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Past exam of the mathematics course of the University of Cambridge / 2013 / ia / Paper 1 / 8B / a / ii

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2013 ia Paper 1 8B a
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ii
By the given matrix similarity, A′=P−1AP for an invertible P. Hence
det(tI−A′)=det{P−1(tI−A)P}=det(P−1)det(tI−A)det(P)=det(tI−A).
(1)
The characteristic polynomials are identical. Therefore the characteristic equations and eigenvalues, including algebraic multiplicities, agree. The conclusion is independent of the choice between det(tI−A) and det(A−tI) conventions.

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