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.
Best Answer
Always try out examples, starting out with the simplest possible examples (it may take some thought as to which examples are the simplest). Does for instance the identity matrix have complex eigenvectors? This is pretty easy to answer, right?
Now for the general case: if $A$ is any real matrix with real eigenvalue $\lambda$, then we have a choice of looking for real eigenvectors or complex eigenvectors. The theorem here is that the $\mathbb{R}$-dimension of the space of real eigenvectors for $\lambda$ is equal to the $\mathbb{C}$-dimension of the space of complex eigenvectors for $\lambda$. It follows that (i) we will always have non-real eigenvectors (this is easy: if $v$ is a real eigenvector, then $iv$ is a non-real eigenvector) and (ii) there will always be a $\mathbb{C}$-basis for the space of complex eigenvectors consisting entirely of real eigenvectors.
As for the proof: the $\lambda$-eigenspace is the kernel of the (linear transformation given by the) matrix $\lambda I_n - A$. By the rank-nullity theorem, the dimension of this kernel is equal to $n$ minus the rank of the matrix. Since the rank of a real matrix doesn't change when we view it as a complex matrix (e.g. the reduced row echelon form is unique so must stay the same upon passage from $\mathbb{R}$ to $\mathbb{C}$), the dimension of the kernel doesn't change either. Moreover, if $v_1,\ldots,v_k$ are a set of real vectors which are linearly independent over $\mathbb{R}$, then they are also linearly independent over $\mathbb{C}$ (to see this, just write out a linear dependence relation over $\mathbb{C}$ and decompose it into real and imaginary parts), so any given $\mathbb{R}$-basis for the eigenspace over $\mathbb{R}$ is also a $\mathbb{C}$-basis for the eigenspace over $\mathbb{C}$.