Block diagonal matrix of identity times scalar.

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Hi,
I currently have a vector a = [1 2]', I wish to create a block diagonal matrix. Each block is identity (3x3) times the corresponding scalar in the vector a.
i.e. with a = [1 2]' I want to produce b = [1 0 0 0 0 0; 0 1 0 0 0 0; 0 0 1 0 0 0; 0 0 0 2 0 0; 0 0 0 0 2 0; 0 0 0 0 0 2]. The catch is a can be a vector of N x 1, thus b is of size 3*N x 3*N
The answer also has to be for loop free. I've tried using blkdiag() and eye() but have dimension issues in my multiplication

Accepted Answer

Philippe Lebel
Philippe Lebel on 29 Nov 2019
Edited: Philippe Lebel on 29 Nov 2019
here is my take.
a=[1,2];
c = round(a(1):1/(length(b(:,1))-1):a(2));
matrix = diag(c);
  3 Comments
Philippe Lebel
Philippe Lebel on 29 Nov 2019
Edited: Philippe Lebel on 29 Nov 2019
here it is:
clear
a=[1,2,3,4];
size_of_sub_matrices = 2;
a = arrayfun(@(x) ones(1,size_of_sub_matrices)*x, a, 'UniformOutput', false);
a = cell2mat(a);
matrix = diag(a)
matrix =
1 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0
0 0 2 0 0 0 0 0
0 0 0 2 0 0 0 0
0 0 0 0 3 0 0 0
0 0 0 0 0 3 0 0
0 0 0 0 0 0 4 0
0 0 0 0 0 0 0 4

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