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  end 

Is there a way to do this in one statement w/ vectorization?

e.g. A(something) = K

The statement A(V,V) = K will not work, it assigns off-diagonal elements, and this is not what I want. e.g.:

>> A = zeros(5); >> V = [1 3 4]; >> A(V,V) = 1  A =   1     0     1     1     0  0     0     0     0     0  1     0     1     1     0  1     0     1     1     0  0     0     0     0     0 

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

>> 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.