# On a special decomposition of a $3\times 3$ matrix

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 $$\tilde{a}_{12}\tilde{a}_{23}\tilde{a}_{31}=-\tilde{a}_{21}\tilde{a}_{32}\tilde{a}_{13}$$?