# Eigenvalue Problem of a hermitian tridiagonal matrix

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.)

• Did you try the 2x2 and 3x3 cases? Apr 28 at 18:06
• I actually don't immediately know the solution to this problem. However, the first idea that comes to mind is "can I explicitly furnish a similarity transformation?" and for that some small examples seem useful.
– Ian
Apr 28 at 18:16
• Thanks for the idea. I calculated the cases 2x2 and 3x3. I am trying to find a general formula now (probably trying a 4x4 matrix). Gonna return as soon as I get a result. Apr 28 at 18:50

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.
• Thanks for the answer. It helped me a lot. After a lot of thinking and computing I got a recursive formula for my characteristic polynomial $\chi(\lambda)$ which holds $\forall n \in \mathbb{N}$. The recursion is $\chi_{A_n}(\lambda) = (\alpha_n - \lambda)\chi_{A_{n-1}}(\lambda) - \overline{\beta}_{k-1}\beta_{k-1}\chi_{A_{n-2}}(\lambda)$ with $\chi_{A_0} := 1$ and $\chi_{A_{-1}} := 1$. This recursion also holds for B (replace the $\overline{\beta}\beta$ with $\vert\beta\vert and \vert \beta \vert$) And since $\overline{\beta}\beta = \vert\beta\vert^2$ the characteristic polynomial is the same. Apr 30 at 18:17
• I meant $\chi_{A_{-1}} = 0$. Apr 30 at 18:27