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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.

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

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  • $\begingroup$ 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? $\endgroup$ – Frederik vom Ende Jan 17 at 7:37
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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.

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  • $\begingroup$ Do you try to say by Heine-Borel theorem instead of Bolzano-Weierstrass theorem? $\endgroup$ – math maniac. Jan 17 at 8:51
  • $\begingroup$ @mathmaniac. Yes. I have made the same mistake several times! Just absent-mindedness. $\endgroup$ – Kavi Rama Murthy Jan 17 at 8:53
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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.

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