 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ia/paper-1/7b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

By the definition of an eigenvalue, there is a nonzero eigenvector

v

with

A v = λ v

. Induction using matrix multiplication gives

A^{r + 1} v = A (A^{r} v) = A (λ^{r} v) = λ^{r} A v = λ^{r + 1} v,

starting from

r = 1

. Thus the same nonzero eigenvector establishes

A^{r} v = λ^{r} v, λ^{r} is an eigenvalue of A^{r} (r \geq 1) .

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