Timeline for Eigenvalues of block-hermitian matrices with zero diagonal blocks
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| S Jul 26, 2017 at 10:10 | history | suggested | Rodrigo de Azevedo | CC BY-SA 3.0 |
Minor edits
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| Jul 26, 2017 at 9:17 | review | Suggested edits | |||
| S Jul 26, 2017 at 10:10 | |||||
| Jul 25, 2017 at 12:59 | comment | added | Surb | As @David said, it is known that the eigenvectors of $D$ are precisely the singular pairs of $C$. | |
| Jul 25, 2017 at 3:37 | review | Close votes | |||
| Jul 25, 2017 at 15:11 | |||||
| Jul 25, 2017 at 3:16 | comment | added | David Handelman | If you square it, the resulting matrix has nonnegative eigenvalues of $CC*$ with multiplicity multiplied by two (since $CC^*$ and $C^*C$ are unitarily conjugate). The corresponding eigenvalues of $D$ are the square roots. This has relatively little to do with the eigenvalues of $C$ itself, rather the singular values. I'm sure you can figure out the details---or was this part of an exercise? Most of the time, when $\dagger$ for conjugacy transpose, it's an exercise. | |
| Jul 25, 2017 at 0:45 | history | edited | Unwieldy Bob |
edited tags
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| Jul 25, 2017 at 0:31 | review | First posts | |||
| Jul 25, 2017 at 1:41 | |||||
| Jul 25, 2017 at 0:28 | history | asked | Unwieldy Bob | CC BY-SA 3.0 |