 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-2/17/c/iii/solution

A = P D P^{- 1}

. In the matrix power, each adjacent pair

P^{- 1} P

cancels to the identity matrix:

A^{n} = (P D P^{- 1}) \dots (P D P^{- 1}) = P D^{n} P^{- 1} .

This also follows by mathematical induction: the formula holds for

n = 1

A = P D P^{- 1}

turns its

n

th instance into the

(n + 1)

st.

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