Let $V = M_n(\mathbb{R})$ be the set of all $n\times n$ matrices with real elements and $V_{-}$ be a subset of Hurwitz stable matrices, i.e. matrices such that all their eigenvalues have strictly negative real part.
Which $\mathbb{R}$-linear invertible maps $f:V \to V$ satisfy $f(V_{-}) = V_{-}$?
For example, the following maps satisfy this condition:
- Map $B \mapsto \alpha A^{-1} B A$ for any $\alpha > 0$, and any invertible matrix $A \in M_n(\mathbb{R})$.
- Map $B \mapsto \alpha A^{-1} B^T A$ for any $\alpha > 0$, and any invertible matrix $A \in M_n(\mathbb{R})$.
Are there any others?
My attempts at answering this:
- One can show that the set of matrices $A$ such that for any $B \in V_{-}$ and any $\varepsilon > 0$ we have $B - \varepsilon A \in V_{-}$ is $\{\alpha I: \alpha > 0\}$. Since $f$ has to preserve this set, we know that $f(I) = \alpha I$.
- For 2x2 matrices we can show that any such $f$ is indeed either of the form #1 or #2 above. In order to do that, one check that the set of vectors pointing into interior of $V_{-}$ is different for matrices on the boundary of $\partial V_{-}$ with different Jordan Normal Form types. More specifically the set $W = \{A \in \partial V_{-}: \partial\{B: \exists \delta > 0 \forall \varepsilon: 0 < \varepsilon < \delta A + \varepsilon B \in V_{-}\}\text{ is an angle formed by 2 hyperplanes}\}$ is the set of matrices similar to $\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$. By construction $f(W) = W$. Hence, $f\left(\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)\right)$ and $f\left(\left(\begin{smallmatrix}0&0\\-1&0\end{smallmatrix}\right)\right)$ are both similar to $\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$. By using this and the fact that their linear combinations with positive coefficients lie on $\partial V_{-}$, one can show that $f = B \mapsto \alpha A^{-1} \tilde f(B) A$ with $\tilde f\left(\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)\right) = \left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ and $\tilde f\left(\left(\begin{smallmatrix}0&0\\-1&0\end{smallmatrix}\right)\right) = \left(\begin{smallmatrix}0&0\\-1&0\end{smallmatrix}\right)$. From here it is relatively easy to show that $\tilde f$ is either identity map, or $B \mapsto X^{-1} B^T X$, where $X = X^{-1} = \left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$.
In order to try to follow the same idea for $n > 2$, one would need to prove that $f$ preserves the "type" of JNF of a restriction of the matrix to its eigenspaces corresponding to $\lambda\colon \text{real}(\lambda) = 0$, perhaps by arguing that $V_{-}$ looks "differently" in the neighbourhood of different points on $\partial{V_{-}}$.
