Schur-stable 2×2 matrices: the Jury conditions and connectedness

Idea. For a real quadratic the location of the roots relative to the unit disc is read from $p(1)$, $p(-1)$ and $\delta$.

Note first that $|\tau|<1+\delta$ is equivalent to $p(1)=1-\tau+\delta>0$ and $p(-1)=1+\tau+\delta>0$, and that it forces $\delta>-1$ (as $1+\delta>|\tau|\ge0$).

($\Rightarrow$) Assume $\rho(A)<1$.
- *Non-real eigenvalues $\lambda,\bar\lambda$.* Then $\delta=|\lambda|^2<1$, $\delta>0$ and $\tau^2<4\delta$ (negative discriminant). Since $(1-\delta)^2\ge0$ we have $4\delta\le(1+\delta)^2$, hence $\tau^2<(1+\delta)^2$, i.e. $|\tau|<1+\delta$.
- *Real eigenvalues $r_1,r_2\in(-1,1)$.* Then $\delta=r_1r_2<1$ (as $|r_1r_2|<1$), $p(1)=(1-r_1)(1-r_2)>0$ and $p(-1)=(1+r_1)(1+r_2)>0$, which gives $|\tau|<1+\delta$.

($\Leftarrow$) Assume $|\tau|<1+\delta$ and $\delta<1$. Then $-1<\delta<1$.
- *Non-real roots.* They satisfy $|\lambda|^2=\delta<1$, so $\rho(A)<1$.
- *Real roots.* We have $p(1)>0$, $p(-1)>0$, and $|\tau|<1+\delta<2$, so the vertex $\tau/2$ of the parabola $p$ lies in $(-1,1)$. The parabola is decreasing on $(-\infty,\tau/2]$, so for $t\le-1$ we get $p(t)\ge p(-1)>0$: no root in $(-\infty,-1]$. Similarly $p$ is increasing on $[\tau/2,\infty)$, so $p(t)\ge p(1)>0$ for $t\ge1$: no root in $[1,\infty)$. Both real roots therefore lie in $(-1,1)$.

Connectedness. The spectral radius is positively homogeneous: $\rho(tA)=t\rho(A)$ for $t\ge0$ (the eigenvalues of $tA$ are $t\lambda_i$). So if $\rho(A)<1$ then $\rho(tA)<1$ for all $t\in[0,1]$, and $t\mapsto tA$ is a path in the set from $A$ to $0$. The set is star-shaped about $0$ and hence path connected.

Check. $A=\begin{pmatrix}0&1\\-0.5&0.9\end{pmatrix}$: $\tau=0.9$, $\delta=0.5$; $|\tau|=0.9<1.5$ and $\delta<1$. The eigenvalues are $0.45\pm0.545i$ with modulus $\sqrt{0.5}\approx0.707<1$. And $A=\begin{pmatrix}2&0\\0&0.1\end{pmatrix}$: $\tau=2.1$, $\delta=0.2$, $|\tau|>1.2$: not Schur stable.

Remark. In the $(\tau,\delta)$-plane the stable region is the open triangle with vertices $(-2,1)$, $(2,1)$ and $(0,-1)$, bounded by $\delta=1$ and the lines $\delta=\pm\tau-1$. These three lines correspond to a pair of complex eigenvalues on the unit circle, and to the eigenvalues $1$ and $-1$.

Lesson. "Roots of a real quadratic inside an interval" reduces to sign conditions on $p$ at the endpoints plus the position of the vertex.