# Block diagonal matrix of identity times scalar.

1 Ansicht (letzte 30 Tage)
Mohammed Kagalwala am 29 Nov. 2019
Bearbeitet: Philippe Lebel am 29 Nov. 2019
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
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### Akzeptierte Antwort

Philippe Lebel am 29 Nov. 2019
Bearbeitet: Philippe Lebel am 29 Nov. 2019
here is my take.
a=[1,2];
c = round(a(1):1/(length(b(:,1))-1):a(2));
matrix = diag(c);
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Mohammed Kagalwala am 29 Nov. 2019
Thank you for your help!
Philippe Lebel am 29 Nov. 2019
Bearbeitet: Philippe Lebel am 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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