Matrix form

Hi,

I have a question about make a matrix which has form of

Let say I have

 cel = spdiags(ones(N,1),1,N,N) - spdiags(ones(N,1),-1,N,N);

N is size in a sense that matrix size is N^3 by N^3, and first N by N part, and then N^2 by N^2 part and then N^3 by N^3 matrices super-diagonals are 1 and -1, main diagonal is 0. I hope it makes sense!

So NxN zero main diag, and from N+1,N+1 starts again same matrix till N^2xN^2, then from N^2+1,N^2+1 starts again same matrix till N^3xN^3

I have tried with kron function, which is obvious function to use, i.e.

 full(kron(speye(N),cel))

but unable to create very same matrix

Let say I have this scheme in 3D, boundary conditions are zero on the boundary

(u_i,j,k+1 - u_i,j,k-1)/(2*h) => which gives my matrix A

and I have vector u = [u_111,u_112,u_121,u_122,u_211,u_212,u_221,u_222] then I have Au = b where b is any vector size of 8...

Thanks in advance!

Answer 1

[Andrei Bobrov]

s = N.^(1:3);
bs = arrayfun(@(x)spdiags(ones(x,1)*[-1 1],[-1 1],x,x),s,'un',0);
out = blkdiag(bs{:});

[Feruz]

Thank you for reply,

But I guess I did explain it not the way I should...

Let say I have this scheme in 3D, boundary conditions are zero on the boundary

(u_i,j,k+1 - u_i,j,k-1)/(2*h) => which gives my matrix A

and I have vector u = [u_111,u_112,u_121,u_122,u_211,u_212,u_221,u_222] then I have Au = b

where b is any vector size of 8...