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.
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 $\mathbb{C}$.
If $A$ is a symmetric $n\times n$ matrix with real entries, then viewed as an element of $M_n(\mathbb{C})$, its eigenvectors always include vectors with non-real entries: if $v$ is any eigenvector then at least one of $v$ and $iv$ has a non-real entry.
On the other hand, if $v$ is any eigenvector then at least one of $\Re v$ and $\Im v$ (take the real or imaginary parts entrywise) is non-zero and will be an eigenvector of $A$ with the same eigenvalue. So you can always pass to eigenvectors with real entries.
If $x$ is an eigenvector correponding to $\lambda$, then for $\alpha\neq0$, $\alpha x$ is also an eigenvector corresponding to $\lambda$. If $\alpha$ is a complex number, then clearly you have a complex eigenvector. But if $A$ is a real, symmetric matrix ( $A=A^{t}$), then its eigenvalues are real and you can always pick the corresponding eigenvectors with real entries. Indeed, if $v=a+bi$ is an eigenvector with eigenvalue $\lambda$, then $Av=\lambda v$ and $v\neq 0$. So $A(a+ib)=\lambda(a+ib)\Rightarrow Aa=\lambda a$ and $Ab=\lambda b$. Thus, because $v\neq 0$ implies that either $a\neq 0$ or $b\neq 0$, you just have to choose.
I also had this doubt once.
The eigenvectors are usually assumed (implicitly) to be real, but they could also be chosen as complex, it does not matter.
Specifically: for a real symmetric matrix $\mathbf{A}$ and a given eigenvalue $\lambda$, we know that $\lambda$ must be real, and this readily implies that we can always find a real $\mathbf{p}$ such that
$$\mathbf{A} \mathbf{p} = \lambda \mathbf{p}$$
But recall that we the eigenvectors of a matrix are not uniquely given (even if the eigenvalue is not degenerated), they lie inside a subspace, so we have quite freedom to choose them: in particular, if $\mathbf{p}$ is eigenvector of $\mathbf{A}$, then also is $\mathbf{q} = \alpha \, \mathbf{p}$ , where $\alpha \ne 0$ is any scalar: real or complex.
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 sum $u+v\cdot i$ 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 $z$.
