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Let $K$ be set of all positive matrices $A \in M_3(\mathbb{C})$ such that all the diagonal entries are $1$. Observe that $K$ is a convex set. An element $A\in K$ is extreme if it is not interior point of a line segment in $K$ i.e. if $\lambda\in (0,1)$ such that $A=\lambda B+(1-\lambda)C$ for some $B,C\in K$, then $B=C=A$. We write $\text{Ext}(K)$ to denote the set of all extreme points of $K$.

Let $K_0$ be the set of all positive matrices in $M_3(\mathbb{C})$ of trace $3$. Then $K\subseteq K_0$. By spectral theorem, the extreme points of $K_0$ are the positive rank one matrices. Now, it is easy to observe that $K\cap\text{Ext}(K_0)\subseteq \text{Ext}(K)$ (but not equal in general). From this we can say that rank one positive matrices with diagonal entries $1$ are extreme point of $K$. Now, I am curious whether there is any other extreme points of $K$ i.e. is there any extreme point of rank more than one.

Note: A matrix $A$ is said to be positive if $\langle v,Av\rangle \ge 0$ for all $v\in\mathbb{C}^3$. Equivalently, $A$ is said to be positive iff $A=B^*B$ for some matrix $B$.

DeltaEpsilon
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  • Is $K_0$ supposed to have some kind of norm bounds? Otherwise there are no extreme points, apart from $0$, since $A = \frac{1}{2} \cdot 0 + \frac{1}{2} \cdot 2A$. – David Gao Mar 20 '25 at 19:05
  • Okay, sorry it should be the set of all positive matrices with trace $3$ – DeltaEpsilon Mar 20 '25 at 19:58
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    I haven't solved the question yet, but I'm leaning towards there being no other extreme points. Two relevant results I can prove: First, no extreme point of $K$ can be of rank $3$. Second, a natural class of extreme points, namely those positive matrices with diagonal entries being $1$ and all other entries having absolute values $1$, all have rank $1$. I just can't prove that there are no extreme points of rank $2$ which have some entries of absolute values strictly smaller than $1$. – David Gao Mar 20 '25 at 21:31
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1 Answers1

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The matrices you are considering are known as correlation matrices (which is also the relevant search term). In "A Note on Extreme Positive Definite Matrices" Math. Ann. 244:65-68 (1979) Christensen and Vesterstrom first proved that in the three-dimensional complex case there are no extreme points beside the ones with rank 1—which settles your question.

If you're also interested in higher-dimensional cases let me list some generalizations and further results on this topic:

  • In "Extreme points of a convex subset of the cone of positive semidefinite matrices" Math. Ann. 253:227-232 (1980) Loewy showed that for any integer $r\leq\sqrt n$ there exists an extreme point of rank $r$. This sharpened a result from the Christensen–Vesterstrom paper which upper bounded the rank of all extreme points by $\sqrt n$ (but they left it as an open problem whether every such rank actually occurs)
  • In "Extremal correlation matrices" Linear Algebra Appl. 134:63-70 (1990) Grone et al. proved that in the real case extreme points of all ranks $k$ with $k(k+1)\leq 2n$ occur. In particular, in the three-dimensional real case there are extreme points other than the rank-1 ones (characterized in Theorem 1 of said paper). This was further generalized by Li and Tam in "A Note on Extreme Correlation Matrices" SIAM J. Matrix Anal. Appl. 15:903-908 (1994), but I feel like this would go beyond the scope of this question.
Frederik vom Ende
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