How to find the eigenvectors of two closely related hermitian tridiagonal matrices Given two tridiagonal hermitian matrices A,B with $a_i\in \mathbb{R}$ and $b_i\in \mathbb{C}$ as follows
\begin{align} 
A=
\begin{pmatrix}
  a_{1} & |b_1| & \cdots & 0  \\
  |b_1| & a_{2} & \cdots & 0 \\
  \vdots  & \ddots  & \ddots & |b_{n-1}|  \\
  0 & 0 & |b_{n-1}| & a_{n}
 \end{pmatrix}
\end{align}
and
\begin{align}
B=
\begin{pmatrix}
  a_{1} & b_1 & \cdots & 0  \\
  \overline{b_1} & a_{2} & \cdots & 0 \\
  \vdots  & \ddots  & \ddots & b_{n-1}  \\
  0 & 0 & \overline{b_{n-1}} & a_{n}
 \end{pmatrix}
\end{align}
is there any way to find the eigenvectors of B in case I already know the eigenvectors of A?
It is relatively easy (if one knows the proper recursion) to see that the eigenvalues of A and B are all the same, but I have a hard time to find the eigenvectors of B by knowing those of A...
All suggestions are warmly appreciated.
bests
EDIT:
I tried some examples with mathematica, but to be honest, I couldn't find any consistent pattern which I could generalize.
I tried to compute it directly by solving $ker(B-\lambda I)$ but I couldn't make use of knowing the eigenvectors of A.
Maybe something is possible using the spectral theorem, the basis transformation S for A is known, so...
$A=S*D*S^h$ and $B=U*D*U^h$ 
but how to find U...I am stuck at the moment
 A: Can't believe that I didn't see this earlier (although it is not the complete solution), the argument goes as follows:


*

*show that $A$ and $B$ are similar to D and therefore similar to each other (equivalence relation)

*use the transformation matrix to express the eigenvectors of $A$ by knowing those of $B$ or vice versa


Since we already know that the matrices are diagonalizable and have the same eigenvalues it is clear that they are similar and therefore we have a transformation matrix $T\in \mathbb{C}^{n,n}$ such that
$$
B=T^{-1}*A*T
$$
now we let $v$ be an eigenvector $A$, such that $A*v=\lambda*v$, then we know that $w:=T^{-1}v$ is an eigenvector of B, since it holds
$$
B*w=T^{-1}*A*T*w=T^{-1}*A*T*T^{-1}v=T^{-1}*Av=\lambda T^{-1}v=\lambda w
$$
and we found a way to express the eigenvectors of B by knowing those of A. 
Clearly, there is one thing still missing and this would be how to find the proper transformation matrix in a general case (if there exists one at all). In fact I think this might be the most tough part.
bests
