Change an element in diagonal of matrix

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mat
mat am 19 Nov. 2015
Beantwortet: Tushar Athawale am 25 Nov. 2015
As example
A=magic(5);
result=diag(A,-1);
How can I change an element of this diagonal array when I don't know their indices?
  1 Kommentar
Kirby Fears
Kirby Fears am 25 Nov. 2015
The size of diag(A,n) will always be (length(A)-abs(n)) by 1.
The indices of your diagonal array are 1 through 4. They correspond to elements (2,1), (3,2), (4,3), and (5,4) of A. This correspondence can be formulated just in terms of n and the size of A.
Does this help? I'm not sure what information you are seeking.

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Tushar Athawale
Tushar Athawale am 25 Nov. 2015
I understand that you want to update the elements stored at a specified diagonal without knowing the element indices.
One way to do this could be to create a new matrix of ones along the specified diagonal with rest of the elements set to 0. This matrix then can be used as a reference to update the elements in original matrix. Try the following code snippet to increment the elements at the specified diagonal by 2:
% out: updated matrix, n: size of the input matrix, shift: diagonal selection
function out = update_diagonal(n,shift)
A=magic(n)
result=diag(A,shift);
% Create matrix of ones along specified diagonal
iden = diag(ones(n-abs(shift),1),shift);
% Modify the elements at location of ones in 'iden'. Here, increment all elements
% along the specified diagonal by 2
A(iden(:,:)~=0) = A(iden(:,:)~=0)+2;
out = A;
I hope this answers your question.

Weitere Antworten (2)

Andrei Bobrov
Andrei Bobrov am 25 Nov. 2015
e.g.:
A=magic(5);
ii = diag(true(4,1),-1);
A(ii) = 100;

Sean de Wolski
Sean de Wolski am 25 Nov. 2015
Bearbeitet: Sean de Wolski am 25 Nov. 2015
The -1st diagonal is 2:n+1:end. You can index into that directly without ever needing diag/tril
n = 5;
x = magic(n);
didx = 2:n+1:numel(x);
% Change third element on -1st diagonal
x(didx(3)) = -100
x =
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 -100 21 3
11 18 25 2 9

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