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Consider the following block matrix

$$A = \left(\begin{matrix} B & T\\ T & 0 \end{matrix} \right)$$

where all submatrices are square and

  • matrix $B = \mbox{diag}\left(b_1 ,0,0,\dots,0,b_n \right)$ with $b_1, b_n > 0$.

  • matrix $T$ is self-adjoint and positive semidefinite.

What can one say about the lowest eigenvalue of this matrix $A$? In particular, how does it depend on the spectrum of $T$ and the entries of $B$? Are there any known results?

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  • $\begingroup$ $A$ may have negative eigenvalues, you want the smallest in absolute value or the most negative one? $\endgroup$
    – Carlo Beenakker
    Commented Feb 19, 2019 at 22:57
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    $\begingroup$ the lowest one (which may be negative)... $\endgroup$
    – Sascha
    Commented Feb 19, 2019 at 23:47

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If $\lambda_\max$ is the greatest eigenvalue of $T$, the least eigenvalue of $A$ is between $-\lambda_\max$ and $\max(b_1, b_n) - \lambda_\max$.

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    $\begingroup$ thank you, however do you mind briefly explaining how you get this? $\endgroup$
    – Sascha
    Commented Feb 20, 2019 at 0:28
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    $\begingroup$ The eigenvalues of $\pmatrix{0 & T\cr T & 0\cr}$ are the eigenvalues of $T$ and $-T$. You're adding a positive semidefinite matrix. Use the min-max theorem. $\endgroup$
    – Robert Israel
    Commented Feb 20, 2019 at 0:34

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