This is a cross-post to the question I asked at MSE.

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.


1 Answer 1


Essentially you're asking if there is a homeomorphism between the monic real polynomials of degree $n$ with roots in the open unit disk and those with roots in the left half plane. Just take $$\prod_{j=1}^n (x - \alpha_j) \mapsto \prod_{j=1}^n \left(x - \frac{\alpha_j+1}{\alpha_j - 1}\right)$$ This should work because the multiset of roots of a polynomial is a continuous function of the coefficients.


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