 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ia/paper-1/7a/b/solution

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

Use the convention

P_{A} (t) = det (t I - A) = \sum_{r = 0}^{n} a_{r} t^{r}

det (A - t I)

instead multiplies it by

(- 1)^{n}

and leaves the requested zero identity unchanged. For each eigenvector

v_{j}

, induction gives

A^{r} v_{j} = λ_{j}^{r} v_{j}

, including

r = 0

. Thus

(\sum_{r = 0}^{n} a_{r} A^{r}) v_{j} = P_{A} (λ_{j}) v_{j} = 0,

because

λ_{j} I - A

has the nonzero kernel vector

v_{j}

and therefore has zero determinant. Part (a) supplies a basis consisting of these

n

eigenvectors. A linear map vanishing on every basis vector vanishes everywhere. Consequently

r = 0 \sum n a_{r} A^{r} = 0.

This proves the Cayley-Hamilton theorem identity directly in the distinct-eigenvalue case assumed here.

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