My proof for "$\Gamma =\{X\in \mathbf{R}^{n\times n} \mid X \succeq 0, \text{Tr}(X)=1\}$ is compact"

Question

This problem comes from:

How to prove the compactness of the set of Hermitian positive semidefinite matrices

In short, we want to prove

$$\Gamma =\{X\in \mathbf{R}^{n\times n} \mid X \succeq 0, \text{Tr}(X)=1\}$$ is compact.

Can I prove this in a terse way:

$\Gamma$ is the intersection of PSD cone, which is convex and closed, and the hyperplane $\{X \mid \text{Tr}(X)=1\}$, so $\Gamma$ is compact.

So it looks like the following graph:

If not, which part I should say more?

Answer 1

You need some kind of further argument to prove boundedness, in my opinion. I propose this one.

Since the matrices are (symmetric) positive-semidefinite, their eigenvalues are positive. Since the trace is $1$, all the eigenvalues are in $[0,1]$. Therefore, if we consider the operator norm on $\Bbb R^{n\times n}$ $$\lVert A\rVert_2=\sup_{v\ne 0}\frac{\lVert Av\rVert}{\lVert v\rVert}$$ your set is bounded with respect to it since $\lVert A\rVert_2\le\max\limits_{\lambda\in\operatorname{Spec A}}\lvert \lambda\rvert$.

Addition: Fact is that you somehow need to prove that your subspace does not cut the cone in a bad way (recall that not only ellypses are conic sections, but parabolas and hyperbolas too!)