# Extreme points of a convex set

Let $S$ denote the set of all complex non-negative definite matrices with all diagonal elements being less that or equal to one. Can we show that any matrix which belongs to the set of all non-zero extreme points of $S$ has all its diagonal elements equal to zero or one?

• Non-negative definite = hermitian with some eigenvalues positive? Or do you mean something else? Apr 13, 2018 at 0:02
• @Igor. Hermitian + all eigenvalues non-negative. Apr 13, 2018 at 0:06
• This cannot be true because the set is the closed convex hull of its extreme points. Or in more elementary style, $0$ is certainly an extreme point. Apr 13, 2018 at 0:25
• What could perhaps be true is that the extreme points have diag entries zero or one. Apr 13, 2018 at 0:26
• Standard terminology is "positive semidefinite" (PSD). Apr 13, 2018 at 18:13

Rank one matrices $xx^\top$ are extreme, now take $x=(1,1/2)$. This gives $\begin{pmatrix} 1&1/2\\1/2&1/4\end{pmatrix}$, a counterexample to your conjecture.
The extremal rays of the PSD cone are the rank one matrices (also known as projections),so those of the form $x x^t.$ - see, for example, https://math.stackexchange.com/questions/678693/positive-semidefinite-cone-is-generated-by-all-rank-one-matrices