Timeline for Matrix transformation that "rotates" a matrix by $45^\circ$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 8, 2017 at 20:21 | comment | added | Stefan Kohl♦ | Ad 3: I think straightforward group theoretic implications are rather unlikely already since your transformation does not preserve invertibility of a matrix. That said, of course a definitive negative answer to the question as stated cannot be given. | |
| Aug 8, 2017 at 19:13 | comment | added | Sylvain JULIEN | Note that the transposition of $A'$ corresponds to the map $(i,j)\mapsto(m+1-i,j)$ in its image. This corresponds to an automorphism of the diedral group with $8$ elements. | |
| Aug 8, 2017 at 18:57 | comment | added | Sylvain JULIEN | You should first extend the matrix $A$ to a matrix $A'$ of size $m$ and attach to each element $a'_{i,j}$ a point $P=P(i,j)$ of coordinates $(x,y)$ in the plane, rotate those points to get $P'(i,j)$ of coordinates $(x',y')$ and get $(k,l)=P^{-1}(x',y')$. | |
| S Aug 8, 2017 at 14:41 | history | suggested | Rodrigo de Azevedo | CC BY-SA 3.0 |
Minor improvements
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| Aug 8, 2017 at 14:14 | review | Suggested edits | |||
| S Aug 8, 2017 at 14:41 | |||||
| Aug 8, 2017 at 12:46 | comment | added | Yaakov Baruch | Notice that since $B$ in general will have rank $m-1=2n-2$, larger than the typical rank of $A$, which is $n$, you cannot generally find matrices $C$ and $D$ such that $CAD=B$; and then clearly never with $C$ and $D$ independent of $A$. So you are looking at just a linear embedding of ${\mathbb F}^{n\times n} \rightarrow {\mathbb F}^{m\times m}$ without immediately obvious further algebraic properties. | |
| Aug 8, 2017 at 11:58 | review | Close votes | |||
| Aug 8, 2017 at 12:22 | |||||
| Aug 8, 2017 at 10:36 | answer | added | Carlo Beenakker | timeline score: 2 | |
| Aug 7, 2017 at 21:03 | history | asked | Mohammad Al-Turkistany | CC BY-SA 3.0 |