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For example, I have a matrix A given by some parameters (a,b,c,d) and two variables (x,y), $$ A=\begin{pmatrix}a&a+bx\\(c-d)y&a(c+d)\end{pmatrix} $$ The condition is that A is a matrix independent of (x,y). Therefore, we find b=0 and c=d.

How should I do it in mathematica? Now I have a complicated 64-by-64 matrix with 32 parameters and 3 variables. So I want a general method to solve this problem.

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  • $\begingroup$ Are the matrix elements simple polynomial functions of $x$ and $y$, or is it more generally complicated than that? $\endgroup$
    – march
    Commented Oct 19, 2023 at 20:08
  • $\begingroup$ If the matrix elements are simple enough, it's possible that something like SolveAlways[ Flatten[A] == Flatten[A /. {x -> 0, y -> 0}] // Thread, {x, y}] would work, although I don't know how fast that will be on a 64 by 64 matrix. $\endgroup$
    – march
    Commented Oct 19, 2023 at 20:11
  • $\begingroup$ They are rational expressions of x and y. If we set all variables to 0, some denominators might be 0 too. $\endgroup$
    – Curious Cat
    Commented Oct 19, 2023 at 20:18
  • $\begingroup$ Can you give a more complicated example, then, one that shows the difficulty in the issue here? Edit your post to include the example, properly formatted in code blocks with proper Mathematica notation. $\endgroup$
    – march
    Commented Oct 19, 2023 at 20:19

1 Answer 1

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Set the derivatives of $A$ with respect to $x$ and $y$ to zero for all values of $x$ and $y$:

A = {{a, a + b x}, {(c - d) y, a (c + d)}};

SolveAlways[Thread[Flatten[D[A, {{x, y}}]] == 0], {x, y}]
(*    {{b -> 0, c -> d}}    *)
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