Proving these two matrix LMIs are the same. For a matrix A $\in R^{nxn}$, the two statements are equivalent:

*

*There exists matrix $P > 0$, such that $A^TP+PA<0$

*There exists matrix $P \geq 0$, such that $A^TP+PA+I\leq0$
Is this true?

Yes, they are equivalent.
Suppose $P>0$ and $A^TP+PA<0$. Let $r=-\lambda_\max(A^TP+PA)$. Then $P_1\ge0$ and $A^TP_1+P_1A+I\le0$, where $P_1=\frac1rP$.
Conversely, suppose $P\ge0$ and $A^TP+PA+I\le0$. Then $A^TP+PA\le-I<0$. Now suppose $Px=0$. Then $x^T(A^TP+PA)x=x^TA^T(Px)+(x^TP)Ax=0$. Since $A^TP+PA$ is negative definite, we must have $x=0$. Hence $P$ is nonsingular and in turn, it is positive definite.
