An n-by-n matrix is diagonalizable if and only if the union of the basis vectors for the eigenspaces is a basis for R^n (or C^n).

Choose a basis for each eigenspace corresponding to an eigenvalue in $\mathbb F$. The matrix is diagonalizable over $\mathbb F$ iff their combined list is a basis of $\mathbb F^n$. Independence between distinct eigenspaces is automatic; having enough eigenvectors is the substantive condition.Copy