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

I am dealing with convex optimization problems. There are some useful theories for optimization problems where real-valued vector parameter, e.g., $x \in \mathbb{R}^n$, is considered. I manage to apply the theories to my semidefinite programming problem. I mean, I verify it by simulations. However, to prove the convergence of the algorithm analytically, I need to verify whether the set of Hermitian positive semidefinite matrices, where $\bf{X}=\bf{X}^{\rm H}$ and $\bf{X} \succeq 0$, is compact given that ${\rm trace}\{\bf{X}\} \leq 1$. Can anyone help me with this or at least point out a direction I should go for? Thank you very much in advance.

Think of it as a subset of $\mathbb R^{n^2}$. You need to show that the set is closed and bounded. Closed is easy. For boundedness, show that each matrix entry is bounded by $1$. You know this is true for the diagonal entries (because they are all non-negative and add up to 1). Now use the fact that each $2\times 2$ submatrix $\left[\begin{matrix} a_{ii} & a_{ij} \\ a_{ij} & a_{jj} \end{matrix}\right]$ is positive semi-definite, and hence its determinant is non-negative.

Let me add one more thing. That the condition $X$ is positive semidefinite is closed under limits is perhaps most easily shown by noting that this is true if and only if all the diagonal minor matrices have non-negative determinant.

• 1) by a subset of $R^n2$, you mean vec{X} \in R^n2 or X can be written as real-valued representation such that X \in R^(n x n). – user111614 Nov 29 '13 at 16:16
• I am not sure what you mean by the latter statement. I definitely mean the former. – Stephen Montgomery-Smith Nov 29 '13 at 16:18