How to assign values to a MATLAB matrix on the diagonal?

Question

Suppose I have an NxN matrix A, an index vector V consisting of a subset of the numbers 1:N, and a value K, and I want to do this:

 for i = V
     A(i,i) = K         


        

Answer 1

I usually use EYE for that:

A = magic(4)
A(logical(eye(size(A)))) = 99

A =
    99     2     3    13
     5    99    10     8
     9     7    99    12
     4    14    15    99

Alternatively, you can just create the list of linear indices, since from one diagonal element to the next, it takes nRows+1 steps:

[nRows,nCols] = size(A);
A(1:(nRows+1):nRows*nCols) = 101
A =
   101     2     3    13
     5   101    10     8
     9     7   101    12
     4    14    15   101

If you only want to access a subset of diagonal elements, you need to create a list of diagonal indices:

subsetIdx = [1 3];
diagonalIdx = (subsetIdx-1) * (nRows + 1) + 1;
A(diagonalIdx) = 203
A =
   203     2     3    13
     5   101    10     8
     9     7   203    12
     4    14    15   101

Alternatively, you can create a logical index array using diag (works only for square arrays)

diagonalIdx = false(nRows,1);
diagonalIdx(subsetIdx) = true;
A(diag(diagonalIdx)) = -1
A =
    -1     2     3    13
     5   101    10     8
     9     7    -1    12
     4    14    15   101

Answer 2

Suppose K is the value. The command

A=A-diag(K-diag(A))

may be a bit faster

>> A=randn(10000,10000);

>> tic;A(logical(eye(size(A))))=12;toc

Elapsed time is 0.517575 seconds.

>> tic;A=A+diag((99-diag(A)));toc

Elapsed time is 0.353408 seconds.

But it consumes more memory.

Answer 3

>> B=[0,4,4;4,0,4;4,4,0]

B =

     0     4     4
     4     0     4
     4     4     0

>> v=[1,2,3]

v =

     1     2     3

>> B(eye(size(B))==1)=v
%insert values from v to eye positions in B

B =

     1     4     4
     4     2     4
     4     4     3

Answer 4

I'd use sub2ind and pass the diagonal indices as both x and y parameters:

A = zeros(4)
V=[2 4]

idx = sub2ind(size(A), V,V)
% idx = [6, 16]

A(idx) = 1

% A =
% 0     0     0     0
% 0     1     0     0
% 0     0     0     0
% 0     0     0     1

Answer 5

I use this small inline function in finite difference code.

A=zeros(6,3);
range=@(A,i)[1-min(i,0):size(A,1)-max(i+size(A,1)-size(A,2),0 ) ];
Diag=@(A,i) sub2ind(size(A), range(A,i),range(A,i)+i );
A(Diag(A, 0))= 10; %set diagonal 
A(Diag(A, 1))= 20; %equivelent to diag(A,1)=20;
A(Diag(A,-1))=-20; %equivelent to diag(A,-1)=-20;

It can be easily modified to work on a sub-range of the diagonal by changing the function range.

Answer 6

>> tt = zeros(5,5)
tt =
     0     0     0     0     0
     0     0     0     0     0
     0     0     0     0     0
     0     0     0     0     0
     0     0     0     0     0
>> tt(1:6:end) = 3
tt =
     3     0     0     0     0
     0     3     0     0     0
     0     0     3     0     0
     0     0     0     3     0
     0     0     0     0     3

and more general:

>> V=[1 2 5]; N=5;
>> tt = zeros(N,N);
>> tt((N+1)*(V-1)+1) = 3
tt =
     3     0     0     0     0
     0     3     0     0     0
     0     0     0     0     0
     0     0     0     0     0
     0     0     0     0     3

This is based on the fact that matrices can be accessed as one-dimensional arrays (vectors), where the 2 indices (m,n) are replaced by a linear mapping m*N+n.