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!

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!

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 appreciated and thanks very much for your help!

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!

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 and thanks very for your help!

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 appreciated and thanks very much for your help!