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I have tried

M = {{(a11 + I a12)^2, (b11 + I b12)^2, (c11 + I c12)^2}, {(b11 + 
  I b12)^2, (d11 + I d12)^2, (e11 + I e12)^2}, {(c11 + 
  I c12)^2, (e11 + I e12)^2, (f11 + I f12)^2}};
A = {{-1, 9, 0}, {8, 5, 9}, {0, 2, 4}}
G = Transpose[A].M.A
Solve[G == Conjugate[M], Flatten[M]]

The last line creating problem that Solve::ivar: (a11+I a12)^2 is not a valid variable. Can we simplify such matrix equality and simply get the simplified M matrix without doing it by hand

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    $\begingroup$ A solution for this system as it is, is given by a11=a12=b11=b12=c11=c12=d11=d12=e11=e12=f11=f12=0 $\endgroup$
    – Cesareo
    Mar 2, 2020 at 20:53
  • $\begingroup$ Thanks @Cesareo but how using mathematica I can solve for those conditions and by not solving individually all the equations $\endgroup$
    – user105697
    Mar 2, 2020 at 22:06

1 Answer 1

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Solve a simpler problem...

mm2 = {{m11, m12, m13},
       {m12, m22, m23},
       {m13, m23, m33}};

gg2 = Transpose[aa].mm2.aa;

Solve[gg2 == Conjugate[mm2], Union@Flatten@mm2];


(* {{m11 -> 0, m12 -> 0, m13 -> 0, m22 -> 0, m23 -> 0, m33 -> 0}} *)

So as @Cesareo notes in the comments, all of the parameters are zero.

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  • $\begingroup$ So according to your code is it can be done for the above system also since mine problem is not taking (a11+i a12) as a variable $\endgroup$
    – user105697
    Mar 2, 2020 at 22:04
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    $\begingroup$ Correct, since we now know (a11+i a12)^2 = 0, we know each of a11 and a12 = 0. $\endgroup$
    – MikeY
    Mar 2, 2020 at 22:08
  • $\begingroup$ ok thanks. Actually my original matrix is more complex but I think I must try your method to my original matrix also $\endgroup$
    – user105697
    Mar 2, 2020 at 22:12

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