Why Matlab/Mupad ignores absolute value when multiplying complex conjugate expressions

Hello everyone,

Recently, I recognized that Matlab (as well as Mupad) ignores to take the absolute value when multiplying a variable with its complex conjugate. I'm currently trying to calculate the eigenvalues of an 4 x 4 Hermitian matrix and I believe that this issue results in wrong eigenvalues. Take for example the matrix (now in Mupad):

assume(x, Type::Real)

M := matrix([[0,conjugate(I*x),0,0], [I*x,0,0,0], [0,0,1,conjugate(I*x)], [0,0,I*x,1]])

then the eigenvalues from Mupad are linalg::eigenvalues(M) => {x, 1-x, -x, x+1}. That's wrong! It should be: {|x| 1-|x|, -|x|, x+1}.

Can anyone help ? Thanks.

Best regards

Bernhard

[Torsten]

You assume x to be real, thus x*conj(x)=x^2.

Best wishes

Torsten.

[Bernhard Ihrig]

Thanks for your comment Torsten. However, the problem with the matrix remains, because it seems that Mupad calculates sqrt(x^2) = x (?). If you calculate the eigenvalues of the example matrix M by hand, you will see what I'm intending to say, hopefully.

Best regards

Bernhard

[Torsten]

As eigenvalues of M I get: x, -x, 1-x and x-1.

Best wishes

Torsten.

Answer 1

[Steven Lord]

If x is -1, then the eigenvalues returned by MuPAD are -1, 2, 1, 0. The eigenvalues you expected are 1, 0, -1, 2 (assuming you meant |x|+1 as your last expression.) They're the same, just in a different order.

More generally, since both |x| and -|x| are eigenvalues in your list of expected eigenvalues and you told MuPAD it can assume x is real, that's the same as including x and -x in the list of eigenvalues. They'll just be listed in a different order if x is negative. Similarly for real x, 1-|x| and 1+|x| are the same as 1-x and 1+x, just in a different order if x is negative.

Now if you assumed x was Type::Complex, and called:

 simplify(linalg::eigenvalues(M))

then the sets {|x|, -|x|} and {x, -x} are NOT necessarily the same but in a different order. In that case, MuPAD knows it cannot eliminate the absolute value signs, and indeed when I make that modification to your assume call the results do include the absolute value signs.

[Bernhard Ihrig]

Hello Steven,

many thanks for your answer. For me it appears still weird, but hopefully I will now get the right eigenvalues.

Best regards

Bernhard