How can I replace the upper off diagonal(super diagonal) and lower off diagonal(sub diagonal) of a matrix?

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Given a symmetric tridiagonal matrix T generated by
n=5;
p=1;
q=1.7;
r=1;
T=full(gallery('tridiagonal',n,p,q,r));
T=[1.8 1 0 0 0;1 1.8 0 0 0;0 1 1.8 1 0;0 0 1 1.8 1;0 0 0 1 1.8]
T = 5x5
1.8000 1.0000 0 0 0 1.0000 1.8000 0 0 0 0 1.0000 1.8000 1.0000 0 0 0 1.0000 1.8000 1.0000 0 0 0 1.0000 1.8000
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How do I change the 1 1 1 1 on both upper and lower off diagonal to 1 0 1 0? What if n is arbitrary, is there any code that fix the off diagonals to a desired vector?

Answers (5)

the cyclist
the cyclist on 13 Aug 2024
% Arbitrary size
N=5;
% Example input
A = rand(N); % NxN random matrix
disp(A)
0.4869 0.1564 0.9023 0.9455 0.0501 0.2761 0.9904 0.0819 0.3533 0.5165 0.9278 0.9213 0.3445 0.1539 0.7664 0.6829 0.5433 0.1951 0.4712 0.8439 0.9469 0.5196 0.3641 0.8529 0.7284
% Vector to set as the sub- or superdiagonal
v = rand(N-1,1); % The length of v should be one less than the number of rows/columns in A
% Set the subdiagonal to the values in v
A(sub2ind(size(A), 2:N, 1:(N-1))) = v;
% Set the superdiagonal to the values in v
A(sub2ind(size(A), 1:(N-1), 2:N)) = v;
% Display the updated matrix
disp(A)
0.4869 0.6336 0.9023 0.9455 0.0501 0.6336 0.9904 0.7321 0.3533 0.5165 0.9278 0.7321 0.3445 0.0237 0.7664 0.6829 0.5433 0.0237 0.4712 0.3475 0.9469 0.5196 0.3641 0.3475 0.7284

David Goodmanson
David Goodmanson on 13 Aug 2024
Edited: David Goodmanson on 13 Aug 2024
Hello Olawale
m = rand(5,5)
m1 = diag(m,1) % original upper diagonal
a = [2 3 4 5]' % new upper diagonal elements
mnew = m - diag(m1,1) + diag(a,1)
and the lower diagonal works the same way with 1 replaced by -1 everywhere.
  1 Comment
John D'Errico
John D'Errico on 13 Aug 2024
Certainly the preferred solution almost always. Far cleaner. Easy on the eyes, to read, to debug. Code that you can follow is important one day in the future, when you will need to change it.
The only reason why an indexing solution would be preferred is if the matrices were large, AND if the time cost was significant, so you were doing this operation sufficiently often to matter.

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Naga
Naga on 13 Aug 2024
Hello Olawale,
To modify the off-diagonal elements of the symmetric tridiagonal matrix T to a specified pattern, you can use indexing to directly set these values. Below is a MATLAB code snippet that demonstrates how to change the off-diagonal elements to 1 0 1 0 for the given matrix T.
for i = 1:length(desired_vector)
if i <= n-1
T(i, i+1) = desired_vector(i); % Upper off-diagonal
T(i+1, i) = desired_vector(i); % Lower off-diagonal
end
end

Matt J
Matt J on 13 Aug 2024
Edited: Matt J on 13 Aug 2024
n=8;
T=diag(repelem(1.8,n));
T(2:2*(n+1):end)=1;
T(n+1:2*(n+1):end)=1
T = 8x8
1.8000 1.0000 0 0 0 0 0 0 1.0000 1.8000 0 0 0 0 0 0 0 0 1.8000 1.0000 0 0 0 0 0 0 1.0000 1.8000 0 0 0 0 0 0 0 0 1.8000 1.0000 0 0 0 0 0 0 1.0000 1.8000 0 0 0 0 0 0 0 0 1.8000 1.0000 0 0 0 0 0 0 1.0000 1.8000
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Matt J
Matt J on 13 Aug 2024
Edited: Matt J on 13 Aug 2024
n=8;
T=diag(rand(n,1))
T = 8x8
0.1935 0 0 0 0 0 0 0 0 0.7146 0 0 0 0 0 0 0 0 0.1206 0 0 0 0 0 0 0 0 0.4606 0 0 0 0 0 0 0 0 0.2613 0 0 0 0 0 0 0 0 0.9348 0 0 0 0 0 0 0 0 0.9571 0 0 0 0 0 0 0 0 0.3898
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diagonal=diag(T);
offdiagonal=rand(n-1,1)' %desired values
offdiagonal = 1x7
0.7561 0.8615 0.9447 0.7633 0.1505 0.2376 0.3172
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B=[ [offdiagonal(:);0], [0;offdiagonal(:)] ];
T=full(spdiags(B,[-1,+1],T))
T = 8x8
0.1935 0.7561 0 0 0 0 0 0 0.7561 0.7146 0.8615 0 0 0 0 0 0 0.8615 0.1206 0.9447 0 0 0 0 0 0 0.9447 0.4606 0.7633 0 0 0 0 0 0 0.7633 0.2613 0.1505 0 0 0 0 0 0 0.1505 0.9348 0.2376 0 0 0 0 0 0 0.2376 0.9571 0.3172 0 0 0 0 0 0 0.3172 0.3898
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