matrix inner product between positive semidefinite matrix and positive definite matrix

Let $$F_0, F_1, \ldots, F_m$$ be a $$n \times n$$ symmetric matrices. We define $$F(x) := F_0 + x_1 F_1 + \cdots + x_m F_m$$ Show that if there does not exist $$x \in \Bbb R^m$$ such that $$F(x)$$ is positive definite, then there does exist a positive semidefinite matrix $$H \neq 0$$ such that $$\sup_x \mbox{tr} \left( F(x) H \right) \leq 0$$

Can you solve it? Contraposition may be useful, but I have no idea.

• can you use SDP duality? Jul 4 '20 at 13:31
• I cannot find the connection between SDP duality and this problem... Jul 4 '20 at 13:56

Recall that $$\langle A,B \rangle = \operatorname{tr}(AB)$$ defines an inner product over the set of symmetric matrices.

The set $$S_1 = \{F(x):x \in \Bbb R^n\}$$ and the set $$S_2$$ of positive definite matrices are both convex, and we are given that these two sets are disjoint. By the hyperplane separation theorem, there exists a non-zero matrix $$H$$ and a constant $$c$$ such that we have $$\langle X,H \rangle \leq c \leq \langle Y, H\rangle$$ for all $$X \in S_1$$ and $$Y \in S_2$$.

Because $$0$$ lies in the closure of $$S_2$$, we have $$c \leq \langle 0,H \rangle = 0$$.

Claim: $$\langle Y,H \rangle \geq 0$$ holds for all positive definite $$Y$$.

Proof of Claim: Suppose that this does not hold. Then, there exists a positive definite $$Y$$ for which $$\langle Y,H \rangle < 0$$. For any positive $$k$$, we note that $$kY$$ is also positive definite, and $$\langle kY,H \rangle = k\langle Y,K \rangle$$. This means that $$\inf_{Y \in S_2} \langle Y,H \rangle = -\infty$$, which contradicts our earlier statement that we always have $$\langle Y,H \rangle \geq c$$ for some (finite) constant $$c$$. $$\square$$

Because $$H$$ is such that $$\langle Y,H \rangle \geq 0$$ holds for all positive definite $$Y$$, it must hold that $$H$$ is positive semidefinite. Thus, there is indeed a non-zero positive semidefinite matrix $$H$$ for which $$\langle X, H \rangle \leq c \leq 0$$ for all $$X \in S_1$$, which was what we wanted.

• Thank you for your answer, but I still have a question. Why S1 and S2 are disjoint? there may exist a positive semidefinite matrix included in both S1 and S2. Jul 4 '20 at 16:19
• @eosa I've fixed it; see my latest edit. $S_2$ should have been the positive definite matrices Jul 4 '20 at 16:25
• Thanks, but sorry I have another question. Why H is positive semi-definite? Jul 5 '20 at 13:34
• I had left something out there; see my latest edit. If $H$ is a matrix such that $\langle Y, H \rangle \geq 0$ for all positive definite $Y$, then $H$ must be positive semidefinite Jul 5 '20 at 13:36
• Thanks, but I have two questions. First, Why H is symmetric? Second, Why ⟨𝑌,𝐻⟩≥0 holds for all positive definite Y? Jul 5 '20 at 13:44