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Exercise 4.2.4

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We have $(CD) q_{k} = CD q_{k} = C λ_{k} (D) q_{k} = λ_{k} (D) C q_{k} = λ_{k} (D) λ_{k} (C) q_{k}$ , on the other hand $(CD) q_{k} = λ_{k} (CD) q_{k}$ , so we have $λ_{k} (CD) = λ_{k} (D) λ_{k} (C)$

niuers

2020-03-20 00:00

Exercise 4.2.4

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