Timeline for Eigenvalues of a block matrix with zero diagonal blocks
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| May 2, 2021 at 19:06 | comment | added | Carlo Beenakker | My surmise is that you need one of the $k_i$'s, say $k_m$ to be larger than the sum of all the others, to have $k_m-\sum_{i\neq m}k_i$ zero eigenvalues. | |
| May 2, 2021 at 17:57 | vote | accept | AdamNie | ||
| May 2, 2021 at 17:56 | comment | added | AdamNie | I think I understand what you meant now, thanks a lot. A further question: if in general I have $k_1<k_2<\ldots<k_n$ diagonal blocks of zero, without the assumption $k_n>\sum_{i\ne n}k_i$, can I say anything at all about the matrix? | |
| May 2, 2021 at 17:43 | comment | added | Carlo Beenakker | this root would cancel with the pole of the inverse $X^{-1}$. | |
| May 2, 2021 at 17:42 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
added 378 characters in body
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| May 2, 2021 at 16:44 | comment | added | AdamNie | I see. What about the first determinant there, wouldn't that have a root of multiplicity k2-k1? | |
| May 2, 2021 at 15:48 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |