Strict Inequality in Homogenous LMI

Question

I'm studying Stephen Boyd's notes for EE 363, here.

In particular, I'm working through lecture 15, slide 9 on strict linear matrix inequalities.

An LMI is an expression of the form $G(x) = G_0 + x_1G_1 + \cdots + x_nG_n \geq 0$, where $G_i$ are symmetric $m \times m$ matrices and $x \in \mathbf{R}^n$. The inequality is understood relative to the positive semidefinite cone. If $G_0 = 0$, then $G(x)$ is homogeneous. That is, $G(\alpha x) = \alpha G(x)$ for $\alpha > 0$.

Consider the set of all $x$ defined by the matrix inequalities $F(x) \geq 0$ and $F_\text{strict}(x) > 0$, where $F$ and $F_\text{strict}$ are homogeneous in $x$. In the slide, he states that these inequalities can be replaced with $F(x) \geq 0$ and $F_\text{strict}(x) \geq I$. My question is how to show this.

I have been unable to prove that this set is equivalent to the one defined by the inequalities $F(x) \geq 0$ and $F_\text{strict}(x) \geq I$. I've unsuccessfully tried using the fact that if $A > 0$, then $A \geq \epsilon I$ for some $\epsilon > 0$ combined with the homogeneity of $F$ and $F_\text{strict}$.

Here's a related question which is essentially the same.

Thanks in advance!

Answer 1

As I said in my comment, any $x$ which satisfies $F(x) \geq 0$ and $F_\text{strict}(x) \geq I$ must satisfy $F(x) \geq 0$ and $F_\text{strict}(x) > 0$ because $I > 0$. I don't think that the converse of this statement holds in general.

However, suppose that we have any $x$ such that $F(x) \geq 0$ and $F_\text{strict}(x) > 0$. For some $\epsilon > 0$, we must have that $F_\text{strict}(x) \geq \epsilon I$, and because $F$ and $F_\text{strict}$ are homogeneous, this implies that $\frac{1}{\epsilon}F_\text{strict}(x) = F_\text{strict}(\frac{1}{\epsilon}x) \geq I$. We also have that $F(\frac{1}{\epsilon}x) \geq 0$. Let $y = \frac{1}{\epsilon} x$, so $F(y) \geq 0$ and $F_\text{strict}(y) \geq I$.

So, in this context, solving the non-strict version (i.e., finding any $x$ satisfying $F(x) \geq 0$ and $F_\text{strict}(x) \geq I$) gives a solution satisfying the strict version (i.e., $F(x) \geq 0$ and $F_\text{strict}(x) > 0$). For any strict version solution there must exist a non-strict solution, $y$, which will itself satisfy the strict version.

We conclude that $F(x) \geq 0$ and $F_\text{strict}(x) > 0$ has a solution if and only if $F(x) \geq 0$ and $F_\text{strict}(x) \geq I$ has one.

Answer 2

You can. You have $F_{strict}(z)>0$ implies that $\exists \epsilon >0$ arbitrary small enough such that $F_{strict}(z)\geq \epsilon I$ and since $F_{strict}(z)$ is homonogeous in z then you get $F_{strict}(\frac{1}{\epsilon}z)\geq I$.

On the other hand, you have $F(z)\geq0$ and $F(z)$ is also homologous in z which implies $F(\frac{1}{\epsilon}z)\geq 0$. Using the exchange of variable $x=\frac{1}{\epsilon}$ you obtain $F(x)\geq 0$ and $F_{strict}(x)\geq I$.

Hence you have proved that $F(x)\geq 0$ and $F_{strict}(x)>0$ implies that $F(x)\geq 0$ and $F_{strict}(x)\geq I$. The implication in the other direction is evident.