Timeline for dyadically recursive matrices: Part II
Current License: CC BY-SA 3.0
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| Jun 1, 2017 at 0:42 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Apr 24, 2017 at 4:02 | answer | added | T. Amdeberhan | timeline score: 1 | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Mar 23, 2017 at 21:00 | comment | added | darij grinberg | @T.Amdeberhan: I wish I had the time to write the long-ish expository answer that this question deserves... As for your second question, see math.stackexchange.com/questions/706513/… (Kaladin's proof is more complicated than necessary). I believe this matrix is something like a multiplication matrix by the quaternion $a+bi+cj+dk$, although I have not checked. The whole thing seems to be connected with the notion of a reduced norm in a central simple algebra, but I am not sure how far this connection is worth pursuing. | |
| Mar 21, 2017 at 15:17 | comment | added | T. Amdeberhan | @darijgrinberg: Can you give me some reference to what you mentioned as "known trick that works for quaterions"? I'm interested what you are doing in quaternions, Thanks. | |
| Mar 17, 2017 at 12:03 | comment | added | T. Amdeberhan | @darijgrinberg: Perhaps you like to make your comments into an answer. | |
| Mar 16, 2017 at 19:31 | comment | added | darij grinberg | ... remains to take the square root. In order to make sure that the sign is right, argue about the evaluation at $a_2=a_3=\cdots=a_n=0$. | |
| Mar 16, 2017 at 19:27 | comment | added | darij grinberg | Ah! There is a known trick that works for quaternions, and that also works here. Your matrices $A_n$ are "almost orthogonal": They satisfy $A_n \left(A_n\right)^T = \left(a_1^2+a_2^2+\cdots+a_n^2\right) I_{2^{n-1}}$. (You can prove this by induction, using the fact that $A_n J_n = J_n \left(A_n\right)^T$, which you can also prove by induction.) Taking determinants, we obtain $\left(\det\left(A_n\right)\right)^2 = \left(a_1^2+a_2^2+\cdots+a_n^2\right)^{2^{n-1}}$. Now, it ... | |
| Mar 16, 2017 at 19:23 | comment | added | darij grinberg | These determinants look like norms of elements of some twisted kind of Cayley-Dickson construction... | |
| Mar 16, 2017 at 19:13 | comment | added | darij grinberg | This also suggests a line of attack. If we can remove the annoying $\left(-1\right)^{x_{k+1}}$ factor, then we are left with a group determinant (for the group $\left(\mathbb{Z} / 2 \mathbb{Z}\right)^{n-1}$), because $k$ depends only on the entrywise XOR of $x_1 x_2 \cdots x_{n-1}$ and $y_1 y_2 \cdots y_{n-1}$. | |
| Mar 16, 2017 at 19:08 | comment | added | darij grinberg | Maybe you should index the rows and the columns of $A_n$ not by numbers from $1$ to $2^{n-1}$, but by length-$\left(n-1\right)$ bitstrings (which correspond to these numbers via base-$2$ representation). The entry in row $x_1x_2\cdots x_{n-1}$ and column $y_1y_2\cdots y_{n-1}$ is then obtained as follows: Let $k$ be the largest index such that $x_k = y_k$. (This is set to be $0$ if no such $k$ exists.) If some $i < k$ satisfies $x_i \neq y_i$, then the entry is $0$. If not, then the entry is $\left(-1\right)^{x_{k+1}} a_{n-k}$, where $x_n = 0$. | |
| Mar 16, 2017 at 17:54 | comment | added | T. Amdeberhan | That's correct, Wolfgang. | |
| Mar 16, 2017 at 17:29 | comment | added | Wolfgang | Well you surely mean $(a_1^2 + \ldots a_n^2)^{2^{n-\color{red}2}}$. :) | |
| Mar 16, 2017 at 15:52 | comment | added | T. Amdeberhan | That's what I am getting too. Perhaps we should seek for a proof then. | |
| Mar 16, 2017 at 15:51 | comment | added | Robert Israel | Looks like $(a_1^2 + \ldots a_n^2)^{2^{n-1}}$. | |
| Mar 16, 2017 at 14:57 | history | asked | T. Amdeberhan | CC BY-SA 3.0 |