At its heart, this is just a basic circulant matrix. So use a tool that will do that. I posted such a tool on the file exchange. But you can just use gallery.
n = 8; % an 8x8 circulant matrix of that form
M = gallery('circul',[0 1,zeros(1,n-3),1])
Or you can view this as a Toeplitz matrix. Toeplitz matrices are a close second cousin to circulant matrices. Your matrix is always symmetric, so a symmetric Toeplitz matrix can be formed by just supplying the first row or column.
M2 = toeplitz([0 1,zeros(1,n-3),1])
Of course there are other ways to do it. Since your matrix is actually quite sparse, you could build it as a sparse banded matrix. The virtue of that is if n is large, (as it often will be in applications like this) then you use efficient sparse storage, as well as efficient computations thereafter with that matrix.
I'll use n=50 here, but typically n might be a number in the thousands or more, if you are really needing to use a sparse matrix. 50 is large enough that you can visualize the banded structure easily, yet not too large that you cannot see the dots. (If I made n=10000, then you might not even see the dots in the corners, or see the line down the middle as actually a pair of bands just above and below the main diagonal.)
n = 50;
M3 = spdiags(ones(n,4),[-n+1, -1, 1, n-1],n,n);
spy(M3)
whos M3
As you can see, M3 is a sparse version of the matrix you want to build.
