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Hello, considering that for real numbers, the intersection of intervals defined by simple inequalities has a quite simple form as $$ \bigcap_i\{x|x\leq a_i\}=\{x|x\leq\min_i\{a_i\}\} $$

However, what is the case if the variables are chosen as Hermitian matrices, and the interval defined by inequality is replaced with the convex cone defined by the generalized inequality?

All variables in following are assumed to be Hermitian matrices.

To be specific, define the generalize inequality $X\preceq A_i$ to denote that $X-A_i$ is negative semi-definite, then $\{X|X\preceq A_i\}$ defines a convex cone in the Hermitian matrix space.

Is there any result about the intersection of these cones? To say, can the following set be simplified? $$ \bigcap_i\{X|X\preceq A_i\} $$

When does there exist such an $A$ to satisfy $\{X|X\preceq A\}=\bigcap_i\{X|X\preceq A_i\}$?

Or how to describe the geometry of the intersection of such cones?

Any suggestion or comment on this question will be appreciated and thanks very much for your help!

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Acknowledgement and more questions about @Suvrit's comment:

Thanks to @Suvrit for your suggestion! Your comment provides a good way to think about this problem. However, I thought about your method but the problem seems to be more complicated than expected.

Take an example for illustration. Denote $\mathcal{C}(A)=\{X|X\preceq A\}$, then if I want to solve \begin{eqnarray} \min_X&&f(X)\\\ \mathrm{s.t.}&&X\in\mathcal{C}(A_1)\cap\mathcal{C}(A_2)\cap\mathcal{C}(A_3) \end{eqnarray} by first solve $\min_{X_1\in\mathcal{C}(A_1)\cap\mathcal{C}(A_2)} f(X_1)$ and then $\min_{X_2\in\mathcal{C}(X_1)\cap\mathcal{C}(A_3)}f(X_2)$, the solution in deed satisfies the constraints due to $$\mathcal{C}(X_1)\cap\mathcal{C}(A_3)\subseteq\mathcal{C}(A_1)\cap\mathcal{C}(A_2)\cap\mathcal{C}(A_3).$$ However, these two sets are not identical, and thus the optimal solution in $\mathcal{C}(X_1)\cap\mathcal{C}(A_3)$ is not guaranteed to be also optimal in $\mathcal{C}(A_1)\cap\mathcal{C}(A_2)\cap\mathcal{C}(A_3)$.

I think the difficulty of this problem results from the complex structure of the intersections of cones $\bigcap_i\mathcal{C}(A_i)$. Do you have some more suggestions about this problem?

Thank you very much for your help!

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    $\begingroup$ Using my answer to your previous question, probably this can be found iteratively. First find $X_1 \preceq A_1$ and $X_1 \preceq A_2$, then find $X_2 \preceq X_1$ and $X_2 \preceq A_3$ and so on. $\endgroup$
    – Suvrit
    Feb 23, 2012 at 2:50
  • $\begingroup$ @Suvrit, thank you for your suggestion. I thought about your method and the question seems to be more complicated than expected. I edited the post and gave a more detailed illustration at the end of the original post. How do you think about it? Thank you very much! $\endgroup$
    – ppyang
    Feb 24, 2012 at 3:17
  • $\begingroup$ In that case, just cycle through the constraints enough number of times (if $f$ is quadratic, then you can use Dykstra's method). It seemed to me that you just want a "feasible" solution that is not "too bad". If I get time, I might think more carefully about your question, otherwise, I hope someone else finds the time to handle it. Best, $\endgroup$
    – Suvrit
    Feb 24, 2012 at 4:51
  • $\begingroup$ I think you mean use Bregman successive projection method to find the optimal solution on the intersection of convex sets, however, I am just interested in whether there is any theoretical result about such kind of problems, rather than to get the solution in practical. You have provided a good way to think about it and thank you for your attention! $\endgroup$
    – ppyang
    Feb 24, 2012 at 5:28
  • $\begingroup$ What does the dual cone of your cone look like? $\endgroup$
    – Felix Goldberg
    Nov 27, 2012 at 16:57

1 Answer 1

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T. Ando has some papers on the structure of the intersection of these cones: Extreme points of an intersection of operator intervals, (1994) Parameterization of minimal points of some convex sets of matrices, Acta Sci Math Szeged 57 (1993), 3-10.

Not sure if this will solve your problem, but it is yet another way for looking at it.

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  • $\begingroup$ Welcome to MO Koenraad --- I hope you find time to participate regularly! I hope to see more matrix inequalities here ;-) $\endgroup$
    – Suvrit
    Nov 27, 2012 at 18:13
  • $\begingroup$ I am reading the papers you suggested and they indeed provide an interesting standpoint for this question. Thank you very much! $\endgroup$
    – ppyang
    Nov 29, 2012 at 2:22

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