Given a matrix $\mathbf{A} \in \mathbb{Z}^{d \times d}$ I need to find an algebraic number $a$ of minimal degree, such that all eigenvalues and eigenvector's coordinates of $\mathbf{A}$ belong to the extension $\mathbb{Q}(a)$.
I would start by finding characteristic polynomial $f$ of the matrix - then $\mathbb{Q}/f$ will contain all eigenvalues. Since all algebraic numbers are eigenvalues of some rational matrix, it's guaranteed that my $a$ is one of the roots of $f$. But how do I ensure that also the coordinates of eigenvectors lie in my extension field?
Or is there a better way to start, or any hints?