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Return only the real nullspace of a complex matrix

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Martin Seltmann
Martin Seltmann 2018-10-24
评论： Torsten 2018-10-24
I need to restrict the computation of the nullspace of a complex matrix A to one over the REAL numbers. How do I do that? null(A,’Real’) did not work.

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采纳的回答

Bruno Luong
Bruno Luong 2018-10-24
编辑：Bruno Luong 2018-10-24
A = randn(2,5)+1i*randn(2,5);
% N is a real orthonormal basis of null(A)
n = size(A,2);
NR=null([[real(A), imag(A)]; [-imag(A), real(A)]; [zeros(n), eye(n)]]);
N=NR(1:n,:);
% Check
norm(A*N)
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Torsten
Torsten 2018-10-24
If x is real valued and A*x = 0, you have (real(A)+i*imag(A))*x = 0. This implies that real(A)*x = 0 and imag(A)*x=0.
Conversely, if real(A)*x=0 and imag(A)*x=0 for a real vector x, then real(A)*x+i*imag(A)*x = A*x = 0.
Best wishes
Torsten.

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Bruno Luong
Bruno Luong 2018-10-24
编辑：Bruno Luong 2018-10-24
Torsten proposition:
NR = null([real(A);imag(A)]);
Torsten is right, it's actually the same null space, and his formula is much simpler.
Because
real(A)*NR = 0
imag(A)*NR = 0
so
A*NR = (real(A)+1i*imag(A))*NR = real(A)*NR + 1i*(imag(A)*NR) = 0
The opposite is also true.
Since
A*NR = 0 in the complex field
because NR is real, therefore the above is equivalent to
real(A)*NR = 0
and because
A*(1i*NR) = 0
as well, this implies
imag(A)*NR = 0
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Bruno Luong
Bruno Luong 2018-10-24
编辑：Bruno Luong 2018-10-24
You should't accept my comment, but Torsten's reply (if he put his code in a separate answer)
Torsten
Torsten 2018-10-24
You were the first who answered ( and your answer is not wrong :-) )
Best wishes
Torsten.

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