Spectral radius balls and star-shaped sets

Idea. Scaling by $t\in[0,1]$ shrinks both $\rho$ and $\|\cdot\|$, so the sets are star-shaped about $0$.

1. The eigenvalues of $tA$ are $t\lambda$ for the eigenvalues $\lambda$ of $A$, so $\rho(tA)=t\rho(A)$ for $t\ge0$. Also $\|tA\|=t\|A\|$. If $\rho(A)<1$ then $\rho(tA)=t\rho(A)<1$ for every $t\in[0,1]$, so the segment from $A$ to $0$ stays in $\{\rho<1\}$. Same for $\{\|A\|<1\}$. Every point is joined to $0$, so both sets are path connected.

2. Suppose $\rho(A)>1$. Then $\rho(tA)=t\rho(A)$ is $<1$ as soon as $t<1/\rho(A)$. In particular $0\in$ the segment $[0,A]$ is *not* in $\{\rho>1\}$ (indeed $\rho(0)=0$), and neither are the points $tA$ with small $t>0$. So the segment from $0$ to $A$ leaves the set $\{\rho>1\}$: it is not star-shaped about $0$ (and $0$ doesn't even belong to it).

Remark. For $n=1$, $\{|a|>1\}=(-\infty,-1)\cup(1,\infty)$ is not even connected. Also note the inclusion $\{\|A\|<1\}\subset\{\rho(A)<1\}$ (since $\rho(A)\le\|A\|$), strict as soon as $n\ge2$ (for example $\begin{pmatrix}0&2\\0&0\end{pmatrix}$ is nilpotent, so $\rho=0$, but has norm $2$).