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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.

Rodrigo de Azevedo
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user1101010
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1 Answers1

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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 $E_1^TC E_1\in\mathcal H'$.

polfosol
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