 Solution

ID: past-exam-of-the-algebra-1-and-differential-equations-1-course-of-the-university-of-oxford/2014/2/c/ii/solution

For

λ \neq = 0

, factor the matrix inside the determinant:

χ_{M} (λ) = det M det (λ M^{- 1} - I) = λ^{2 m} det (M^{- 1} - λ^{- 1} I) = λ^{2 m} det (λ^{- 1} I - M^{- 1}) = λ^{2 m} χ_{M} (λ^{- 1}) .

det M = 1

; the third uses the even matrix size; the last uses part (i). Consequently

χ_{M} (λ) = λ^{2 m} χ_{M} (λ^{- 1}) .

 New to topics? Read the docs here!