Solution
= Solution
Successive substitution in the <matrix> <recurrence relation> gives $V_k=V_0Q^k$. In the <matrix power> of $Q=EDE^{-1}$, every adjacent $E^{-1}E$ cancels, so $Q^k=ED^kE^{-1}$. Consequently
$$
\boxed{V_k=V_0ED^kE^{-1}},\qquad k\geq0.
$$
Here $D^k=\operatorname{diag}(\lambda_1^k,\ldots,\lambda_n^k)$ for $k\geq1$, and $D^0=I$. The formula also gives $V_0$ correctly at $k=0$.