sufficient condition for checking whether a point $(-{\lambda}, \mu) \in
    {{\mathcal{Z}}}_{{\mathbb{R}}}$. To show a point $(-{\lambda}, \mu) \not\in {{\mathcal{Z}}}_{{\mathbb{R}}}$ it suffices to show that $(0,0) \not\in A$. This can be done utilizing information about the convex hull of $A$ and using the techniques described in [@HS]. In particular, let $K = K_0$ be the convex hull of $A$ and let $K_n = T_p(K_{n-1}) \cup T_m(K_{n-1})$. It is easy to see that $A \subset K_n$ for all $n$. Hence if there exists an $n$ such that $(0,0) \not\in K_n$ then $(0,0) \not\in A$. A precise description of $K$ is given in Section \[sec:unique\]. See Figure \[fig:Z2\] for illustration.

![Points in ${{\mathcal{Z}}}_{\mathbb R}$ (red) and points not in ${{\mathcal{Z}}}_{\mathbb R}$ (black)[]{data-label="fig:Z2"}](Z20.eps){width="275pt" height="275pt"}

The Jordan block case
---------------------

Consider the polynomial ($n=8, N=2$) $$P(x) = x^8 - \frac{8}{7 \nu} x^7 + \frac{1}{7 \nu^8}.$$ A quick check shows that $P(1/\nu) = P'(1/\nu) = 0$. Furthermore, for all $\nu \geq 0.831458513$ then we have $$\left| \frac{8}{7 \nu} \right| + \left|\frac{1}{7 \nu^8}\right| \leq 2.$$

In this case, the matrix from the Proof is $$B=\begin{pmatrix}
0 & \frac{1}{7 \nu^8} & \frac{2}{7 \nu^7} & \frac{3}{7 \nu^6} & \frac{4}{7 \nu^5} &
    \frac{5}{7 \nu^4} & \frac{6}{7 \nu^3} & \frac{7}{7 \nu^2} 