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Let $A\in\mathbb{R}^{n\times n}$ be an upper triangular matrix with strictly positive diagonal entries.

Is it possible to find:

  • an orthogonal matrix $T\in\mathbb{R}^{n\times n}$ ($T^\top T=I_n$),
  • a skew-symmetric matrix $S\in\mathbb{R}^{n\times n}$ ($S^\top=-S$),
  • an integer $r$, $1\le r\le n$, and
  • diagonal matrices $D_1\in\mathbb{R}^{r\times r}$, $D_2\in\mathbb{R}^{(n-r)\times (n-r)}$ with strictly positive diagonal entries,

such that $A$ can be written as $$\label{eq:decomp} A=T\left(\left[\begin{array}{c|c}D_1 & 0 \\\hline 0 & 0\end{array}\right] + S\left[\begin{array}{c|c}I_r & 0 \\\hline 0 & D_2\end{array}\right] \right)T^\top \ \ \ \ \ ? $$

After some lengthy computations, I managed to prove this result for $n=2$ (see Remark 2 below). Dealing with the general case ($n\ge 3$) however seems out of reach for me at this moment. So any help is very welcome.


Remark 1 ($A+A^\top>0$). If $A+A^\top$ is positive definite, 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 the desired decomposition.


Remark 2 ($n=2$). Let $n=2$ and explicitly write $A$ as: $$ A=\begin{bmatrix}a & c \\ 0 & b \end{bmatrix}, $$ where $a,b>0$, $c\in\mathbb{R}$. 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 Remark 1). Otherwise, by virtue of the Schur-Horn Theorem, there exists an orthogonal matrix $T\in\mathbb{R}^{2\times 2}$ such that $$ T^\top A_sT = \begin{bmatrix}a+b & \sqrt{\frac{c^2}{4}-ab} \\ \sqrt{\frac{c^2}{4}-ab} & 0 \end{bmatrix}. $$ 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}a+b & \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}a+b & 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 $r=1$, $$ S=\begin{bmatrix} 0 & -\sqrt{\frac{c^2}{4}-ab}+\frac{c}{2} \\ \sqrt{\frac{c^2}{4}-ab}-\frac{c}{2} & 0 \end{bmatrix},\ \ D_1=a+b>0, \ \ D_2= \frac{\sqrt{\frac{c^2}{4}-ab}+\frac{c}{2}}{-\sqrt{\frac{c^2}{4}-ab}+\frac{c}{2}}>0, $$ we can write \eqref{eq:2x2} as $$ A=T\left(\begin{bmatrix}a+b & 0 \\ 0 & 0 \end{bmatrix} + S \begin{bmatrix}1 & 0 \\ 0 & D_2 \end{bmatrix} \right)T^\top.$$ Hence, we have obtained the desired decomposition.

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