if A is similar to B with P^-1 A P = B and x is an eigenvector of A, prove that P^-1x is an eigenvector of B
The Attempt at a Solution
so far I did this:
A = PBP^-1
Ax = PBP^-1x
P^-1Ax = BP^-1x
B^-1P^-1Ax = P^-1x
is that sufficient to prove that P^-1x is the eigenvector of B? ??
I don't think so .. if yes please explain how.. or else please give a proper "proof" .
I guess the reason I don't think this is right is becaues I am not able to bring it to the format of Ax = tx where t is an eigenvalue and x is the eigenvector. Any help would be appreciated.
Final exam on tuesday ... :(