Let A and B be similar matrices. Prove that there is a nice relationship between the eigenvectors of A and the eigenvectors of B
I suppose you could say that the relationship between the eigenvectors is that they share the same eigenvalue. If you're looking for how to prove that the eigenvalues are the same, I have written a short proof here:
det(A-xI)=det(inv(P)BP-xI)
=det(inv(P)BP-inv(P)xIP)
=det(inv(P)(B-xI)P)
=det(B-xI)
Thus, as the characteristic polynomials are the same, the eigenvalues are the same.
I've put a formatted version up: Maths homework help
If A and B are similar matrices, then there exist an invertible matrix P such that $\displaystyle A= P^{-1}BP$.
Let v be an eigenvector of A corresponding to eigenvalue $\displaystyle \lambda$. Then $\displaystyle Av= (P^{-1}BP)v= \lambda v$. Multiplying on both sides by P, $\displaystyle (BP)v= P\lambda v$ so $\displaystyle B(PV)= \lambda (PV)$.
That is, if v is an eigenvalue of A corresponding to eigenvalue $\displaystyle \lambda$ then Pv is an eigenvalue of B also corresponding to eigenvalue $\displaystyle \lambda$.
Of course, we can think of P as a "change of basis" matrix. That is, A and B represent the same linear transformation, written in different bases. The eigenvectors of A and B are representations of the same vector, written in different bases.