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?
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2$\begingroup$ Non-negative definite = hermitian with some eigenvalues positive? Or do you mean something else? $\endgroup$– Igor RivinCommented Apr 13, 2018 at 0:02
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$\begingroup$ @Igor. Hermitian + all eigenvalues non-negative. $\endgroup$– MathbuffCommented Apr 13, 2018 at 0:06
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$\begingroup$ 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. $\endgroup$– Christian RemlingCommented Apr 13, 2018 at 0:25
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$\begingroup$ What could perhaps be true is that the extreme points have diag entries zero or one. $\endgroup$– Christian RemlingCommented Apr 13, 2018 at 0:26
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2$\begingroup$ Standard terminology is "positive semidefinite" (PSD). $\endgroup$– Igor RivinCommented Apr 13, 2018 at 18:13
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2 Answers
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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.
answered Apr 13, 2018 at 19:38
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$\begingroup$ @ Dima and Igor. I am talking about extreme points of the set of all non-negative definite matrices. I am considering a special convex set of non-negative matrices with diagonal entries all less than or equal to one. So your argument is not correct. $\endgroup$– MathbuffCommented Apr 14, 2018 at 12:26
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2$\begingroup$ the matrix I give is in your set, and we claim it is extreme $\endgroup$ Commented Apr 14, 2018 at 12:36
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$\begingroup$ @ Dima and Igor. I am really sorry. you are absolutely right. $\endgroup$– MathbuffCommented Apr 14, 2018 at 14:03
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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
The answer to your question follows immediately.
answered Apr 13, 2018 at 18:17
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1$\begingroup$ no, the question is whether on the boundary one always get matrices with only 0 or 1 on the diagonal. E.g. for x=(1,1/2) one gets an extreme matrix which is not like this. $\endgroup$ Commented Apr 13, 2018 at 19:33