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4 In response to Suvrit's comment, I added some results of further thinking and more questions.

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:

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!

3 added 66 characters in body; edited title

# ArethereresultsaboutWhatis the propertygeometry of the intersection of some cones defined by generalized inequalities?

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 intersection 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!

2 added 9 characters in body

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 intersection 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 $\{X|X\preceq A\}=\bigcap_i\{X|X\preceq A_i\}$?

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

1