Timeline for Cartesian dissimilarity of a function $\ f:A^3\to A^3\ $ and its inverse
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 14, 2020 at 18:27 | vote | accept | Wlod AA | ||
| May 14, 2020 at 0:13 | comment | added | Wlod AA | Actually, without the assumption about $\ a+c=1,\ $ the above matrix product equals to $\ (a+c)\cdot\Bbb I_3.\ $ This certainly has potential for number theory and algebraic geometry; in particular, the determinant of the above left matrix is equal to $\ (a+b)^2.$ | |
| May 13, 2020 at 23:18 | comment | added | Wlod AA | @YCor, visually, you can make your matrices more pleasing(?), i.e. more symmetric (just visually) by applying $\ a\ c\ $ in place of $\ a\ b.\ $ But then, poet Julian Tuwim said that "symmetry is an idiot's aesthetics." $$\left[\begin{array}{ccc} -c & c & a\\ 1 & -1 & 1\\ c & a & -a\end{array}\right]\cdot \left[\begin{array}{ccc} 0 & a & 1\\ 1 & 0 & 1\\ 1 & c & 0\end{array}\right]\ =\ \Bbb I_3,\ $$ where a+c=1. | |
| May 13, 2020 at 22:28 | comment | added | Wlod AA | Aleksei and YCor, this is so nice! | |
| May 13, 2020 at 16:35 | comment | added | Aleksei Kulikov | @YCor actually, your example seems even better because it actually doesn't need $a$ to be invertible, only nonzero! So apparently there's an example in any (commutative) ring with at least three elements... | |
| May 13, 2020 at 16:32 | comment | added | YCor | ...Namely $\begin{pmatrix}0 & a & b\\ 1 & 0 & 1\\ 1 & 1 & 0\end{pmatrix}$ for $a+b=1$, $a$ invertible, $b\neq 0$, the determinant being 1 and inverse being $\begin{pmatrix}-1 & a & a\\ 1 & -b & b\\ 1 & b & -a\end{pmatrix}$. | |
| May 13, 2020 at 16:26 | comment | added | Aleksei Kulikov | @YCor there are no examples for $n = 0,1$ for vacuous reasons :) | |
| May 13, 2020 at 16:25 | comment | added | YCor | Yes you're right, sorry. The condition of depending on all coordinates is weak enough to not be affected by projecting on the 2-part. So there are examples for $n$ iff $n\ge 3$ (modulo double-checking my computation) [corrected: there are none for $n=0,1$]. | |
| May 13, 2020 at 16:02 | comment | added | Aleksei Kulikov | @YCor sorry, I do not understand at all what you are talking about. My example in the edit of the post works for all $n \ge 3$, including $2 \mod 4$. For the $3\times 3$ matrix that you mentioned to exist we need $a, b$ with $a-b = 1$, $a$ invertible and $b\ne 0$, there is no need for $b$ to be invertible as well. | |
| May 13, 2020 at 15:59 | comment | added | YCor | Invertible ensures the non-vanishing of the coefficient. I'll post a partial answer to clarify my points without skipping details. | |
| May 13, 2020 at 15:57 | comment | added | Aleksei Kulikov | @YCor I'm not sure what you mean by this comment. We don't need $b$ (in your comment's notation) to be invertible! Only nonzero. | |
| May 13, 2020 at 15:54 | comment | added | YCor | If the commutative finite ring has invertible elements $a,b$ with $a-b=1$, then you can get a $3\times 3$ invertible matrix with zero diagonal, whose inverse has only nonzero entries. For given $n$, there exists a ring of cardinal $n$ with this property iff $n\notin 4\mathbf{Z}+2$ (by Chinese theorem one boils down to $n=q\neq 4$ prime-power in which case a field structure works). So powers of 2 are settled. For $6\le n\in 4\mathbf{Z}+2$ however we can't expect to solve the problem with an abelian group structure and an endomorphism, since projecting to the 2-part yields a contradiction. | |
| May 13, 2020 at 15:47 | history | edited | Aleksei Kulikov | CC BY-SA 4.0 |
edited body
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| May 13, 2020 at 15:45 | comment | added | Aleksei Kulikov | @YCor oh well, yes, you're right, I will fix this issue | |
| May 13, 2020 at 15:44 | comment | added | YCor | In the question it's required that $g_1$ doesn't depend on $x$, $g_2$ doesn't depend on $y$, $g_3$ doesn't depend on $z$. | |
| May 13, 2020 at 15:37 | history | edited | Aleksei Kulikov | CC BY-SA 4.0 |
added 374 characters in body
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| May 13, 2020 at 15:27 | comment | added | Aleksei Kulikov | Actually, my answer works for all rings $R$ with at least one invertible element $b\ne 1$ (just put $a = b -1$), in particular, for all $\mathbb{Z}/\mathbb{Z}_N$, $N \ge 3$. | |
| May 13, 2020 at 15:20 | history | answered | Aleksei Kulikov | CC BY-SA 4.0 |