# Convex hull of the set of stable matrices?

Take a matrix $A \in \mathbb{R}^{n \times n}$, and define the spectral radius $\rho(A) = |\lambda_{\text{max}}(A)|$. $A$ is stable if $\rho(A) < 1$, or since it is relevant for my particular application, I'll call it $\gamma$-stable if for $\gamma \in [0, 1)$, $\rho(A) \le \gamma$.

The set of all $\gamma$-stable matrices $S = \{A\ |\ \rho(A) \le \gamma\}$, is not convex (unless $A$ is symmetric). I am interested in the convex hull of $S$, and have conjectured that $\textbf{conv}\ S = \{A\ |\ ||A|| \le \gamma\} \overset{\Delta}{=} C$. Since $\rho(A) \le ||A||$, and $C$ is convex, it is certainly true that $\textbf{conv}\ S \subseteq C$, but I'm unsure of the other direction.

EDIT: I confused the inequality, $C \subseteq S$, not the other way around. The question still stands, is there a known characterization of $\mathbf{conv}\ S$?

Is there a known characterization of $\textbf{conv}\ S$? I have not been able to find one myself, and surprisingly, my searches are coming up empty handed.

• it is certainly true that $\textbf{conv}\ S \subseteq C$: hmmm... – Omnomnomnom Jul 8 '17 at 17:42
• I suspect that the convex hull of the nilpotent matrices alone yields all trace-zero matrices – Omnomnomnom Jul 8 '17 at 17:44
• I would conjecture that $\operatorname{co} S = \{ A | \, |\operatorname{tr} A | \le n \gamma \}$. – copper.hat Jul 9 '17 at 21:30
• @copper.hat It does not seem possible to write $\begin{bmatrix}2 & 0\\0 & 0\end{bmatrix}$ as a convex combination of $1$-stable matrices. Such matrices will necessarily have trace $2$, both eigenvalues $1$ and, hence, be triangular with identities on the main diagonal. – A.Γ. Jul 11 '17 at 20:44
• @A.Γ.: The matrix you wrote is in the (real) closure of $\operatorname{co} S$, see the aside in the answer below. – copper.hat Jul 11 '17 at 23:59

The matrix $T_n = \begin{bmatrix} \gamma & n \\ 0 & \gamma \end{bmatrix}$ is $\gamma$-stable for all $n$ but $\|T_n\| \ge n$ for all $n$.
Hence $S \not\subset C$.
If $M(t) = (1-2t)\begin{bmatrix} 2 &1 \\ -1 & 0 \end{bmatrix}+ t\begin{bmatrix} 0 &-{1 \over t} \\ 0 & 0 \end{bmatrix} + t\begin{bmatrix} 0 &0 \\ {1 \over t} & 0 \end{bmatrix} = \begin{bmatrix} (1-2t)2 &-2t \\ 2t & 0 \end{bmatrix}$, then $M(t) \in \operatorname{co} S$ for all $t \in (0,1)$, hence $\begin{bmatrix} 2 &0 \\ 0 & 0 \end{bmatrix} \in \overline{\operatorname{co}} S$.