Are $A^TP+PA<0$, $P>0$ and $A^TP+PA\leq-I$, $P\geq I$ equivalent?

Question

Consider the LMI, where $A$ is a Hurwitz matrix:

$A^TP+PA<0$, $P>0$, minimize trace(P)

According to Stephen Boyd's book, the inequalities are homogeneous in $P$ and hence can by replaced with the nonstrict inequalities:

$A^TP+PA\leq-I$, $P\geq I$, minimize trace(P)

I do not understand why this is equivalent. Apparently the solution $P$ changes.

Answer 1

If $P$ is a solution to $A^\top P + P\,A \leq -I,\ P \geq I$ then scaling $P$ by some arbitrary small positive constant $\gamma$ the strict inequalities are also always satisfied, since

$$ \begin{array}{c} A^\top \gamma\,P + \gamma\,P\,A \leq -\gamma\,I < 0, \\ \gamma\,P \geq \gamma\,I > 0. \end{array} $$

So the solution to the nonstrict inequalities can always give you a solution to the strict inequalities.

Answer 2

Because $A^T P + P A$ and $P$ are homogenous functions of $P$, they are equivalent in the following sense:

1. Any solution of the non-strict inequalities ($A^T P + P A \leq -I$ and $P \geq I$) is necessarily a solution of the strict inequalities ($A^T P + P A < 0$ and $P > 0$).
2. For any solution of the strict inequalities there exists a solution of the non-strict inequalities (which itself will satisfy the non-strict inequalities)

We conclude that the LMI $A^T P + P A < 0$ and $P > 0$ has a solution if and only if $A^T P + P A \leq -I$ and $P \geq I$ has one.

See this answer for a detailed explanation of why homogeneity is needed to show this.