I want to make an identity matrix

Question

Qonitat 2023-3-5

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https://ww2.mathworks.cn/matlabcentral/answers/1923450-i-want-to-make-an-identity-matrix

编辑： John D'Errico 2023-3-5

i want to make a matrix that look like this

How could i achieve this efficiently .Thanks in advace

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Answer 1

the cyclist 2023-3-5

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在 MATLAB Online 中打开

I've made some assumptions about the pattern, but I expect it is what you want.

This should be highly efficient. The algorithm is a bit obfuscated.

% Input 
N = 4;
% Start with all 0s
M = zeros(N,N*(N+1));
% Fill in the 1s
M(1:N*N+1:end-N*N) = 1;
% Fill in the -1s
M(:,N*N+1) = -1;
% Display the result
disp(M)
     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0    -1     0     0     0
     0     0     0     0     1     0     0     0     0     0     0     0     0     0     0     0    -1     0     0     0
     0     0     0     0     0     0     0     0     1     0     0     0     0     0     0     0    -1     0     0     0
     0     0     0     0     0     0     0     0     0     0     0     0     1     0     0     0    -1     0     0     0

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Answer 2

John D'Errico 2023-3-5

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https://ww2.mathworks.cn/matlabcentral/answers/1923450-i-want-to-make-an-identity-matrix#answer_1185645

编辑：John D'Errico 2023-3-5

在 MATLAB Online 中打开

These are starting to look like homework questions.

Did you not read the answer to the last question you asked? You could not use that idea to build this matrix?

A = kron([eye(4),zeros(4,1)],[1 0 0 0])
A = 4×20
     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     1     0     0     0     0     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     1     0     0     0     0     0     0     0     0     0     0     0
     0     0     0     0     0     0     0     0     0     0     0     0     1     0     0     0     0     0     0     0

That gets you almost there, with the first 16 columns correct, and a total of 20 columns. Then it would be simple enough to stuff in the -1 elements in column 17.

A(:,17) = -1

A = 4×20

1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 -1 0 0 0

spy(A)