Let B be an n x n invertible matrix. Define Ф: Mnxn (F) -> Mnxn (F) by Ф (A) = B^-1 AB. Prove that Ф is an isomorphism
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Originally Posted by leungsta Let B be an n x n invertible matrix. Define Ф: Mnxn (F) -> Mnxn (F) by Ф (A) = B^-1 AB. Prove that Ф is an isomorphism Just check three conditions: (i)it is one-to-one, (ii)it is onto, (iii)$\displaystyle \phi (AB) = \phi (A) \phi (B)$.
Originally Posted by ThePerfectHacker Just check three conditions: (i)it is one-to-one, (ii)it is onto, (iii)$\displaystyle \phi (AB) = \phi (A) \phi (B)$. Quick note: Make sure it is a different B from the one defined in the question. It is also a ring isomorphism.
Originally Posted by whipflip15 Quick note: Make sure it is a different B from the one defined in the question. It is also a ring isomorphism. Thank you for correcting the notational mistake.
Originally Posted by ThePerfectHacker Just check three conditions: (i)it is one-to-one, (ii)it is onto, (iii)$\displaystyle \phi (AB) = \phi (A) \phi (B)$. This is correct. Another way for showing is to proof as the matrix is invertible. That is a constraint for isomorphism. If you have a linear map which is bijective then this linear map always has a inverse map. greetings Herbststurm
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