Timeline for Eigenvalues of block matrix
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Dec 3, 2020 at 0:16 | comment | added | Fedor Petrov | @Trb2 the conjugation does not change the eigenvalues. Or what do you mean? | |
| Dec 2, 2020 at 15:51 | comment | added | Trb2 | @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? | |
| Dec 2, 2020 at 15:48 | history | edited | Rodrigo de Azevedo | CC BY-SA 4.0 |
added 54 characters in body
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| Dec 2, 2020 at 15:16 | comment | added | Richard Stanley | 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. | |
| Dec 2, 2020 at 14:10 | history | edited | Rodrigo de Azevedo | CC BY-SA 4.0 |
edited tags
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| Dec 2, 2020 at 13:32 | comment | added | Trb2 | @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! | |
| Dec 2, 2020 at 11:13 | comment | added | Federico Poloni | 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? | |
| Dec 2, 2020 at 10:11 | review | First posts | |||
| Dec 2, 2020 at 12:13 | |||||
| Dec 2, 2020 at 10:09 | history | asked | Trb2 | CC BY-SA 4.0 |