Question

Let A, B be n × n matrices. The following are two incorrect proofs that ABhas the same non-zero eigenvalues as BA. For each, state two things wrong with the proof: (i) We will prove that AB and BA have the same characteristic equation. We have that

det(AB − λI) = det(ABAA−1 − λAA−1) = det(A(BA − λI)A−1)

= det(A) + det(BA − λI) − det(A)

= det(BA − λI)

Hence det(AB − λI) = det(BA − λI), and so AB and BA have the same eigenvalues

(ii) Suppose not. We will prove by contradiction. Let λ,μ be non-zero eigenvalues of AB,BA respectively such that λ ̸= μ. Then we have that for an eigenvector ⃗v

AB⃗v=λ⃗v, BA⃗v=μ⃗v=⇒(AB−BA)⃗v=(λ−μ)⃗v= ⇒ 0 ⃗v = ( λ − μ ) ⃗v

=⇒ λ − μ = 0

=⇒ λ = μ
Hence we have arrived at a contradiction. So AB and BA have the same eigenvalues.

Answer 1