Timeline for How to judge a determinant equals non-zero in cryptography?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 12, 2018 at 8:09 | vote | accept | p1gd0g | ||
| Oct 12, 2018 at 8:09 | answer | added | p1gd0g | timeline score: 0 | |
| Oct 4, 2018 at 11:26 | answer | added | Vadym Fedyukovych | timeline score: 0 | |
| Oct 3, 2018 at 18:13 | comment | added | p1gd0g | @rikhavshah Thanks anyway. Please contact me #p1gd0g if you have telegram. | |
| Oct 3, 2018 at 17:11 | comment | added | rikhavshah | That's not necessarily true. $|A|=0$ means there are either 0 or infinitely many solutions. Without more information about their construction, I can't say much more. | |
| Oct 3, 2018 at 16:51 | comment | added | p1gd0g | @rikhavshah Sounds interesting, but if $|A|=0$, there is not a solution. | |
| Oct 3, 2018 at 16:16 | comment | added | rikhavshah | They do say "up to" $n$ solutions. Perhaps they say this so that they don't exclude the case of $\det(A)=0$. | |
| Oct 3, 2018 at 13:28 | comment | added | p1gd0g | @eins6180 I mean I send the paper to you privately. All universities can access the paper, then the discuss is still helpful for others. | |
| Oct 3, 2018 at 13:07 | comment | added | eins6180 | I'm sorry but this is really not how it works. If you're uncomfortable to provide a free link to the paper you need to provide more context. But having this discussion in private prevents everyone else to contribue and learn from it. | |
| Oct 3, 2018 at 10:16 | comment | added | p1gd0g | @eins6180 Thanks for concerning. I'm sorry I'm not sure if it is legal to send you the paper. May I have your contact? Whichever is OK. | |
| Oct 3, 2018 at 7:23 | comment | added | eins6180 | From the context you provide it's not clear to me that the matrix is invertible, and I don't have access to the paper since it's behind a paywall. So we need either more context please or a free link to the paper. | |
| Oct 3, 2018 at 6:48 | history | edited | p1gd0g | CC BY-SA 4.0 |
More precise.
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| Oct 3, 2018 at 2:40 | comment | added | p1gd0g | @SEJPM Yes, I know it. My question is how to prove $A\neq 0$? The authors didn't give an explaination. | |
| Oct 2, 2018 at 17:06 | comment | added | SEJPM | Let $A$ be an $n\times n$ matrix. Then the system $Ax=b$ has exactly one solution vector $x$ with $n$ entries iff $\det A \neq 0$, that is iff $A$ is invertible. This is a basic result of linear algebra. | |
| Oct 2, 2018 at 15:15 | history | asked | p1gd0g | CC BY-SA 4.0 |