Finding eigenvectors of similar matrices Suppose I have two matrices $B$ and $ABA^{-1}$ where $A = [v_{1}\, v_{2}\, v_{3}]$ with $v_{i}$ column vectors such that $\{v_{1}, v_{2}, v_{3}\}$ form an orthonormal basis of $\mathbb{R}^{3}$ (not the standard basis). The eigenvalues of $B$ and $ABA^{-1}$ are the same. If I know the eigenvectors of $B$, is there an easy way to immediately know the eigenvectors of $ABA^{-1}$?
The example I'm thinking of takes $B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1/2 & \sqrt{3}/2\\ 0 & -\sqrt{3}/2 & 1/2\end{pmatrix}$.
 A: It is pretty straightforward. Let $v$ be an eigenvector of $B$, i.e., $Bv = \lambda v$ for some $\lambda$ (the corresponding eigenvalue). Then
$$A B v = A (\lambda v) = \lambda A v,$$
so
$$\lambda (A v) = A B v = A B (A^{-1} A) v = (A B A^{-1}) (A v),$$
which proves that $Av$ is an eigenvector of $A B A^{-1}$ associated with the same eigenvalue $\lambda$.
A: Here the change of basis matrix from the old to new is $A^{-1}$. So the eigen vector $v$ with the coordinate vector $X$ will be the vector with coordinate vector $(A^{-1})^{-1}X=AX$
Reason: If the $X$ is the coordinate vector of $v$ w.r.t basis $B_1$ and $Y$ w.r.t $B_2$ and $P$ is the change of basis matrix then $Y=P^{-1}X$
We have $B_2=B_1P$
$B_2Y=v=B_1X=B_2P^{-1}X \Rightarrow Y=P^{-1}X$
And if $A$ is the matrix of some linear transformation w.r.t $B_1$ then the linear transformation w.r.t $B_2$ is $P^{-1}AP$ because if we let $B$ be the matrix the linear transformation w.r.t $B_2$ then $v\to B_1AX=B_2BY=B_1PBP^{-1}X\Rightarrow A=PBP^{-1}\Rightarrow P^{-1}AP=B$
