Author 
Topic: Rank of Redheffer Matrix (Read 1244 times) 

Aryabhatta
Uberpuzzler
Gender:
Posts: 1321


Rank of Redheffer Matrix
« on: Apr 11^{th}, 2007, 4:54pm » 
Quote Modify

Let RH_{n} be the nxn RedHeffer matrix defined as: RH_{n}(i,j) = 1 if i divides j or j = 1 RH_{n}(i,j) = 0 otherwise. For instance RH_{4} is [1 1 1 1] [1 1 0 1] [1 0 1 0] [1 0 0 1] Prove that the rank of RH_{n} is atleast n1.

« Last Edit: Apr 11^{th}, 2007, 5:51pm by Aryabhatta » 
IP Logged 



Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948


Re: Rank of Redheffer Matrix
« Reply #1 on: Apr 11^{th}, 2007, 6:48pm » 
Quote Modify

The lowerright (n1)x(n1) submatrix is upper unitriangular. 2) Show that det(RH_{n}) = _{k=1}^{n} (k). Hence RH_{n} is invertible except for n=2, 39, 40, 58, 65, 93, .... 3) Is the above set infinite?


IP Logged 



Aryabhatta
Uberpuzzler
Gender:
Posts: 1321


Re: Rank of Redheffer Matrix
« Reply #2 on: Apr 12^{th}, 2007, 12:39am » 
Quote Modify

on Apr 11^{th}, 2007, 6:48pm, Eigenray wrote: 3) Is the above set infinite? 
 Is this known?


IP Logged 



Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948

No idea. You'd think they would say one way or the other on Mathworld, but Sloane usually says when a sequence is not known to be infinite. It seems likely though. Here's a graph of M(n)/n. It looks roughly like a random walk. But it's not as a random as it looks  on the right is a graph of A(A(A(M))), where A(F)(n) = 1/n _{k=1}^{n} F(k) is the averaging operator. The position of the nth maxima fits Ce^{.45 n} pretty well.


IP Logged 



