How do I create a circulant matrix of shift 3 towards the left ?

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Derick Wong
Derick Wong 2013-12-16
评论： Derick Wong 2013-12-17
Hi,
Does anyone know how do I create a circulant matrix like the one I attached. The first 2 row is for the x-axis, center 2 row is for the y-axis and the last 2 row is for the z-axis. The shift from each row is 3. The x-axis started from 1st column. The y-axis started from 2nd column and the z-axis started from 3rd column.
  1 个评论
Jos (10584)
Jos (10584) 2013-12-16
Given your description, I fail to see how the 3rd row is looking in A. Or has A only six rows? And should it be flexible somehow?

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Jos (10584)
Jos (10584) 2013-12-16
编辑：Jos (10584) 2013-12-16
Not very flexible:
A = eye(6)
A = A([1 4 2 5 3 6],:) % permute as desired
A(:,10) = 0 % add some zero columns
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Jos (10584)
Jos (10584) 2013-12-16
You could help by stating what the third line of A should be:
A = [1 0 0 0 0 0 0 0 . . . line 1
0 0 0 1 0 0 0 0 . . . line 2
WHAT IS HERE!
0 1 0 0 0 0 0 0 line 32
Derick Wong
Derick Wong 2013-12-17
Is abit tedious when I did it. But it surely works. A= eye(93,93); A=A([1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73 76 79 82 85 88 91 2 5 8 11 14 17 20 23 26 29 32 35 38 41 44 47 50 53 56 59 62 65 68 71 74 77 80 83 86 89 92 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60 63 66 69 72 75 78 81 84 87 90 93],:);
Many thanks!

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更多回答（2 个）

Matt J
Matt J 2013-12-16
编辑：Matt J 2013-12-16
Using FUNC2MAT
A=func2mat(@(x)circshift(x,3),zeros(93,1));
Jos (10584)
Jos (10584) 2013-12-16
It does look like a row-permuted circulant
N = 31 ;
dn = 3 ;
A = eye(dn*N) ; % circulant
ix = ones(N,d) ; % make a permution index
ix(1,:) = 1+dn*(0:N-1) ;
ix = cumsum(ix,1) ;
ix = ix.' ;
A = A(ix,:) % row-permutation

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