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.
댓글 수: 14
Stephan
2021년 5월 12일
Please give a small example of what you want as result
Mücahit Özalp
2021년 5월 12일
편집: Mücahit Özalp
2021년 5월 12일
Walter Roberson
2021년 5월 12일
I do not know what it means for a square matrix to be a subdiagonal?
Mücahit Özalp
2021년 5월 12일
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
2021년 5월 12일
the cyclist
2021년 5월 12일
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
2021년 5월 12일
편집: Mücahit Özalp
2021년 5월 12일
Jan
2021년 5월 12일
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
2021년 5월 12일
Mücahit Özalp
2021년 5월 12일
Mücahit Özalp
2021년 5월 12일
Mücahit Özalp
2021년 5월 14일
편집: Mücahit Özalp
2021년 5월 14일
Matt J
2021년 5월 14일
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.
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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
댓글 수: 1
Mücahit Özalp
2021년 5월 12일
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