How to prove the set of all symmetric matrices with eigenvalues in $(0,2)$ is connected?

Question

Prove that space $X$ of all symmetric matrices in $GL_2(\mathbb R)$ with both the eigenvalues belonging to the interval $(0,2),$ with the topology inherited from $M_2(\mathbb R) $ is connected.

Space of all symmetric matrices in $M_2(\mathbb R)$ is path -connected.

I was not able to show why $det(\lambda A+(1-\lambda)B)\neq 0$ where $\lambda \in (0,1)$ and $A,B\in GL_2(\mathbb R), A=A^T,B=B^T.$

Also how to use the eigenvalues from $(0,2)$ to prove the connectedness.

I hope my doubts are clear to you.

Any help is appreciated. Thank you.

Answer 1

The space $SO_2(\mathbb{R})$ is connected. Now let$$\Lambda=\left\{\begin{bmatrix}\lambda_1&0\\0&\lambda_2\end{bmatrix}\,\middle|\,\lambda_1,\lambda_2\in(0,2)\right\}.$$The set $\Lambda$ is connected too. So, the range of the map$$\begin{array}{ccc}SO_2(\mathbb{R})\times\Lambda&\longrightarrow&GL_2(\mathbb{R})\\(P,D)&\mapsto&P^{-1}DP\end{array}$$is connected too. But this range is $X$.

Answer 2

We can prove much more (in $S_n(\mathbb{R})$ , the set of $n\times n$ real symmetric matrices).

$\textbf{Proposition 1}$. The set $Z_I=\{A\in S_n(\mathbb{R});spectrum(A)\subset I\}$ is convex (where $I$ is some interval of $\mathbb{R}$).

$\textbf{Proof}$. We use the fact that if $A\in S_n(\mathbb{R})$, then $\sup_{||x||_2=1}x^TAx=\sup(spectrum(A)),\inf_{||x||_2=1}x^TAx=\inf(spectrum(A))$.

Then $\lambda x^TAx+(1-\lambda)x^TBx\in I$ when $A,B\in Z_I,||x||_2=1,\lambda\in [0,1]$. $\square$

$\textbf{Proposition 2}$. If $I=[a,b],a<b$, then $Z_I$ is homeomorphic to the compact ball $B_{n(n+1)/2}$.

$\textbf{Proof}.$ $Z_I$ is also closed (the eigenvalues are continuous functions of the entries) and bounded ($||A||_2=\sup_{\lambda\in spectrum(A)}(|\lambda|))$.

Finally, $Z_I$ is convex compact and, therefore, homeomorphic to the compact ball in $\mathbb{R}^k$, where $k$ is the number of degrees of freedom of $S_n(\mathbb{R})$.

Answer 3

One-line proof: every $A\in X$ is path-connected to $I$ by the line segment $\{tA+(1-t)I: 0\leq t\leq 1\}\subseteq X$.

Your original approach also works. Let $A,B\in X$ and $0\leq t\leq 1$. Then $A,B$ and in turn $tA+(1-t)B$ are positive definite and hence all eigenvalues of $tA+(1-t)B$ are positive. Also, $$ \|tA+(1-t)B\|_2 \leq t\|A\|_2+(1-t)\|B\|_2 < 2t+2(1-t)=2. $$ Therefore all eigenvalues of $tA+(1-t)B$ lie inside the interval $(0,2)$. Consequently, $X$ is path-connected because $\{tA+(1-t)B: 0\leq t\leq 1\}\subseteq X$ for any $A,B\in X$.