Return only the real nullspace of a complex matrix

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.

[Torsten]

It's not clear what you mean. You want to know if kern(A) contains purely real vectors ?

[Martin Seltmann]

Yes, and extract only those.

Answer 1

[Bruno Luong]

    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)

[Torsten]

Shouldn't this simply be

NR = null(real(A);imag(A));

?

[Martin Seltmann]

But would that last option not just return all vectors that lie in the nullspace of both real(A) and imag(A) ?

[Torsten]

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.

[Martin Seltmann]

Of course, no clue why I was confused there. Thanks!

Answer 2

[Bruno Luong]

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

[Martin Seltmann]

You are right. Of course NR being real forces BOTH real(A)*NR AND imag(A)*NR to vanish. Thank you very much!

[Bruno Luong]

You should't accept my comment, but Torsten's reply (if he put his code in a separate answer)

[Torsten]

You were the first who answered ( and your answer is not wrong :-) )

Best wishes

Torsten.