If all eigenvalues of a matrix $A \in \Bbb{R}^{nxn}$ are real, then I can choose each eigenvector so that is not purely complex.

Question

Suppose $A \in \Bbb{R}^{n\times n}$. I was trying to reduce to an absurd (i.e. suppose always $v \in \Bbb{C}^{n}$) but I think I am not getting anywhere.

Answer 1

If $\lambda\in\mathbb R$ is an eigenvalue of $A$, then the eigenvectors for $\lambda$ are the solutions to the linear system $(A-\lambda I)v=0$. This is a system consisting only of real entries, so the solutions will always be real. This can be seen for instance by finding the solutions via row reduction.