show that the matrix $A$ has stable eigenvalues?

Question

Assume that for the system $\dot x=Ax$ there exist $P,Q>0$ and suppose $\mu>0$ such that $A^TP+PA+2\mu P=-Q$. I want to prove all eigenvalues of $A$ have real part less than $-\mu$

Answer 1

Assume, for simplicity, that $A\in\mathbb{R}^{n\times n}$ is diagonalizable and let $e_i$ be the eigenvector associated with the eigenvalue $\lambda_i$, $i=1,\ldots,n$.

Then, we have that $A^TP+PA+2\mu P=-Q$ implies

$$e_i^*(A^TP+PA+2\mu P)e_i=-e_i^*Qe_i,\ i=1,\ldots,n.$$

Since $Ae_i=\lambda_ie_i$ (and hence $e_i^*A^T=\lambda_i^*e_i^*$), this is equivalent to saying that

$$2\Re[\lambda_i]e_i^*Pe_i=-e_i^*Qe_i-2\mu e_i^*Pe_i,\ i=1,\ldots,n$$

which implies that

$$2\Re[\lambda_i]e_i^*Pe_i<-2\mu e_i^*Pe_i,\ i=1,\ldots,n$$

Dividing both sides by $2e_i^*Pe_i>0$ yields

$$\Re[\lambda_i]<-\mu,\ i=1,\ldots,n$$

which proves the desired result.

The non-diagonalizable case is analogous except that we only consider the eigenvectors and not the generalized eigenvectors.

Answer 2

Write the formula as $B^TP+PB = -Q$, with $B=A+\mu I$. Presumably you know that if there is a $P=P^T$ and $Q>0$ such that the formula holds then all solutions of $\dot{x} = Bx$ are asymptotically stable and hence the real parts of all eigenvalues of $B$ are strictly negative. Now conclude the desired result.