Suppose $A \in M_{n \times m}(\mathbb R)$ and $B \in M_n(\mathbb R)$ are fixed with $m < n$. Denote two sets $\mathcal E, \mathcal F$ by \begin{align*} \mathcal E &= \{X \in M_{m \times n}(\mathbb R): \max_i \text{Re} \lambda_i(B+AX) < 0 \}, \\ \mathcal F &= \{X \in M_{m \times n}(\mathbb R): \rho(B+AX) < 1 \}, \end{align*} where $\lambda_i$ denotes the eigenvalues and $\rho(\cdot)$ denotes spectral radius of a square matrix. In other words, $\mathcal E$ contains matrix $X$ such that $B+AX$ has eigenvalues on the left open half plane of $\mathbb C$ and $\mathcal F$ contains matrix $X$ such that $B+AX$ has spectral radius bounded by $1$. Let us further assume $\mathcal E$ and $\mathcal F$ are both nonempty. My question is: could we determine whether they are homeomorphic or not?
I know the transformation $A \mapsto (A-I)^{-1}(A+I)$ is a diffeomorphism between the square matrices with spectral radius bounded by $1$ and those whose eigenvalues on the left half plane of $\mathbb C$. But it seems to me this map does not necessarily work for $\mathcal E$ and $\mathcal F$. If $X \in \mathcal F$, then $(B+AX-I)^{-1} (B+AX+I) = - \sum_{j=0}^{\infty} (B+AX)^j(I + B+AX) := C$. I am not sure $C-B$ can be represented by $AY$ for some $Y$. Or there are some other maps. An example to show the non-existence would also be nice.
Before voting this answer down, give me a chance to improve the answer or to delete it if it is complete nonsense.
EDIT: Example is now working as far as I see.
This being said, let's consider the simplest example. Take $m=1$, $n=2$, $B=\begin{pmatrix}\frac{1}{2} &0\\ 0& -\frac{1}{2} \end{pmatrix}$, $A=\begin{pmatrix}1 \\ 0\end{pmatrix}$. Take an arbitrary $X=\begin{pmatrix}x & y\end{pmatrix}$. We have $AX=\begin{pmatrix}x & y\\ 0 & 0\end{pmatrix}$. Now the Eigenvalues of $B+AX$ are the roots of the characteristic polynomial, i.e. $$p(\lambda)=\det\begin{pmatrix}\frac{1}{2}+x-\lambda & y \\ 0 & -\frac{1}{2}-\lambda\end{pmatrix}=(\frac{1}{2}+x-\lambda)(-\frac{1}{2}-\lambda)$$ with roots $\frac{1}{2}+x$ and $-\frac{1}{2}$.
The maximal real part of all eigenvalues is $\max\{Re(\frac{1}{2}+x),-\frac{1}{2}\}=\max\{\frac{1}{2}+x,-\frac{1}{2}\}$, since we assumed the matrices to be real. As you can see the answer does not depend on the choice of $y$. We may write the first set $$\mathcal E=\{(x,y)\in \mathbb R^2\mid \max\{\frac{1}{2}+x,-\frac{1}{2}\}<0\}.$$ If $x\leq -1$, the maximum always is $-\frac{1}{2}$ and strictly smaller than $0$. If $-1<x<-\frac{1}{2}$, the maximum is $x+\frac{1}{2}<0$. Thus, $$\mathcal E=\{(x,y)\in \mathbb R^2\mid x<-1\}=(-\infty,-\frac{1}{2})\times \mathbb R.$$
Now for the spectral radius. $\rho(B+AX)=\max\{|\frac{1}{2}+x|,|-\frac{1}{2}|\}=\max\{\sqrt{\frac{1}{4}+x^2},\frac{1}{2}\}$. The maximum is $\frac{1}{2}$ iff $\sqrt{\frac{1}{4}+x^2}\leq \frac{1}{2}$ which is the case iff $\frac{1}{4}+x^2\leq \frac{1}{4}$ i.e. for $x=0$.
$\sqrt{\frac{1}{4}+x^2}<1$ iff $\frac{1}{4}+x^2<1$ iff $x^2<\frac{3}{4}$. Since for $x\geq 0$ the maximum is $\frac{1}{2}$, $\sqrt{\frac{1}{4}+x^2}$ is the maximum and $<1$ precisely for $x<0$.
Together this yields $$\mathcal{F}=(-\infty, 0]\times \mathbb R$$
Now we would have to find a continuous map between an open interval and an interval that is closed on one side. This is not possible because the inverse image of the open interval would have to be open again by continuity.
Thus these sets are not homeomorphic in general.