. A real symmetric matrix is a special case of Hermitian matrices, so it too has orthogonal eigenvectors and real eigenvalues, but could it ever have complex eigenvectors?
My intuition is that the eigenvectors are always real, but I can't quite nail it down.

REPLY [6 votes]: I have a shorter argument, that does not even use that the matrix $A\in\mathbf{R}^{n\times n}$ is symmetric, but only that its eigenvalue $\lambda$ is real. Imagine a complex eigenvector $z=u+ v\cdot i$ with $u,v\in \mathbf{R}^n$. We simply have $(A-\lambda I_n)(u+v\cdot i)=\mathbf{0}\implies (A-\lambda I_n)u=(A-\lambda I_n)v=\mathbf{0}$, i.e., the real and the imaginary terms of the product are both zero. We obtained that $u$ and $v$ are two real eigenvectors, and so,
the complex eigenvector $z$ is merely a combination of other real eigenvectors. One can always multiply real eigenvectors by complex numbers and combine them to obtain  complex eigenvectors like $z$.