Hurwitz (stable) matrices form a connected set

Idea. The set is star-shaped about $-I$.

Step 1: eigenvalues of the deformation. For $t\in[0,1]$ let $A_t=(1-t)A-tI$. If $Av=\lambda v$ with $v\in\mathbb{C}^n\setminus\{0\}$, then
$$A_tv=\big((1-t)\lambda-t\big)v,$$
so $(1-t)\lambda-t$ is an eigenvalue of $A_t$. Conversely every eigenvalue of $A_t$ has this form: for $t<1$, $A=\frac1{1-t}(A_t+tI)$ so the eigenvalues of $A$ are $\frac{\mu+t}{1-t}$ for $\mu\in\operatorname{Spec}(A_t)$, i.e. $\mu=(1-t)\lambda-t$. For $t=1$, $A_1=-I$ with unique eigenvalue $-1$.

Step 2: stability is preserved. For $t\in[0,1)$,
$$\operatorname{Re}\big((1-t)\lambda-t\big)=(1-t)\operatorname{Re}\lambda-t<0$$
since $1-t>0$, $\operatorname{Re}\lambda<0$ and $t\ge0$. For $t=1$ the eigenvalue is $-1<0$. So $A_t$ is Hurwitz for all $t\in[0,1]$.

Step 3: conclusion. $t\mapsto A_t$ is a continuous (affine) path in the set of Hurwitz matrices from $A$ to $-I$. Every Hurwitz matrix is joined to $-I$, so the set is path connected.

Remark. The Hurwitz set is open (eigenvalues depend continuously on the matrix) and *convex?* No: it is star-shaped about $-I$ but not convex in general for $n\ge2$. A convex combination of two Hurwitz matrices need not be Hurwitz, a well-known fact in control theory.