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

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Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2014 / 2 / c / iii / Solution by

Codex  0 2026-10-31

Let

P = χ_{M}

, and suppose

P (z) = (z - λ)^{d} R (z)

with

R (λ) \neq = 0

. The reciprocal identity gives, for

x \neq = 0

,

P (x) = x^{2 m - d} (1 - λ x)^{d} R (1/ x) .

Near

x = λ^{- 1}

the factor

x^{2 m - d} R (1/ x)

is nonzero. Also

1 - λ x = - λ (x - λ^{- 1})

has a simple zero there. Hence

λ^{- 1} is a root of multiplicity exactly d .

This proves equality of the algebraic multiplicities, not merely that the inverse is another eigenvalue. Invertibility excludes zero eigenvalues throughout.

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