Is the set $\{A \in \mathcal{M}(n \times n; \mathbb C ): \rho(A) < 1\}$ open?

Question

Let $F = \{A \in \mathcal{M}(n \times n; \mathbb C ): \rho(A) < 1\}$, where $\rho$ denotes the largest eigenvalue modulus. I want to know whether $F$ is open or closed.

My idea is to consider $F_i = |\cdot| \circ \lambda_i : \mathcal{M}(n \times n) \to [0, \infty)$ where $|\cdot|$ denotes the modulus function and $\lambda_i$ denotes the eigenvalue function. Then \begin{align*} F = \bigcap_{j=1}^n F_i^{-1} \left( [0, 1) \right). \end{align*} Since $[0, 1)$ is open in the subspace $[0, \infty)$, so $F$ is open.

Answer 1

Consider $g(A) = \det(I - A)$. This is a continuous function from $\mathbb M_{n\times n}(\mathbb C)$ to $\mathbb C$. $\rho(A) < 1$ if and only if $g(z A) \ne 0$ for all $z$ in the closed unit disk $\overline{D} = \{z \in \mathbb C: |z| \le 1\}$.

Since $\overline{D}$ is compact, if $\rho(A) < 1$, $g(\overline{D} A)$ is a compact set that does not contain $0$; thus there is $\epsilon > 0$ such that $|g(z A)| > \epsilon$ for all $z \in \overline{D}$. Using uniform continuity of $g$ on the compact set $S = \{zB: \; |z| \le 1, \|B - A\| \le 1\}$, there is $\delta > 0$ such that $|g(z A) - g(z'B)| < \epsilon$ for all $z', B$ with $|z'-z| < \delta$ and $\|A-B\| < \delta$. In particular, if $|B-A| < \delta$, $|g(zB)| > |g(zA)| - \epsilon > 0$. This shows that your set is open.

Answer 2

The idea is reasonable. Here is an outline of the proof. The details will be left for you to fill in.

Consider $\|\cdot\|_\mathbb{M}$ any matrix norm, $\|\cdot\|_\mathbb{P}$ any polynomial norm and the topologies induced by these norms.

Start by showing that the function $p : (\mathbb{M}_{n \times n}(\mathbb{C}),\|\cdot\|_\mathbb{M}) \to (\mathbb{P}(\mathbb{C}), \|\cdot\|_\mathbb{P}) $ that associates each matrix $A$ to its characteristic polynomial $p(A)$ is continuous.

Next, show that $\rho : (\mathbb{P}(\mathbb{C}), \|\cdot\|_\mathbb{P}) \to (\left[0, \infty\right[, |\cdot|) : f \mapsto \max \{|r| : f(r) = 0\}$ is continuous using the fact that the roots of a complex polynomial with complex coefficients varies continuously as a function of its coefficients.

Then finally show that $F = p^{-1}( \rho ^{-1} ( \left[0, 1\right[ ))$.

Answer 3

For any fixed $A$, let $J=P^{-1}AP$ be the Jordan form of $A$. You may shrink the super-diagonal of $J$ at will via similarity transform. In fact, if $D:=\operatorname{diag}(1,t,t^2,\ldots,t^{n-1})$, the super-diagonal entries of $D^{-1}JD$ are $t$ times their counterparts in $J$. When $t\to0^+$, the super-diagonal of $D^{-1}JD$ approaches zero too.

So, if $\rho(A)<1$ and $t>0$ is small, we have $\|A\|_?<1$, where $\|B\|_?:=\|D^{-1}P^{-1}BPD\|_\infty$. Since all norms on a finite-dimensional vector space induce the same topology, $\mathcal O=\{B: \|B\|_?<1\}$ is open in the usual topology. Yet, our new norm is also submultiplicative. Therefore $\rho(B)\le\|B\|_?<1$ on $\mathcal O\ni A$ and our conclusion follows.