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Let $A$ be a complex Hermitian matrix of order $n$. Let $$B=\begin{bmatrix}A & I_n\\ I_n & A^T\end{bmatrix},$$ where $I_n$ is the identity matrix of order $n$ and $A^T$ is the transpose (not the conjugate transpose) of $A$. Then whether we can find the eigenvalues of $B$ in terms of eigenvalues of $A$. What are the eigenvalues of $B$?

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  • $\begingroup$ @G_o_pi_i_e \begin{align*} |\lambda I_{2n} - B| & = \left|\begin{matrix} \lambda I_n - A & -I_n \\ -I_n &\lambda I_n - A^{T} \\ \end{matrix}\right| = |(\lambda I_n - A)(\lambda I_n - A^{T})| - I_n \\ & = |\lambda I_n - A| |\lambda I_n - A^{T}| - I_n. \end{align*} So, the eigenvalues of $B$ seem to have some connections with $A$ and $A^T$. $\endgroup$ – bing Apr 6 '17 at 3:40
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    $\begingroup$ @bing: The determinant is a scalar, you can't end up with a $I_n$ as part of the equation... $\endgroup$ – copper.hat Apr 6 '17 at 3:41
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    $\begingroup$ @bing I think copper.hat is right. You made a mistake while computing the determinant of $B$. It is a block matrix and these rules won't work. $\endgroup$ – G_0_pi_i_e Apr 6 '17 at 7:37
  • $\begingroup$ The above progress I wrote is wrong (as copper.hat and G_0_pi_i_e kindly point out). Such as a block matrix \begin{align*} \left|\begin{matrix} 2 & 0 & 1 & 0 \\ 0 & 2 & 0 & 1 \\ 1 & 0 & 3 & 0 \\ 0 & 1 & 0 & 3 \end{matrix}\right| = 25 \neq \left|\begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix}\right| \left|\begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix}\right| - 1. \end{align*} $\endgroup$ – bing Apr 6 '17 at 8:08

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