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Given scalars $\alpha, \beta \in \mathbb{R}$, a symmetric positive definite matrix $A \in \mathbb{R}^{n\times n}$ and a flat matrix $B \in \mathbb{R}^{m\times n}$, where $m < n$, can I say something about the eigenvalues of the following block matrix?

$$T := \begin{bmatrix} \alpha A & \alpha B^T \\ \beta B & O_m \end{bmatrix} $$

Can I maybe give bounds on the eigenvalues of $T$ as a function of $\alpha, \beta$?

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    $\begingroup$ If $\alpha$ were equal to $\beta$, I would point you to the standard Benzi-Golub-Liesen review paper on saddle-point matrices, but the fact that this is not symmetric makes the theory in there not really applicable, I am afraid. Are you sure you cannot somehow reduce to the symmetric case by scaling a later step of your problem? $\endgroup$
    – Federico Poloni
    Commented Dec 2, 2020 at 11:13
  • $\begingroup$ @FedericoPoloni Thanks for the answer and references! Yes, to scale at a later step is maybe a solution. I will give it a try, thanks! $\endgroup$
    – Trb2
    Commented Dec 2, 2020 at 13:32
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    $\begingroup$ We can conjugate $T$ by the diagonal matrix $\mathrm{diag}(\sqrt{\beta}I,\sqrt{\alpha}I)$ to obtain a symmetric matrix of the same form. $\endgroup$
    – Richard Stanley
    Commented Dec 2, 2020 at 15:16
  • $\begingroup$ @RichardStanley with $\text{diag}(\sqrt{\alpha}I,\sqrt{\beta}I)$, right? The eigenvalues of the symmetric matrix are then known, if $B$ has full rank (Benzi-Golub-Liesen review paper), but the eingenvalues after the conjugation still not, right? $\endgroup$
    – Trb2
    Commented Dec 2, 2020 at 15:51
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    $\begingroup$ @Trb2 the conjugation does not change the eigenvalues. Or what do you mean? $\endgroup$
    – Fedor Petrov
    Commented Dec 3, 2020 at 0:16

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