# A decomposition for triangular matrices with positive diagonal entries

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.