 Solution

A diagonalizable matrix over the field $F$ admits an invertible matrix $P$ with entries in $F$ such thatEquivalently, the vector space $F^n$ has a basis of eigenvectors of $A$. Indeed, $AP=PD$ says that each column of $P$ is an eigenvector with the corresponding diagonal entry as its eigenvalue. Requiring the basis and eigenvalues to lie in $F$ makes the choice of field essential.