For a $2 \times 2$ real matrix, let us concern an affine parametrization over the entries. In other words, we want the entries $x_1, x_2, x_3, x_4$ \begin{align*} \pmatrix{ x_1 & x_2 \\ x_3 & x_4} \end{align*} to be in affine spaces. Furthermore, they do not need to be in the same affine spaces and we allow to specify some entries to be constants. For example, we may say our parametrized family to be \begin{align*} \pmatrix{ 0 & 1-t \\ t & 0} \end{align*} As such, the off-diagonal entries are parametrized by an affine line. Or we may say \begin{align*} \pmatrix{ 1 & 1-t_1 \\ t_2 & 0} \end{align*} In this case, the diagonal entries are constant and off-diagonal entries are parametrized by $2$ parameters $t_1, t_2$.

Let us denote the set $\mathcal E = \{A \in M_2(\mathbb R): \max_i \text{Re}(\lambda_i(A)) < 0\}$. Now let us choose a parametrization $\star$. More concretely, let us assume $\star$ is describe by $j$ parameters $t_1, \dots, t_j$ with $1 \le j \le 4$. Define the set $\mathcal F = \{(t_1, \dots, t_j): \star(t_1, \dots, t_j) \in \mathcal E\}$. I am wondering what kind of parametrization could give us the most number of connected components of $\mathcal F$.

For example, if $\star$ is parametrized by $4$ parameters, by translation and rescaling, $\mathcal F$ are all the real matrices in $\mathcal E$. Clearly this is a connected set.

On the other hand, if we consider following parametrized family \begin{align*} \pmatrix{-1 & t \\ -t & 0} \end{align*} Then there are two connected components $(-\infty, 0)$ and $(0, \infty)$.

We may also consider \begin{align*} \pmatrix{-1 & t_1 \\ t_2 & 0} \end{align*} Then the set $\mathcal F$ has $2$ connected components \begin{align*} (-\infty, 0) \times (0, \infty) \\ (0, \infty) \times (-\infty, 0) \\. \end{align*} These observations are direct consequence of Routh-Hurwitz criterion.

My question is: can we have more with some other parametrization?


No, two is the maximum.

By Routh-Hurwitz, $\mathcal E$ is the set of matrices with negative trace and positive determinant. So $\mathcal F$ is the intersection of the sets $\{x\mid T(x)<0\}$ and $\{x\mid D(x)>0\}$ where $T=\mathrm{tr}\circ \star$ and $D=\mathrm{det}\circ\star.$ All I will use is that $T$ is affine and $D$ is quadratic.

For $j=1$ we can add a dummy variable to the parameterization to reduce to the case $j=2.$ For $j>2$ consider $p_1,p_2,p_3\in \mathcal F.$ Define $\phi(s,t)=p_1+s(p_2-p_1)+t(p_3-p_1)\in\mathbb R^j.$ By the $j=2$ case proved below, the set $\phi^{-1}(\mathcal F)=(\star\circ \phi)^{-1}(\mathcal E)$ has at most two path components, i.e. there is a path $\gamma$ between two of the points $(0,0),(1,0),(0,1).$ The image $\phi\circ \gamma$ connects two of $p_1,p_2,p_3.$ This shows that $\mathcal F$ has at most two path components.

So assume $j=2.$ By checking the classification of plane conics, the set $\{x\mid D(x)>0\}$ is either the union of at most two convex sets, or is path-connected. In the former case, intersecting with $\{x\mid T(x)<0\}$ still leaves at most two convex sets.

So assume $\{x\mid D(x)>0\}$ is path-connected and suppose for contradiction that $\mathcal F$ has at least three path-components. Pick points $x_1,x_2,x_3$ in distinct path components. For each $i\neq j$ there is a path from $x_i$ to $x_j$ in $\{x\mid D(x)>0\},$ which must pass through $\{x\mid T(x)=0\}$ at some point. This proves that for each $i$ there is a path from $x_i$ to some point in the set $\{x\mid T(x)=0, D(x)>0\}.$ But the set $\{x\mid T(x)=0, D(x)>0\}$ has at most two path components (it's the region on a line where a certain quadratic is positive), so we can travel along it and compose paths to get a path $\gamma:[0,1]\to \{x\mid T(x)\geq 0, D(x)>0\}$ with $\gamma(0)=x_i$ and $\gamma(1)=x_j,$ for some $i\neq j.$

We can modify this path to lie in $\mathcal F$ as follows. Write $T$ in the form $T(x)=c-x\cdot n$ with $n\cdot n=1,$ so $T(x)<0$ if and only if $x\cdot n>c.$ Then for small enough $\epsilon>0$ the path $\gamma'(t)=\gamma(t)+(\max(c+\epsilon-\gamma(t)\cdot n,0))n$ lies in $\mathcal F$ and still has the same endpoints.

  • $\begingroup$ Thanks for answering my question. Could you elaborate your third paragraph? "It suffices to consider the case $j=2$...". I cannot understand how we change other paramerterization to two parameters. Thanks. $\endgroup$ – user1101010 Feb 18 at 19:51
  • $\begingroup$ @user1101010: I've added more detail for the $j>2$ case, in case that was the part that was unclear $\endgroup$ – Dap Feb 20 at 14:23

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