Finding eigenvector and eigenvalue for a hypothetical diagonalized matrix? 
If there are some $A, B,$ and $S$ (all of which are $n \times n$) such that $A = S(B)S^{-1}$. Suppose $A$ has eigenvector $v$ and eigenvalue $\lambda$. How can we solve for an eigenvector of $B$ and find a corresponding eigenvalue?

My Attempt
If $B = S^{-1}(A)S$, then the column vectors of $S$ give an eigenbasis for $A$.
By that logic, the column vectors of $S$ in our case give an eigenbasis for $B$ and the diagonal entries of $A$ will be the eigenvalues. I don't really know where to go from here. Thank you!
 A: Regarding your attempt:

My thought process so far: If $B = S^{-1}A S$, then the column vectors of S give an eigenbasis for A. 

This only holds when $A$ is a diagonal matrix.  Although this might look like "diagonalization", the fact that $B = S^{-1}AS$ doesn't necessarily have anything to do with diagonal matrices.  If you want a way to think about the relationship between $B$ and $A$, then we can note that $B$ represents the transformation $x \mapsto Ax$ relative to a non-standard basis.
Setting all that aside, here is a very direct approach for your problem. We know that $A$ "has eigenvector $v$ and eigenvector $\lambda$".  In other words, $v$ is a non-zero vector and $A v = \lambda v$.  In order to try to say something about $B$, we can plug in:
$$
(S^{-1}BS)v = \lambda v
$$
That doesn't look very helpful yet, but we can rewrite the equation by multiplying both sides by $S$ (from the left) to get
$$
BSv = \lambda Sv.
$$
Notice that the term "$Sv$" appears on both sides. If we group it off, the equation reads $B(Sv) = \lambda (Sv)$.  Since $Sv$ is non-zero (why?), that means that the vector $Sv$ is an eigenvector of $B$!
