# A parametrization of stable matrices

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}.

• Is $r$ fixed? <character limit> – Federico Poloni Sep 20 '18 at 5:56
• @FedericoPoloni: no, $r$ is a parameter. – Ludwig Sep 20 '18 at 5:58
• Some info on the number of (real) eigenvalues of $A$? Would you also allow $A$ to have no real eigenvalues at all, as technically they would then also all be strictly negative? – Dirk Sep 20 '18 at 9:21
• @DirkLiebhold: All the eigenvalues of $A$ ($n$ eigenvalues) are real and strictly negative. – Ludwig Sep 20 '18 at 13:45