Proving these two matrix LMIs are the same.

Question

For a matrix A $\in R^{nxn}$, the two statements are equivalent:

1. There exists matrix $P > 0$, such that $A^TP+PA<0$
2. There exists matrix $P \geq 0$, such that $A^TP+PA+I\leq0$

Is this true?

Answer 1

I think that the second point should be: There exists matrix $P\ge I$, such that $A^TP+PA+I\le0$. The implication 2. implies 1. is immediate. So, we will assume that 1. holds. This is equivalent to say that $A$ is Hurwitz stable.

Therefore, there exists a $Q>0$ such that $A^TQ+QA+I=0$. Moreover, this $Q$ is given by $$Q=\int_0^\infty\exp(A^Ts)\exp(As)\mathrm{d}s.$$

If $Q\ge I$, then we are done. On the other hand, let $\alpha=1/\min_i(\lambda_i(Q))$ where the $\lambda_i(Q)$'s are the eigenvalues of $Q$. Since, we do not have $Q\ge I$, then $\min_i(\lambda_i(Q))<1$ and $\alpha>1$. Therefore, $P:=\alpha Q\ge I$.

Now, we have $$A^TP+PA=\alpha(A^TQ+AQ)=-\alpha I\le -I,$$

which proves the result.

Answer 2

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.