Creating a tridiagonal matrix
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Aaron Atkinson
el 11 de Nov. de 2019
Comentada: Ana Sarai
el 22 de Feb. de 2025
I am currently trying to create a 500*500 matrix in matlab with diagonals a=-1, b=4, c=2. My teacher has said that the best way to go about it is using loops, but is there a coded in function to use?
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Stephen23
el 11 de Nov. de 2019
Editada: Stephen23
el 20 de Mzo. de 2022
"My teacher has said that the best way to go about it is using loops"
Why on earth would they say that? Here are some non-loop aproaches:
1- See @giannit's comment: https://www.mathworks.com/matlabcentral/answers/490368-creating-a-tridiagonal-matrix#comment_1027546
>> N = 10;
>> a = -1;
>> b = 4;
>> c = 2;
>> M = diag(a*ones(1,N)) + diag(b*ones(1,N-1),1) + diag(c*ones(1,N-1),-1)
M =
-1 4 0 0 0 0 0 0 0 0
2 -1 4 0 0 0 0 0 0 0
0 2 -1 4 0 0 0 0 0 0
0 0 2 -1 4 0 0 0 0 0
0 0 0 2 -1 4 0 0 0 0
0 0 0 0 2 -1 4 0 0 0
0 0 0 0 0 2 -1 4 0 0
0 0 0 0 0 0 2 -1 4 0
0 0 0 0 0 0 0 2 -1 4
0 0 0 0 0 0 0 0 2 -1
3- indexing is reasonably simple:
>> M = zeros(N,N);
>> M( 1:1+N:N*N) = a;
>> M(N+1:1+N:N*N) = b;
>> M( 2:1+N:N*N-N) = c
M =
-1 4 0 0 0 0 0 0 0 0
2 -1 4 0 0 0 0 0 0 0
0 2 -1 4 0 0 0 0 0 0
0 0 2 -1 4 0 0 0 0 0
0 0 0 2 -1 4 0 0 0 0
0 0 0 0 2 -1 4 0 0 0
0 0 0 0 0 2 -1 4 0 0
0 0 0 0 0 0 2 -1 4 0
0 0 0 0 0 0 0 2 -1 4
0 0 0 0 0 0 0 0 2 -1
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Stephen23
el 30 de Oct. de 2020
"It's not clear to me how you're indexing a matrix using just one argument."
The second example uses linear indexing:
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