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.


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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.