Are $A^TP+PA<0$, $P>0$ and $A^TP+PA\leq-I$, $P\geq I$ equivalent? 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.
 A: 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.
A: Because $A^T P + P A$ and $P$ are homogenous functions of $P$, they are equivalent in the following sense:


*

*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 $P > 0$).

*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.
