Homework on matrix and convex set

Question

Suppose that $A,B\in\mathbb{R}^{n\times n}$ and both symmetric. Define $$ H=\{\sigma\in\mathbb{R} \mid A+\sigma B \text{ is semi-positive definite}\} $$ Assume that there exist $\sigma_1\in\mathbb{R}$ such that $A+\sigma_1B$ is positive definite, $a=\inf_{x\in H} x$, $b=\sup_{x\in H} x$. Prove that $a<b$, and for any $\sigma\in(a,b)$, $A+\sigma B$ is positive definite.

Answer 1

Note that the eigenvalues of a matrix depend continuously on its entries. So, if there exist a $\sigma_1$ such that the smallest eigenvalue $A + \sigma_1 B$ is positive, then there exists an $\delta > 0$ such that the lowest eigenvalue $A + \sigma B$ is positive for any real $\sigma$ with $|\sigma - \sigma_1| < \delta$.

This is enough to show that $a < b$.

We may then show that $H$ is a convex set by noting that $$ A + (s\sigma_1 + t\sigma_2)B = s(A + \sigma_1B) + t(A + s \sigma_2B) $$ for any $s,t \geq 0$ satisfying $s + t = 1$.

---

If we wanted to avoid continuous dependence of eigenvalues on entries, then it suffices to note that $$ \sigma \mapsto \min_{\|x\| = 1} x^*(A + \sigma B)x $$ is a continous function. This is a straightforward analytic proof.

---

Showing that if $\sigma \in (a,b)$, then $A + \sigma B$ is positive definite:

Suppose that there exists a $\sigma_0 \in (a,b)$ such that $A + \sigma_0 B$ is non-invertible. Then $A+aB,A+\sigma_0B,A+bB$ are three co-linear points that lie on the boundary of the set of positive semidefinite matrices.

Since $A+aB,A+\sigma_0B,A+bB$ are distinct co-linear points on the border of a convex set, we may deduce that all points on the line segment from $A + aB$ to $A + bB$ lie on the border. That is, $A + \sigma B$ is positive semidefinite for every $\sigma \in (a,b)$. This, however, contradicts the assumption that $A + \sigma_1 B $ is positive definite.