Let $A\in\mathbb{R}^{3\times 3}$ be a diagonalizable matrix with strictly positive eigenvalues. (Note that $A$ is not required to be symmetric.)

Let $A_S$ be the symmetric part of $A$, that is $$ A_S = \frac{1}{2}(A+A^\top). $$ Suppose that $A_S$ possesses only one positive eigenvalue. In this case, since the vector of eigenvalues of $A_S$ always majorizes the vector $d=[\mathrm{tr}(A)\ 0\ 0]$ we can find an orthogonal matrix $T\in\mathbb{R}^{3\times 3}$ such that $\tilde{A}_S=TA_S T^\top$ has diagonal $d$ (Schur-Horn Theorem).

By applying the same orthogonal transformation $T\in\mathbb{R}^{3\times 3}$ to $A$, we obtain $$\tag{$\ast$} \label{eq:ast} \tilde{A}=T A T^\top = \begin{bmatrix}\mathrm{tr}(A) & \tilde{a}_{12} & \tilde{a}_{13} \\ \tilde{a}_{21} & 0 & \tilde{a}_{23} \\ \tilde{a}_{31} & \tilde{a}_{32} & 0 \end{bmatrix}, $$ where $\tilde{a}_{ij}$ are suitable real numbers.

With reference to the form in Eq. \eqref{eq:ast}:

  1. Does there exist an orthogonal $T$ such that $\tilde{A}$ is sign skew-symmetric, that is $\mathrm{sign}(\tilde{a}_{ij})= - \mathrm{sign}(\tilde{a}_{ji})$ for all $i\ne j$?
  2. If so, is it possible to find an orthogonal $T$ which further yields $\tilde{a}_{12}\tilde{a}_{23}\tilde{a}_{31}=-\tilde{a}_{21}\tilde{a}_{32}\tilde{a}_{13}$?

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