 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/ib/paper-1/9g/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

The

n

spanning vectors form a basis. Write

T^{n} x = - a_{0} x - a_{1} T x - \dots - a_{n - 1} T^{n - 1} x .

In this cyclic basis the matrix is the companion matrix

C = 010 ⋮ 0 001 ⋮ 0 \dots \dots \dots ⋱ \dots 000 ⋮ 1 - a_{0} - a_{1} - a_{2} ⋮ - a_{n - 1} .

Expanding

det (t I - C)

along its last column gives

χ_{T} (t) = t^{n} + a_{n - 1} t^{n - 1} + \dots + a_{1} t + a_{0} .

For example, the term containing

a_{j}

has minor

t^{j}

and the signs from the subdiagonal cancel the cofactor sign; the diagonal term supplies

t^{n}

. The defining relation says

χ_{T} (T) x = 0

. Every polynomial in

T

commutes with

T

, so

χ_{T} (T) T^{j} x = T^{j} χ_{T} (T) x = 0 (0 \leq j < n) .

It therefore vanishes on a basis, proving

χ_{T} (T) = 0.

This direct argument for a cyclic vector does not assume the Cayley-Hamilton theorem.

 New to topics? Read the docs here!