Extreme points of a set (space) of PSD matrices

Question

Let us consider the set of all $n$-by-$n$ PSD matrices (i.e. those $A$ satisfying $x^*Ax \geq 0$ $\forall x\in\mathbb{C}^n$) with equal diagonal elements ($A_{ii}=d$, $\forall i=1,2,\cdots,n$), which, to my understanding forms a vector space over $\mathbb{R}$ of dimension $n^2-n+1$ (or convex subset of $\mathbb{R}^{n^2-n+1}$). I was wondering about the extreme points of this set. I am looking actually for a possible decomposition of any such non-zero $A$ in terms of these extreme points.

Answer 1

If you meant that $d$ is not fixed, then the only extreme point is the zero matrix. The set is

$$C=\{X \in \mathbb{C}^{n\times n}: C\geq 0, \; \exists\; d\in \mathbb{C}: \;C_{ii}=d, i=1,\ldots,n. \}$$ This is a convex cone. Since it is contained in the cone of PSD matrices, it is also pointed, i.e, $C\cap (-C)={0}.$ The only extreme point of a convex pointed cone is $0.$ To see this, assume $X\neq 0.$ then

$$X= \frac{\frac{1}{2}X +\frac{3}{2}X}{2}.$$ Since $\frac{1}{2}X$ and $\frac{3}{2}X$ belong to $C,$ this means that $X$ it is not an extreme point.