Is there a homeomorphism between the sets of Schur stable and Hurwitz stable matrices in companion forms? The set of Schur stable matrices is
\begin{align*}
\mathcal S = \{A \in M_n(\mathbb R): \rho(A) < 1\},
\end{align*}
where $\rho(\cdot)$ denotes the spectral radius of a matrix and the set of Hurwitz stable matrices is 
\begin{align*}
\mathcal H = \{A \in M_n(\mathbb R): \max_{i=1, \dots, n} \text{Re}(\lambda_i(A)) < 0\},
\end{align*}
i.e., matrices with eigenvalues lying on the open left half plane.
Let $\mathcal C$ denote all matrices in companion form. Let $\hat{\mathcal S} = \mathcal S \cap \mathcal C$ and $\hat{\mathcal H} = \mathcal H \cap \mathcal C$. Now I would like to determine whether there is a homeomorphism between the sets $\hat {\mathcal S}$ and $\hat{\mathcal H}$. Let us exclude the trivial case $n=1$.
If we consider the sets $\mathcal S$ and $\mathcal H$ only, there is a diffeomorphism $f: \mathcal S \to \mathcal H$ given by
\begin{align*}
A \mapsto (A-I)^{-1}(A+I).
\end{align*}
But apparently this doesn't work for $\hat {\mathcal S}$ and $\hat{\mathcal H}$ since the inversion and multiplication will not necessarily yield a matrix in companion form. 

 A: You are halfway there. Let $\mathcal H'$ and $\mathcal S'$ be subsets of $\mathcal H$ and $\mathcal S$ that only contain Jordan canonical matrices in the form of:
$$M=\text{diag}\left(\lambda_1,\lambda_2,\dots,\lambda_n\right)$$
where $|\lambda_i|\ge|\lambda_{i+1}|\;$ ($i=1,\dots,n-1$),
and


*

*if $|\lambda_i|=|\lambda_{i+1}|$ then $\Re\lambda_i\ge\Re\lambda_{i+1}$

*if $|\lambda_i|=|\lambda_{i+1}|$ and $\Re\lambda_i=\Re\lambda_{i+1}$ then $\Im\lambda_i\ge\Im\lambda_{i+1}$
Since every matrix in $\hat{\mathcal H}$ is neither upper/lower-triangular, then there is a unique matrix of this form for each $A\in\hat{\mathcal H}$. Because for every two distinct matrices in $\hat{\mathcal H}$, their characteristic polynomials are distinct, so that their sets of eigenvalues are not equal. Thus there exists a unique bijection that maps every matrix in $\hat{\mathcal H}$ to a matrix in $\mathcal H'$, and vice versa.
In a same manner, there exists a bijection between $\mathcal S'$ and $\hat{\mathcal S}$.
Now we show that there is also a bijection between $\mathcal H'$ and $\mathcal S'$. Let $f$ be a complex function defined as:
$$f(z)=\frac{z+1}{z-1}$$
which is a Möbius transformation that maps the left-half of the complex plane onto the unit disk, i.e. it's a bijection between these two subspaces.
Now let $g:\mathcal H'\to\mathcal S''$ be given by
$$A\mapsto(A-I)^{-1}(A+I)$$
since $A$ is diagonal, then every diagonal element, $\lambda_i$ of $A$ is mapped to $f(\lambda_i)$ and the resulting matrix is also diagonal. Note that $\mathcal S''$ is not necessarily equal to $\mathcal S'$ as the ordering might change. But the ordering of diagonals in $B\in\mathcal S''$ can be fixed by a unitary matrix $E$ (whose columns are a permutation of $n$-dimensional unit vectors), so that $E^TBE\in\mathcal S'$.
In other words, we can define a bijection $p:\mathcal H'\to\mathcal S'$ as
$$A\mapsto E^T(A-I)^{-1}(A+I)E$$
whose reverse mapping is:
$$B\mapsto E(B+I)(B-I)^{-1}E^T$$
To clarify more, if $B\in\hat{\mathcal S}$ then $C=(B+I)(B-I)^{-1}\in\mathcal H''$ and we can use another unitary matrix $E_1$ so that $E_1^TC E_1\in\mathcal H'$.
