Timeline for Full expansion of $\det(I+\varepsilon A)$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Oct 27, 2023 at 18:57 | comment | added | Jan Nienhaus | Yes, this doesn't require anything. The matrix should be square, otherwise this works in full generality, even over arbitrary coefficient fields. The proof is just multiplying $1+tA$ by $t^{-1}$ and noting that this matrix's determinant is, by definition, the characteristic polynomial of $A$ evaluated at $-t^{-1}$. | |
| Oct 26, 2023 at 15:20 | comment | added | tommy1996q | A question: the formula you provide holds in full generality, right? No hypothesis on triangulability of the matrix or something like that? | |
| Oct 24, 2023 at 9:34 | comment | added | tommy1996q | I think this might settle it. $\sigma_1$ is the trace, while for the other $\sigma_i$s, I can bound the eigenvalues in terms of the $L^\infty$ norm of the matrix (Gershgoring theorem if I am not mistaken). If I make the entries of $A$ scale like, say, $\varepsilon ^{-1/2}$, the higher order terms in the sum scale as $\varepsilon$ or as $o (\varepsilon)$. | |
| Oct 24, 2023 at 9:28 | vote | accept | tommy1996q | ||
| Oct 24, 2023 at 1:49 | history | edited | Michael Hardy | CC BY-SA 4.0 |
added 1 character in body
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| Oct 23, 2023 at 20:58 | history | answered | Jan Nienhaus | CC BY-SA 4.0 |