# Is there a matrix $X$ that conserves orthgonality through this function $f_X$?

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.

• Yes, I forgot to mention I am looking for non-zero solutions. I edited my question accordingly. May 5, 2020 at 10:51
• $X=cI$ is also a solution if $|c|<1$. May 5, 2020 at 11:23
• 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. May 5, 2020 at 12:03
• @Ronan One can prove that $X$ must be full rank. Have you investigated diagonal $X\ne cI$ ? May 11, 2020 at 12:02