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

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.

Successive substitution in the matrix recurrence relation gives $V_k=V_0{Q}^k$. In the matrix power of $Q=ED{E}^{-1}$, every adjacent $E^{-1}E$ cancels, so $Q^k=E{D}^k{E}^{-1}$. ConsequentlyHere $D^k=\mathrm{diag}({\lambda }_1^k,\dots ,{\lambda }_n^k)$ for $k\ge 1$, and $D^0=I$. The formula also gives $V_0$ correctly at $k=0$.