For about real eigenvector. Let $A$ be a real $n×n$ matrix. If $A$ has a complex eigenvalue, then is there any possibility to have a real eigenvector corresponding to this complex eigenvalue?
And my second question is if there is no real eigenvector for  this complex eigenvalue, then is it not violating the definition of eigenvector to be always in $\mathbb{R^n}$?
 A: Let $A$ be a real matrix, and let $v$ be its real eigenvector corresponding to a complex eigenvalue $\lambda$. Then, $Av = \lambda v$. However, $Av$ is a product of a real matrix and a real vector, thus a real vector, whereas $\lambda v$ is a complex vector. Therefore, either $\lambda$ is real or $v$ is complex.
There are two ways to look at this case. One could say that, since we are working with real vector spaces, complex eigenvectors are not eigenvectors at all, as well as complex eigenvalues. This way, some matrices may have no eigenvalues at all (consider the $2\times 2$ matrix of plane rotation by $90^\circ$).
Another option is to view the matrix as acting on a complex vector space (that is, the matrix can in general be complex, it is just this particular matrix that happens to have real entries). Then, the eigenvalues and eigenvectors can be complex (and in fact any matrix has one, thanks to the fundamental theorem of algebra).
A: I assume that by a complex number here you are referring a number with imaginary part non-zero. Consider the eigenvalue- eigenvector equation
$Av=\lambda v.........(1)$
For the possibilities you mentioned in question the LHS is a real vector and RHS will be a complex vector (with imaginary part non-zero) which makes impossible for (1) to hold.
Though in usual sense for a complex eigenvalue or eigenvector, this is possible if the vector space you are dealing with has the underlying field $\mathbb C$.
