The boundary of the positive definite cone

Idea. The boundary is "closure minus interior". We already know both for PD.

Step 1: closure and interior. In $\mathrm{Sym}_n$ the set PD is open (so $\operatorname{int}\mathrm{PD}=\mathrm{PD}$) and $\overline{\mathrm{PD}}=\mathrm{PSD}$ (problem on cones above). Hence
$$\partial\,\mathrm{PD}=\overline{\mathrm{PD}}\setminus\operatorname{int}\mathrm{PD}=\mathrm{PSD}\setminus\mathrm{PD}.$$

Step 2: identify $\mathrm{PSD}\setminus\mathrm{PD}$. A symmetric matrix has real eigenvalues $\lambda_1\le\dots\le\lambda_n$ and
- $A\in\mathrm{PSD}\iff\lambda_1\ge0$,
- $A\in\mathrm{PD}\iff\lambda_1>0$.
So $A\in\mathrm{PSD}\setminus\mathrm{PD}$ iff $\lambda_1=0$. For a PSD matrix all eigenvalues are $\ge0$ and $\det A=\prod\lambda_i$, so $\det A=0$ iff some $\lambda_i=0$ iff $\lambda_1=0$. Thus
$$\mathrm{PSD}\setminus\mathrm{PD}=\{A\ge0:\det A=0\}.$$

Example ($n=2$). The boundary of PD is the set of $\begin{pmatrix}a&b\\b&c\end{pmatrix}$ with $a,c\ge0$, $ac=b^2$: one nappe of a quadratic cone in the three-dimensional space $\mathrm{Sym}_2\cong\mathbb{R}^3$: the boundary of the Lorentz cone $\{a+c\ge\sqrt{(a-c)^2+4b^2}\}$.

Remark.
- Note the condition $A\ge0$ cannot be dropped: $\operatorname{diag}(1,0,-1)$ has $\det=0$ but is *not* a boundary point of PD, since it is not even in $\mathrm{PSD}$.
- The boundary consists of PSD matrices of rank $<n$: the faces of the cone PSD, parametrised by their ranges.