Block diagonal matrix of identity times scalar.

2 visualizaciones (últimos 30 días)
Mohammed Kagalwala
Mohammed Kagalwala el 29 de Nov. de 2019
Editada: Philippe Lebel el 29 de Nov. de 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

Respuesta aceptada

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

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Operating on Diagonal Matrices en Help Center y File Exchange.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by