How to show that two Hermitian tridiagonal matrices are similar?

Question

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 show that the matrices have the same eigenvalues? I want to use the property that they are similar but I don't now how to do that.

Thanks in advance for any help!

Answer 1

Suppose that $b_j = |b_j|e^{i \theta_j}$ for $j = 1,2,...,n-1$ and perform the unitary similarity $B \mapsto DB\overline{D} = A$, in which $D$ is the $n \times n$ diagonal unitary matrix $$ D = \operatorname{diag}(1,e^{i\theta_1},e^{i(\theta_1 + \theta_2)},e^{i(\theta_1 + \theta_2 + \theta_3)},...) $$