Re: simpler way to get a particular banded matrix?

*To*: mathgroup at smc.vnet.net*Subject*: [mg98145] Re: simpler way to get a particular banded matrix?*From*: rip pelletier <bitbucket at comcast.net>*Date*: Tue, 31 Mar 2009 04:19:35 -0500 (EST)*References*: <gqn910$2o$1@smc.vnet.net> <gqq3qa$7m3$1@smc.vnet.net>

In article <gqq3qa$7m3$1 at smc.vnet.net>, Raffy <raffy at mac.com> wrote: > On Mar 29, 12:46 am, rip pelletier <bitbuc... at comcast.net> wrote: > > Hi, > > > > The following command constructs a matrix (typically denoted M0, it > > arises in filter banks in general, in Burrus, Gopinath, & Guo "intro to > > wavelets and wavelet transforms: a primer" in particular). > > > > I'm sure there's a way to do this without typing out all 5 bands. Here's > > what i did: > > > > SparseArray[Band[{1,1}]->H,{6,6}]+ > > SparseArray[Band[{2,3}]->H,{6,6}]+ > > SparseArray[Band[{3,5}]->H,{6,6}]+ > > SparseArray[Band[{-2,-3},{1,1},{-1,-1}]->rH,{6,6}]+ > > SparseArray[Band[{-3,-5},{1,1},{-1,-1}]->rH,{6,6}] > > > > where > > > > H={h0,h1,h2,h3,h4,h5}; > > rH=Reverse[H]; > > > > I get what I want: > > > > h0 0 0 0 0 0 > > h2 h1 h0 0 0 0 > > h4 h3 h2 h1 h0 0 > > 0 h5 h4 h3 h2 h1 > > 0 0 0 h5 h4 h3 > > 0 0 0 0 0 h5 > > > > Any easier ways to get this? > > > > TIA and vale, > > rip > > > > -- > > NB eddress is r i p 1 AT c o m c a s t DOT n e t > > To error on the side of readability, I'd probably go with: > > mat[n_Integer] := With[ > {vRow = Join[ > ConstantArray[0, n - 1], > Table[ToExpression["h" <> ToString[i]], {i, n - 1, 0, -1}], > ConstantArray[0, n - 1]] > }, > Table[Take[vRow, {2 n + 1, 3 n} - 2 i], {i, n}] > ]; Thanks. I'm not sure it's the kind of simplification i'm looking for, but I'll play with it. Vale, Rip -- NB eddress is r i p 1 AT c o m c a s t DOT n e t