$K = \{A \in \Bbb M_n (\Bbb R)\ |\ A=A^T, \mathrm {tr} (A) = 1, x^TAx \ge 0\ \text {for all}\ x \in \Bbb R^n \}$ is compact or not?

Question

Let $K \subseteq \Bbb M_n (\Bbb R)$ be such that $$K = \{A \in \Bbb M_n (\Bbb R)\ |\ A=A^T, \mathrm {tr} (A) = 1, x^TAx \ge 0\ \text {for all}\ x \in \Bbb R^n \}.$$ Is $K$ compact in $\Bbb M_n (\Bbb R)$?

Please help me in this regard.

Thank you very much.

Answer 1

If you are familiar with spectral radius you can argue as follows: the given set is obviously closed. To show that it is bounded note that eigen values of any matrix $A$ in this set are between $0$ and $1$ so the spectral radius is $\leq 1$. For a positive definite matrix, the spectral radius is same as the norm. So we have $\|A\| \leq 1$ for all $A$ in this set which makes the set bounded. By Heine - Borel Theorem it is compact.

Answer 2

As a physicist i'm not a specialist on this topic, but I figured that $K$ is isomorph to $\partial\Delta_n\times SO(n)$, where $\Delta_n$ is the $n$-simplex. Both of these factors are compact, and so, I guess, is $K$. In case this argument is wrong, please let me know.