Let $A\in\mathbb{R}^{n\times n}$ be a diagonalizable matrix with real and strictly negative eigenvalues. Furthermore, suppose that $\mathrm{tr}(A)=-1$.
My question. I'm wondering whether it is possible to write any $A$ as above in the following particular form: $$\tag{#}\label{eq:decomp} A=T\left(\left[\begin{array}{c|c}D_1 & 0 \\\hline 0 & 0\end{array}\right] + S\left[\begin{array}{c|c}-I_r & 0 \\\hline 0 & D_2\end{array}\right] \right)T^\top $$ where $T\in\mathbb{R}^{n\times n}$ is an orthogonal matrix ($T^\top T=I_n$), $S\in\mathbb{R}^{n\times n}$ is a skew-symmetric matrix ($S^\top=-S$), $D_1\in\mathbb{R}^{r\times r}$ $D_2\in\mathbb{R}^{(n-r)\times (n-r)}$, $1\le r\le n$, are diagonal matrices with strictly negative entries. Note that $r$ is a further integer parameter that can vary between $1$ and $n$.
Notice that each $A$ as above is orthogonally similar to an (upper) triangular matrix (Schur form). The latter has $n(n+1)/2$ "degrees of freedom" (one for each non-zero entry). The decomposition in \eqref{eq:decomp} has the same degrees of freedom ($n$ for the non-zero diagonal entries in $D_1$ and $D_2$ and $n(n-1)/2$ for the entries in the skew-symmetric matrix $S$). This simple observation suggests that, in principle, the decomposition in \eqref{eq:decomp} could be valid.
For $n=2$, I've managed to prove that it is always possible to write $A$ as in \eqref{eq:decomp} (see remark below). However, for $n\ge 3$ proving this fact (or finding a counterexample) seems quite hard. Thus, I would be very grateful to receive some feedback from the MO community. Thank you.
A special case. If $A+A^\top$ is negative definite (i.e., $A+A^\top<0$) then by decomposing $A$ as $$ A=\underbrace{\frac{1}{2}(A+A^\top)}_{=:A_s} + \underbrace{\frac{1}{2}(A-A^\top)}_{=:A_{as}}, $$ we can select $r=n$, $T$ and $D_1$ to be the eigenvector and eigenvalue (resp.) matrix of $A_s$ (i.e., $A_s =TD_1T^\top$), and $S=-T^\top A_{as}T$. This choice yields a decomposition as in \eqref{eq:decomp}.
$2\times 2$ case. Let $n=2$, and (wlog) suppose that $A$ is in its Schur (upper) triangular form: $$ A=\begin{bmatrix}a & c \\ 0 & b \end{bmatrix}, $$ where $a,b<0$, $a+b=-1$, $c\in\mathbb{R}$. Consider $$ A_s=\frac{1}{2}(A+A^\top)=\begin{bmatrix} a & \frac{c}{2} \\ \frac{c}{2} & b \end{bmatrix}, $$
If $ab-c^2/4>0$, then $A_s<0$ and we are done (in view of the above remark). Otherwise, by virtue of the Schur-Horn Theorem, there exists an orthogonal matrix $T\in\mathbb{R}^{2\times 2}$ such that $$ T^\top A_sT = \begin{bmatrix}-1 & \sqrt{\frac{c^2}{4}-ab} \\ \sqrt{\frac{c^2}{4}-ab} & 0 \end{bmatrix}. $$ Next, consider $$ A_{as}=\frac{1}{2}(A-A^\top)=\begin{bmatrix} 0 & \frac{c}{2} \\ -\frac{c}{2} & 0 \end{bmatrix} $$ and notice that, under the previous orthogonal $T$, $A_{as}$ is still skew-symmetric and so it remains unchanged (up to a $\pm 1$). Thus, we have \begin{align} A=A_s+A_{as}&=T\left(\begin{bmatrix}-1 & \sqrt{\frac{c^2}{4}-ab} \\ \sqrt{\frac{c^2}{4}-ab} & 0 \end{bmatrix} + \begin{bmatrix} 0 & \frac{c}{2} \\ -\frac{c}{2} & 0 \end{bmatrix}\right)T^\top\\ &=T\left(\begin{bmatrix}-1 & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & \sqrt{\frac{c^2}{4}-ab}+\frac{c}{2} \\ \sqrt{\frac{c^2}{4}-ab}-\frac{c}{2} & 0 \end{bmatrix}\right)T^\top.\tag{1}\label{eq:2x2} \end{align} Finally, by picking $r=1$, $$ S=\begin{bmatrix} 0 & \sqrt{\frac{c^2}{4}-ab}-\frac{c}{2} \\ -\sqrt{\frac{c^2}{4}-ab}+\frac{c}{2} & 0 \end{bmatrix},\quad D_2= \frac{\sqrt{\frac{c^2}{4}-ab}+\frac{c}{2}}{\sqrt{\frac{c^2}{4}-ab}-\frac{c}{2}}<0, $$ we can write \eqref{eq:2x2} as $$ A=T\left(\begin{bmatrix}-1 & 0 \\ 0 & 0 \end{bmatrix} + S \begin{bmatrix}-1 & 0 \\ 0 & D_2 \end{bmatrix} \right)T^\top.$$ Hence, we have obtained a decomposition as in \eqref{eq:decomp}.
