 Solution

ID: past-exam-of-the-rings-and-modules-course-of-the-university-of-oxford/2023/3/a/iv/solution

In the basis

(v, A v)

, the linear map given by

A

sends

v

to the second basis vector and sends

A v

A^{2} v = - v

. Equivalently,

A E = (A v, - v) = E (01 - 1 0) .

Multiplying on the left by the matrix inverse

E^{- 1}

gives

E^{- 1} A E = (01 - 1 0) .

This change of basis produces the companion matrix found in part (a)(ii).

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