Can eigenvectors of a symmetric circulant matrix be real? Since eigenvectors of a symmetric circulant matrix are related to the Fourier basis, I expect that they are complex.
But I do not find such statement in mathematical books, and so I think my supposition is wrong.
Can you help me to understand the correct reasoning?
 A: First of all, the eigenvalues and thus eigenvectors of  whatever symmetric real matrix are real. That the matrix happens to be circulant does not change this. But let's go through it for circulant specifically: 
The eigenvalues of a circulant matrix (with first row $(c_0, \dots, c_{n-1})$) are given by (see the Wikipedia page)
$$\lambda(w)=c_0 +c_1 w + c_2 w^2 + \dots +   c_{n-1} w^{n-1}$$
where $w$ is some $n$-th root of unity. 
For the matrix to be symmetric one needs $c_{n-i} = c_i$.
One can then rewrite 
$$c_i w^i  + c_{n-i} w^{n-i}= 2c_i \Re(w^i).$$
Thus for odd $n$ one gets:
$$c_0 +  2c_1 \Re w + 2c_2\Re w^2 +   \dots +   2c_{(n-1)/2} \Re(w^{(n-1)/2})$$
which is always real (assuming a real matrix of course).
And for even $n$ one gets 
$$c_0 + c_{n/2}w^{n/2} + 2c_1 \Re w + 2c_2\Re w^2 +   \dots +   2c_{(n-1)/2} \Re(w^{(n-2)/2})$$
which is also always real as $w^{n/2}$ is $\pm 1$. 
The eigenvectors then can also be chosen to be real. 
Let me further add that while $(1, w , \dots, w^{n-1})$ is always an eigenvector for the eigenvector $\lambda(w)$, as explained on the site above. One has  $\lambda(w)= \lambda(w^{-1})$  so that one has the eigenvectors $(1, w , \dots, w^{n-1})$  and $(1, w^{-1} , \dots, w^{-(n-1)})$. The sum of the two, twice the real part, is then a real eigenvector (same for the imaginary part).
