 Solution

ID: 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/ii/solution

Successive substitution in the matrix recurrence relation gives

V_{k} = V_{0} Q^{k}

. In the matrix power of

Q = E D E^{- 1}

, every adjacent

E^{- 1} E

cancels, so

Q^{k} = E D^{k} E^{- 1}

. Consequently

V_{k} = V_{0} E D^{k} E^{- 1}, k \geq 0.

Here

D^{k} = diag (λ_{1}^{k}, \dots, λ_{n}^{k})

for

k \geq 1

, and

D^{0} = I

. The formula also gives

V_{0}

correctly at

k = 0

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