Elementary matrices in Matlab

I am very new to MATLAB, and I am trying to create a numerical scheme to solve a differential equation. However I am having trouble implementing matrices. I was wondering if anyone can help with constructing a following NxN matrix?

I am sure there is a better way to implement, but the following works

N=10
tod=zeros(N)
for k=1:(N-1)
    tod(k, k+1)=-2
end
for k=1:(N-2)
    tod(k, k+2)=1
end
tod_2=zeros(N)
for k=1:(N-1)
    tod_2(k, k+1)=2
end
for k=1:(N-2)
    tod_2(k, k+2)=-1
end
tran=transpose(tod_2)
Final=tran+tod

Answer 1

[DGM]

(EDIT to fix misread)

Here's one way:

n = 8;
z = zeros(1,n-3);
R = toeplitz([0 2 -1 z],[0 -2 1 z])

R = 8×8

0 -2 1 0 0 0 0 0 2 0 -2 1 0 0 0 0 -1 2 0 -2 1 0 0 0 0 -1 2 0 -2 1 0 0 0 0 -1 2 0 -2 1 0 0 0 0 -1 2 0 -2 1 0 0 0 0 -1 2 0 -2 0 0 0 0 0 -1 2 0

[Steven Lord]

The diag and spdiags functions may also be of interest for creating these types of matrices.

[Voss]

This answer does not produce the requested matrix.

[Stephen23]

@DGM: as Benjamin points out the requested matrix is not symmetrical, but this is very easy to achieve with a small modification to your answer:

n = 8;
v = zeros(1,n-3);
R = toeplitz([0,2,-1,v],[0,-2,1,v])

R = 8×8

0 -2 1 0 0 0 0 0 2 0 -2 1 0 0 0 0 -1 2 0 -2 1 0 0 0 0 -1 2 0 -2 1 0 0 0 0 -1 2 0 -2 1 0 0 0 0 -1 2 0 -2 1 0 0 0 0 -1 2 0 -2 0 0 0 0 0 -1 2 0

[DGM]

Oof . I totally overlooked that.

[Stephen23]

@DGM: add it to your answer, no one will look in these comments.

Answer 2

[John D'Errico]

So many ways to do this. DGM showed a great way using toeplitz. But there are others. For example, you could use diag.

n = 5;
R = diag(ones(n-2,1),2) - 2*diag(ones(n-1,1),1); R = R + R'

R = 5×5

0 -2 1 0 0 -2 0 -2 1 0 1 -2 0 -2 1 0 1 -2 0 -2 0 0 1 -2 0

Or use spdiags, creating the matrix directly. Since your matrix is banded, if n is at all large, then you truly want to learn to use sparse matrices.

n = 10;
R = spdiags(ones(n,1)*[1 -2 -2 1],[-2 -1 1 2],n,n);

R is now a sparse matrix, so it will offer great advantages in computing when n grows large.

spy(R)

full(R)

ans = 10×10

0 -2 1 0 0 0 0 0 0 0 -2 0 -2 1 0 0 0 0 0 0 1 -2 0 -2 1 0 0 0 0 0 0 1 -2 0 -2 1 0 0 0 0 0 0 1 -2 0 -2 1 0 0 0 0 0 0 1 -2 0 -2 1 0 0 0 0 0 0 1 -2 0 -2 1 0 0 0 0 0 0 1 -2 0 -2 1 0 0 0 0 0 0 1 -2 0 -2 0 0 0 0 0 0 0 1 -2 0

I could probably have used tools like meshgrid, tril and triu.

[Voss]

This answer does not produce the requested matrix.

Answer 3

[Voss]

Here's one way you can do it (and it produces the correct matrix too):

N = 10;
R = zeros(N);
R(2:N+1:end-N) = 2;
R(3:N+1:end-2*N) = -1;
R(N+1:N+1:end-1) = -2;
R(2*N+1:N+1:end-2) = 1;
R

R = 10×10

0 -2 1 0 0 0 0 0 0 0 2 0 -2 1 0 0 0 0 0 0 -1 2 0 -2 1 0 0 0 0 0 0 -1 2 0 -2 1 0 0 0 0 0 0 -1 2 0 -2 1 0 0 0 0 0 0 -1 2 0 -2 1 0 0 0 0 0 0 -1 2 0 -2 1 0 0 0 0 0 0 -1 2 0 -2 1 0 0 0 0 0 0 -1 2 0 -2 0 0 0 0 0 0 0 -1 2 0