# 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

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