Is the set $\{\zeta: \rho(A(\zeta))< 1\}$ connected for matrices under parameterization of first $m$ rows? $\newcommand{\eqqcolon}{=\mathrel{\vcenter{:}}}$
Fix $n, m \in \mathbb N$ with $n > m$. Let $\zeta \in \mathcal{M}(m \times n; \mathbb C)$ and we fix a $\zeta_0 \in \mathcal M( (n-m) \times n; \mathbb C)$ with rank $n-m$. Let us define a parameterization by
\begin{align*}
F : \zeta \mapsto \begin{pmatrix}
\zeta \\
\zeta_0 
\end{pmatrix} \eqqcolon A(\zeta).
\end{align*}
Let $S = \{ \zeta: \rho( A(\zeta) )<1\}$ where $\rho(\cdot)$ denotes the spectral radius and assume $S \neq \emptyset$. I am interested to know whether $S$ is connected?
 A: As a test case, for $n=2, m=1$ one can see it is connected. 
Consider a matrix $A=\left[\begin{array}{cc}a & b \\c & d\end{array}\right]$, and assume that $c, d$ are given so $\zeta_0=[c,d]$. 
Case 1: $c=0$, and the spectral radius is $\max \{ |a|,|d|\}$, so clearly the set with $\rho(A) <1$ is connected.
Case 2: $c\neq 0$. Supppose we have matrices $A_0 = \left[\begin{array}{cc}a_0 & b_0 \\c & d\end{array}\right], A_1=\left[\begin{array}{cc}a_1 & b_1 \\c & d\end{array}\right]$ with $\rho(A_i)<1$. Then $tr(A)=a+d$, so $a=tr(A)-d$, and $det(A)=ad-bc$, so $b=((tr(A)-d)d-det(A))/c$. Thus each matrix in $im(F)$ is determined by its trace and determinant, and hence by its (generalized) eigenvalues. So if the eigenvalues of $A_i$ are $\lambda_i, \mu_i$, so $\rho(A_i)=\max\{|\lambda_i|,|\mu_i|\}<1$, we may connect the pair by a path $(\lambda_t,\mu_t)$ with $\max\{|\lambda_t|,|\mu_t|\}<1$, and hence get a unique path of matrices $A_t \in im(F)$ with $tr(A_t)=\lambda_t+\mu_t, det(A_t)= \lambda_t\mu_t$ according to the above formula with $\rho(A_t)<1, 0\leq t\leq 1$. Hence $S$ is connected. 
More generally, the coefficients of the characteristic polynomial of $A(\zeta)$ is a linear function of the first row of $\zeta$. If this linear function is invertible, then one may first interpolate eigenvalues keeping the maximum absolute value $<1$, then interpolate the first rows of $\zeta$ to match the eigenvalues. Since invertible linear functions are generic, I would expect this strategy to work in general, but some case-by-case analysis will still be needed. 
