How to compute the eigenvalues of the following matrix of order 3n, $X= \begin{pmatrix} I & m^2 A & m A \\ m^2 A & I & m A \\ m A & m A & I \end{pmatrix} $, where $I$, $A$ are $n \times n$ matrices. $I$ is identity matrix, $A$ is Hermitian and $A\geq 0$ with ${\rm Tr}[A]=1$ and $m=\frac{1}{n}$. Assume that the eigenvalues of $A$ are $\{\lambda_i; i=1,...,n\}$. Please help. Thanks.

I have tried to find the determinant of the matrix, $X$ by using the following lemma. "For a square block matrix, $S=\begin{pmatrix} S_{11} & S_{12} & S_{13} \\ S_{21} & S_{22} & S_{23} \\ S_{31} & S_{32} & S_{33}\end{pmatrix} $, the ${\rm det}(S)={\rm det}\Big([S_{11}-S_{13}S_{33}^{-1}S_{31}]-[S_{12}-S_{13}S_{33}^{-1}S_{32}][S_{22}-S_{23}S_{33}^{-1}S_{32}]^{-1}[S_{21}-S_{13}S_{33}^{-1}S_{31}]\Big){\rm det}(S_{22}-S_{23}S_{33}^{-1}S_{32}){\rm det}(S_{33})."$ But somehow this doesn't help. I want the eigenvalues of $X$ as a function of the eigenvalues of $A$. Will it be possible?

closed as off-topic by Jyrki Lahtonen, amWhy, José Carlos Santos, ArsenBerk, Trevor Gunn Nov 9 at 2:13

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