Is it true that for a given real matrix, $A$, the eigenvector corresponding to real eigenvalues of $A$ will have entirely real values?
I know that of course eigenvectors are the same up to a constant scaling factor, which could be a complex value. So I am asking if such eigenvectors can always be rescaled to totally real eigenvectors.
I can see the other way: given a real eigenvector of a real matrix, the corresponding eigenvalue must be real, but I'm having trouble with this direction.
Best Answer
Let $Av=\lambda v$, where $A\in \mathbb R^{n\times n}$, $\lambda\in \mathbb R$ and $v\in \mathbb C^n$. Observe, that $(A-\lambda I)\in \mathbb R^{n\times n}$. By row manipulations, we can calculate $\operatorname{ker}(A-\lambda I)$, which is a subspace of $\mathbb R^n$, because with every row manipulation there doesn't occur a complex but not real number. So, you are right, we get $v\in \mathbb R^n$.