Timeline for Jordan decomposition of powers of the Shift Matrix [closed]
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 6, 2018 at 22:31 | vote | accept | G Cab | ||
| Jan 6, 2018 at 15:26 | history | closed |
David Handelman Pedro Lauridsen Ribeiro abx Stefan Kohl♦ Jan-Christoph Schlage-Puchta |
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| Jan 6, 2018 at 2:24 | answer | added | LSpice | timeline score: 2 | |
| Jan 6, 2018 at 1:53 | review | Close votes | |||
| Jan 6, 2018 at 15:26 | |||||
| Jan 5, 2018 at 23:39 | comment | added | G Cab | @LSpice: I mean that C is an off-diagonal matrix, but where $n-1$ ones are replaced by zeros. We can express the position of the $1/0$ by a permutation of a vector with $n$ zeros and $h-n$ ones. Concerning your 2nd comment, it looks very interesting: could you pls. explicitate in an answer ? many thanks | |
| Jan 5, 2018 at 23:26 | comment | added | LSpice | Modulo errors in calculation, $C_h(n)$ will have blocks of size $\lfloor(h - 1)/n\rfloor, \cdots, \lfloor(h - n)/n\rfloor$, one for each residue class modulo $n$. The permutation underlying $P_h(n)$ will send $n i + j$ to $i + \sum_{k = 1}^{j - 1} \lfloor(h - k)/n\rfloor$ (or its inverse, depending on how you match matrices and permutations). | |
| Jan 5, 2018 at 23:13 | comment | added | LSpice | What do you mean when you say that $C_h(n)$ "is [also] expressible as a permutation"? It's certainly not a permutation matrix, since it's not invertible. | |
| Jan 5, 2018 at 22:55 | history | asked | G Cab | CC BY-SA 3.0 |