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

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 $$\Bbb M_n (\Bbb R)$$?

Thank you very much.

## closed as off-topic by GNUSupporter 8964民主女神 地下教會, Adrian Keister, callculus, Paul Frost, José Carlos SantosFeb 4 at 22:01

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• What you describe here is the set of all (real $n\times n$) density matrices which is in fact compact. What have you tried so far regarding this problem? – Frederik vom Ende Jan 17 at 7:37

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.
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.