Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 17 / c / i / Solution

The eigenvector equations $Q{\mathbf{e}}_i={\lambda }_i{\mathbf{e}}_i$ combine column by column into $QE=ED$. Since $|E|\ne 0$, multiplication by the inverse matrix gives the diagonalizationThe nonzero determinant of $E$ ensures that its columns form an eigenbasis.