1. ## Isomorphism...

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

2. 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)$\displaystyle \phi (AB) = \phi (A) \phi (B)$.

3. 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.

4. 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.

5. 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