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I am trying to construct a symbolic matrix with the following property- $$ \left(D_{3\times3}(q)\right)^T = D_{3\times3}(-q) $$ Note that for $q\in \mathbb{R}$, $D(q)$ would be hermitian.

So far i have tried to bend a construct from a related question

dmat[q_] :=Table[If[i <= j, Dm[i, j][q], Dm[j, i][-q]], {i, 1, 3}, {j, 1, 3}]
Assuming[{Apply[And, Diagonal[dmat[q]] == Diagonal[dmat[-q]]]},Simplify[Transpose[dmat[q]] == dmat[-q]]]

The proper output, however, eludes me. I get the output

{{-Dm[1, 1][-q] + Dm[1, 1][q], 0, 0}, {0, -Dm[2, 2][-q] + Dm[2, 2][q],0}, {0, 0, -Dm[3, 3][-q] + Dm[3, 3][q]}} == {{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}

The error, obviously lies in my inability to express

{Apply[And, Diagonal[dmat[q]] == Diagonal[dmat[-q]]]}

properly. Any fix to this?

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I would make use of Mathematica patterns in the assumptions:

Assuming[{Dm[a_, a_][x_] == Dm[a_, a_][-x_]}, Simplify[Transpose[dmat[q]] == dmat[-q]]]
(* True *)

This tells Simplifythat a matrix element with equal indices (i.e. a diagonal element) is an even function.

The problem with your code is that with Apply, you are replacing the equal sign with an && operator

Apply[And, Diagonal[dmat[q]] == Diagonal[dmat[-q]]]

(* {Dm[1, 1][q], Dm[2, 2][q], Dm[3, 3][q]} && {Dm[1, 1][-q], Dm[2, 2][-q], Dm[3, 3][-q]} *)

The working code that is closest to your approach would be

MapThread[Equal, {Diagonal[dmat[q]], Diagonal[dmat[-q]]}]

which gives

{Dm[1, 1][q] == Dm[1, 1][-q], Dm[2, 2][q] == Dm[2, 2][-q], Dm[3, 3][q] == Dm[3, 3][-q]}

EDIT

Another possible solution is to define the diagonal elements as functions of Abs[q].

dmat[q_] := Table[If[i == j, Dm[i, i][Abs[q]], 
            If[i <= j, Dm[i, j][q], Dm[j, i][-q]]], {i, 1, 3}, {j, 1, 3}]

This way, you won't need any further assumption:

Simplify[Transpose[dmat[q]] == dmat[-q]]
(* True *)
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  • $\begingroup$ Regarding the EDIT: $q\in\mathbb{C}$ in general. So defining the diagonal elements in terms of Abs[q] wouldn't work. In fact even for real q, my gut feeling is that restricting your matrix to a function of Abs[q] is too restrictive, as opposed to $f(q)=f(-q)$. $\endgroup$ Commented Jan 26, 2015 at 17:17
  • $\begingroup$ @DebanjanBasu you are right, I thought I read that $q \in \mathbb{R}$ in the question but it was just a note. However, if we restrict to $\mathbb{R}$, I think that an even function is equivalent to a function of $|q|$. I can't think of a counterexample... $\endgroup$ Commented Jan 26, 2015 at 22:44
  • $\begingroup$ @Pincopalino I think you are right. I convinced myself by making a power series expansion of $f(q)$ around $q=0$ and comparing powers of $f(\pm q)$. $\endgroup$ Commented Jan 28, 2015 at 12:33

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