Question about eigenvectors of real matrix with real eigenvalues I have two related questions:
Can a real matrix with real eigenvalues have complex eigenvectors?
Is it always the case that a real matrix with real eigenvalues is diagonalisable?
 A: 
Can a real matrix with real eigenvalues have complex eigenvectors?

Of course, if you have an eigenvector with real entries, then you can get one with complex entries simply by multiplying by $i$, for example.
But I think what you really mean to ask is:

If a real matrix has a real eigenvalue, can it be that all eigenvectors for that eigenvalue have complex (that is, non-real) entries?

For that, the answer is "no". If $\lambda$ is an eigenvalue, then the eigenvectors are the solutions to $(A-\lambda I)X=0$. And since $A-\lambda I$ has determinant 0, we can find a real solution by doing plain old Gaussian elimination on it, which involves nothing more dangerous than division by a nonzero number and therefore doesn't take us out of $\mathbb R$.
A: *

*If $A$ is real and $\lambda$ is real then the eigenvectors are the solutions to the real linear equation $(\lambda I-A)x=0$ which can be found as real.

*No. For example,
$$
\left[\matrix{0 & 1\\0 & 0}\right]
$$
has real eigenvalues (both are $0$) and is not diagonalizable.

