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Is there a procedure to find the eigenvalues of $\textbf{M}$? ‎ $$\begin{eqnarray} ‎\textbf{M}=\left[‎ ‎\begin {array}{ccccc}‎ ‎\textbf{A} & \textbf{B} & & &\\‎ ‎\textbf{B}^T & \textbf{ A} & \textbf{B} & &\\‎ ‎&\ddots &\ddots & \ddots &\\‎ ‎& & & & \textbf{B} \\‎ ‎& & & \textbf{B}^T & \textbf{A} ‎\end {array}‎ ‎\right]‎, ‎\end{eqnarray}‎‎ $$

where $\textbf{B}^T$ is transpose of matrix $\textbf{B}$.

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    $\begingroup$ Presumably, $A$ is symmetric. $\endgroup$
    – Denis Serre
    Commented Oct 8, 2012 at 20:28
  • $\begingroup$ How does this problem differ from the usual symmetric eigenvalue computation? Indeed, you should be able to calculate the eigenvalues faster than $n^3$ time because of the "tridiagonal" structure. $\endgroup$
    – Christopher A. Wong
    Commented Oct 9, 2012 at 22:16
  • $\begingroup$ Think about a 2 by 2 block matrix---seems to get its exact eigenvalues, knowing the eigenvalues of A and B does not really help that much (to get bounds, yes, but to get exact ones...) $\endgroup$
    – Suvrit
    Commented Nov 6, 2012 at 17:58
  • $\begingroup$ If the matrix ${\bf B}$ is symmetric, we can use Theorem 4.1 in the paper: DOI: 10.1080/09720529.2020.1854939. In this case, the eigenvalues of matrix ${\bf M}$ are the same as eigenvalues of the matrix ${\bf A} + 2\cos\left(\frac{j \pi}{N+1}\bf{B}\right) $ where $N$ is the number of matrices $\bf{A}$ in the matrix $\bf{M}$. $\endgroup$
    – Sakurai.JJ
    Commented Jul 17 at 8:07

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