Let $\mathcal{M}_2(\mathbb{C})$ be the set of square matrices of dimension 2, and $\mathcal{S}_2(\mathbb{C})$ the subset of self-adjoint matrices. The function $f_X: \mathcal{S}_2(\mathbb{C}) \rightarrow \mathcal{S}_2(\mathbb{C})$ is defined by $$ f_X(A) = \sum_{n=0}^\infty X^n A (X^\dagger)^n $$ where $X \in \mathcal{M}_2(\mathbb{C})$, and where $X^\dagger$ is the Hermitian adjoint of $X$. The problem is the following:

Is there a non-trivial matrix $X$ such that, $\forall A, B \in \mathcal{S}_2(\mathbb{C})$, if $\mathrm{Tr}\left[AB\right] = 0$, then $\mathrm{Tr}\left[f_X(A)f_X(B)\right] =0$?

Any hint would be appreciated!

EDIT: As mentionned in the comments, $X=0$ and $X = cI$ with $|c| < 1$ are obvious solutions. Also, if $X^d = cI$, then $f_X(A) = \sum_{n=0}^\infty c^n \sum_{m=0}^{d-1} X^m A (X^\dagger)^m = \frac{1}{1-c} \sum_{m=0}^{d-1} X^m A (X^\dagger)^m$. Furthermore, the eigenvalues $\lambda_i$ of $X$ are such that $\lambda_i^d = c$, so one can diagonalize $X = PDP^{-1}$ but I did not manage to go further when substituting in the trace.

  • $\begingroup$ Yes, I forgot to mention I am looking for non-zero solutions. I edited my question accordingly. $\endgroup$
    – Ronan
    May 5, 2020 at 10:51
  • $\begingroup$ $X=cI$ is also a solution if $|c|<1$. $\endgroup$
    – user1551
    May 5, 2020 at 11:23
  • $\begingroup$ Thanks for your comment. This indeed works, but is there other solution than $X = cI$ and $X=0$? I would especially be interested in the mathematics behind the resolution of this problem. $\endgroup$
    – Ronan
    May 5, 2020 at 12:03
  • $\begingroup$ @Ronan One can prove that $X$ must be full rank. Have you investigated diagonal $X\ne cI$ ? $\endgroup$
    – River Li
    May 11, 2020 at 12:02


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