Tridiagonal Matrix with subdiagonal and main diagonal is also matrix
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I have two matrices A and B. I want A to be main diagonal and B to be my subdiagonals. How do I create such a matrix? By the way sizes of A and B changes but they are square matrices.
Specifically, a1=4,b1=-1
A =diag(a1*ones(1,N-1)) + diag(b1*ones(1,N-2),1) + diag(b1*ones(1,N-2),-1)
B=(-1)*eye(N-1)
These are my A and B matrices. I need to define a (N-1)*(N-1) times (N-1)*(N-1) matrix . For example for N=1000 or N=5000 I should be able to change the N value.
14 Comments
Stephan
on 12 May 2021
Please give a small example of what you want as result
Mücahit Özalp
on 12 May 2021
Edited: Mücahit Özalp
on 12 May 2021
Walter Roberson
on 12 May 2021
I do not know what it means for a square matrix to be a subdiagonal?
Mücahit Özalp
on 12 May 2021
I am also confused by your terminology. Let's take a look at a very small example:
N = 5;
a1 = 4;
b1 = -1;
A =diag(a1*ones(1,N-1)) + diag(b1*ones(1,N-2),1) + diag(b1*ones(1,N-2),-1)
B=(-1)*eye(N-1)
It seems that you are saying that A and B, as defined above, are the inputs to the new matrix you want. Can you tell us what the output should be? Don't just describe it. Please write the output, ideally in MATLAB syntax.
Mücahit Özalp
on 12 May 2021
the cyclist
on 12 May 2021
I was not asking you to make the general algorithm. I was asking you what the output would be for that one small example.
But I think what you just posted makes it clearer.
Mücahit Özalp
on 12 May 2021
Edited: Mücahit Özalp
on 12 May 2021
Jan
on 12 May 2021
You mention the general form "[(N-1)^2] x [(N-1)^2]" and post a 12x12 matrix as example, which does not match this format.
You mention: "B=(-1)*eye(N-1)" but the example contains multiple different full 3x3 matrices.
Please post small examples e.g. for N==3 including all inputs and a manually constructed output. Maybe an N=4 example is required.
Mücahit Özalp
on 12 May 2021
Mücahit Özalp
on 12 May 2021
Mücahit Özalp
on 12 May 2021
Mücahit Özalp
on 14 May 2021
Edited: Mücahit Özalp
on 14 May 2021
Matt J
on 14 May 2021
I don't think so, but if you post a new question on it (ideally with a demo), others on the forum may have some thoughts.
Accepted Answer
More Answers (1)
Here's another way, probably much faster.
N=1000;
a1=4;b1=-1;
A =diag(a1*ones(1,N-1)) + diag(b1*ones(1,N-2),1) + diag(b1*ones(1,N-2),-1);
B=(-1)*eye(N-1);
E0=speye(N);
E1=E0(2:end,1:end-1);
E0=E0(1:end-1,1:end-1);
result=kron(E0,A) + kron(E1,B)+kron(E1.',B);
whos result
1 Comment
Mücahit Özalp
on 12 May 2021
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