Let $\mathcal E = \{A \in M_n(\mathbb R): \rho(A) < 1\}$ where $\rho(\cdot)$ is the spectral radius and $\mathcal U$ be an affine space in $M_n(\mathbb R)$. If we assume $\mathcal E \cap \mathcal U \neq \emptyset$, how many connected components could the intersection have?

In proving $\mathcal E$ is connected, I know we can use a path $(1-t)A + t 0$ but if $B$ is in the intersection, $(1-t)B$ could not be guaranteed in the intersection.

  • $\begingroup$ Intersections of convex sets are convex $\endgroup$ – Bananach Nov 15 '18 at 21:23
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    $\begingroup$ $\mathcal E$ is not convex. $\endgroup$ – user1101010 Nov 15 '18 at 21:29
  • $\begingroup$ Just as a curious observer could you provide an example of two matrices whose path leaves $\mathcal{E}$? $\endgroup$ – user25959 Nov 15 '18 at 21:32
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    $\begingroup$ @user25959 You can take $\pmatrix{a & 10\\ 0 & a}$ and $\pmatrix{b & 0 \\10 & b}$. Make suitable choices of $a, b$. $\endgroup$ – user1101010 Nov 15 '18 at 22:05
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    $\begingroup$ @user25959 he is right, and $a=b=0$ already suffice to show it. I got confused with the spectral radius vs. operator norm $\endgroup$ – Bananach Nov 16 '18 at 6:16

This is an elaboration on a comment, it provides only a lower bound on the amount of components you can get and I only look at a one-dimensional affine subspace (also known as a line).

The first step is to look at the matrix $$A(t;a):=\begin{pmatrix}0&a\,t\\a\,(1-t)& 0\end{pmatrix},$$ which has spectral radius $|a|\sqrt{|t-t^2|}$. The values of $t$ for which the spectral radius is $≤1$ are $$\mathcal D(a):=[B_1(a), B_2(a)]\,\cup\,[B_3(a),B_4(a)].$$ (To be concrete, $B_1(a)=\frac{a-\sqrt{a^2+4}}{2a}$, $B_2(a)= \frac{a-\sqrt{a^2-4}}{2a}$, $B_3(a) = \frac{a+\sqrt{a^2-4}}{2a}$, $B_4(a)=\frac{a+\sqrt{a^2+4}}{2a}$, but these values don't matter.) What matters is that you have two intervals where the gap between them becomes arbitrarily small as $a\to2$. Further the length of the intervals does not grow unboundedly as $a\to2$.

Now look at the matrix $$\begin{pmatrix} A(c_1 t-d_1; a_1) & 0 \\ 0 & A(c_2 t-d_2; a_2)\end{pmatrix}.$$ The values for $t$ where this matrix has spectral radius $≤1$ is $$(c_1\cdot\mathcal D(a_1)+d_1)\cap (c_2\cdot \mathcal D(a_2)+d_2).$$

Manipulating $a_1,c_1,d_1$ you have basically arbitrary freedom to tune the following things on $c_1\cdot\mathcal D(a_1)+d_1$:

  1. The gap between the two intervals can be made arbitrarily small (tune parameter $a$).
  2. The gap can be shifted to wherever you want (tune parameter $d$).
  3. The length of the intervals can be made as large as you like.

Note that adjusting the length of the intervals alters the length of the gap and the position of the gap. Clearly you can correct the position again, but it is also important to note that you can correct the length of the gap (by making it smaller), since this procedure will not shorten the intervals beyond a certain factor.

All this was just to give a sound background to the pictorial story I'm going to tell.

Here is an abstraction of what you can let the two domains from before look like: two domains

The intersection of these two looks like:

intersection of two domains

You see $3$ components. Now if you are looking at $M_{2n}(\Bbb R)$ you can take a look at the line $$\begin{pmatrix} A(c_1 t-d_1;a_1) &... &0\\ 0 &\ddots &\vdots \\ 0& ... &A(c_nt -d_n;a_n) \end{pmatrix}$$ and the intersection of this line with $\mathcal E$ will by the intersection of $n$ domains with the given parameters. An example of what you can tune for $M_8(\Bbb R)$ (ie $4$ domains) is the following picture:

four domains and their intersection

This story is a lower bound of $\lfloor n/2\rfloor$ on the amount of components achievable with a line.


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