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Exercise 2.4.16

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We can check that $Φ$ is linear since

Φ (A + cD) = B^{- 1} (A + cD) B = B^{- 1} (AB + cDB)

= B^{- 1} AB + c B^{- 1} DB = Φ (A) + cΦ (D) .

And it’s injective since if $Φ (A) = B^{- 1} AB = O$ then we have $A = BO B^{- 1} = O$ . It’s also be surjective since for each $D$ we have that $Φ (BD B^{- 1}) = D$ .

jephianlin

2011-06-27 00:00

Exercise 2.4.16

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