I want to prove that the eigenvalues of the following complex, hermitian matrices are the same $$ A=\begin{pmatrix} \alpha_1&\beta_1\\ \overline{\beta_1}&\alpha_2&\beta_2\\ &\overline{\beta_2}&\ddots&\ddots\\ &&\ddots&\ddots&\ddots\\ &&&\ddots&\ddots&\beta_{n-1}\\ &&&&\overline{\beta_{n-1}}&\alpha_n \end{pmatrix}, B=\begin{pmatrix} \alpha_1&\vert\beta_1\vert\\ \vert\beta_1\vert&\alpha_2&\vert\beta_2\vert\\ &\vert\beta_2\vert&\ddots&\ddots\\ &&\ddots&\ddots&\ddots\\ &&&\ddots&\ddots&\vert\beta_{n-1}\vert\\ &&&&\vert\beta_{n-1}\vert&\alpha_n \end{pmatrix} $$ I already know that $\forall i \in {1,...,n}: \alpha_i \in \mathbb{R}$ and that the eigenvalues have to be real, since A, B are hermitian.
Any help is welcome. (The question is from a book I want to study but I have no clue how to start.)
The easiest argument I can see is that both matrices have identical characteristic polynomials. This comes from the fact that for each term in the Leibniz formula for the determinant giving the characteristic polynomial is associated to a permutation, and the two terms associated to a given permutation for both matrices are identical. If a permutation moves any index by more than $1$ place up or down, then the term involves a position outside the three diagonals, giving a factor$~0$; such terms contribute nothing to either characteristic polynomial. So we only need to consider permutation that only move indices by at most 1 place; such permutations are a product of a number (maybe $0$) of disjoint adjacent transpositions. Any fixed point of the permutation gives a factor from the main diagonal of the matrix, and this factor is the same for both matrices. The remaining factors are can be grouped by the adjacent transpositions of the permutation, and then give a product $\beta_i\overline{\beta_i}$ for the first matrix, and $|\beta_i|^2$ for the second; these numbers are equal. Hence the characteristic polynomials are the same.