Timeline for Is it impossible for determinants of these matrices to both be negative?
Current License: CC BY-SA 4.0
24 events
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| Aug 17, 2022 at 13:08 | history | bounty ended | CommunityBot | ||
| Aug 17, 2022 at 4:02 | vote | accept | BAYMAX | ||
| Aug 14, 2022 at 6:15 | comment | added | BAYMAX | Thanks, I am still thinking about how Schur complement of $B+I+A^{-1}$ still gives information about the determinant of $AB+A+I$? I am reading about Schur complements to understand this but seems like I can see that Schur complement of $D$ in $M$ is $A - BD^{-1}C$, where $M = \begin{bmatrix} A & B\\ C & D\end{bmatrix}$ more discussion here: chat.stackexchange.com/rooms/138505/… | |
| Aug 13, 2022 at 13:57 | history | edited | Fred Hucht | CC BY-SA 4.0 |
$n/2 \mapsto n/2-1$ in (3)
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| Aug 13, 2022 at 13:27 | comment | added | Fred Hucht | True, a typo, it should read $n/3-1$. | |
| Aug 13, 2022 at 13:15 | comment | added | BAYMAX | Ok, I was worried because in equation (3), when k=n/3, inside the summation we would have a_n+j but we only have a_1,..., a_n in matrix A | |
| Aug 13, 2022 at 12:41 | history | edited | Fred Hucht | CC BY-SA 4.0 |
renamed index $i\mapsto j$ for $\alpha,\beta$
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| Aug 13, 2022 at 10:56 | comment | added | Fred Hucht | I‘ll send you a MMA file later. Note that $i=1,2,3$ only, I‘ll rename the index $i$ of $\alpha,\beta,\gamma,\delta$ to $j$ to avoid confusion. | |
| Aug 13, 2022 at 9:54 | comment | added | BAYMAX | Also should $$\alpha_i = -\delta_{3,i} + \sum_{k=0}^{n/3} a_{3k+i}$$ be actually $$\alpha_i = -\delta_{3,i} + \sum_{k=0}^{n/3} a_{3k+i (\rm{mod} \, n)} $$ ? | |
| Aug 13, 2022 at 9:39 | comment | added | BAYMAX | I was trying to check $\alpha, \beta$ as mentioned in eqn (2), (3) using Mathematica but it seems somewhere its is wrong: here: imgur.com/ghuln8m, can you please advise? | |
| Aug 12, 2022 at 11:57 | comment | added | Fred Hucht | As I wrote, the proof of (9) remains open and might be performed geometrically or algebraically. Maybe you can work it out. | |
| Aug 12, 2022 at 11:09 | comment | added | BAYMAX | Nice idea! I wonder if you proved in the answer: $$ D(\alpha,\alpha,\beta)>0 \land D(\alpha,\beta,\beta)>0 \Rightarrow D(\alpha,\delta,\beta)>0 \lor D(\beta,\delta,\alpha)>0. $$?? because then it is a way of saying that $D(\alpha,\delta,\beta)<0$ and $D(\beta,\delta,\alpha)<0$ is not possible.. Is it intuitive that the implication $D(\alpha,\alpha,\beta)>0 \land D(\alpha,\beta,\beta)>0 \Rightarrow D(\alpha,\delta,\beta)>0 \lor D(\beta,\delta,\alpha)>0$ is true? as I just want to see if in the current case it is true? because then other cases would be similar and conjecture becomes true | |
| Aug 12, 2022 at 8:52 | history | edited | Fred Hucht | CC BY-SA 4.0 |
Cosmetic
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| Aug 12, 2022 at 8:38 | comment | added | Fred Hucht | @BAYMAX: See my updated answer. | |
| Aug 12, 2022 at 8:27 | history | edited | Fred Hucht | CC BY-SA 4.0 |
Added edit 12.08.22
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| Aug 12, 2022 at 3:52 | comment | added | BAYMAX | Will it always be the case that $P_{a}(-1) > 0$ if all $|\lambda_{i}| <1$? is it because $P_{a}(\lambda) \approx (-\lambda)^n$ [but then this holds for large $|\lambda|$, is in our case $\lambda=-1$ be considered large] | |
| Aug 11, 2022 at 17:59 | comment | added | Fred Hucht | I again thought about $P_a(-1)$. I think one can show due to complex conjugate pairs, that $P_a(-1)>0$ if all $|\lambda_i|<1$, as the point $-1$ lies left of all eigenvalues. Same of course holds for $P_b(-1)$. Therefore, one has to show that $P_{a,b}(-1)>0$ contradicts D_a < 0 and D_b < 0. This should be possible. | |
| Aug 11, 2022 at 9:15 | comment | added | Fred Hucht | Maybe one can show that if both Eqs. (2) are negative, then one zero of $P$ must be outside the interval $[-1,1]$, which can be seen from the sign of $P(\pm1)$. Remember that $\{\alpha_i,\beta_i\}$ are independent real variables, and that $$P_a(\lambda)=\prod_{i=1}^n (\lambda_i-\lambda).$$ Note that I have just corrected typos in (2) and (5). | |
| Aug 11, 2022 at 8:54 | history | edited | Fred Hucht | CC BY-SA 4.0 |
Fixed typos in (2) and (5) by interchanging indices $2 \leftrightarrow 3$.
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| Aug 11, 2022 at 8:07 | comment | added | BAYMAX | I will try to understand your answer carefully by working out the steps you mentioned..but I can't see why both can't be negative simultaneously. | |
| Aug 11, 2022 at 6:23 | history | edited | Fred Hucht | CC BY-SA 4.0 |
Re-introduced $\nu$, changed last sentence.
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| Aug 10, 2022 at 23:17 | history | edited | Fred Hucht | CC BY-SA 4.0 |
deleted 2 characters in body
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| Aug 10, 2022 at 23:03 | history | edited | Fred Hucht | CC BY-SA 4.0 |
minor
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| Aug 10, 2022 at 22:54 | history | answered | Fred Hucht | CC BY-SA 4.0 |