## 5 Concluding Remarks
\begin{split}&\det(sI_{n}-A)=\det\Bigl{(}sI_{n}-A_{0}- \widetilde{A}\Bigr{)}=\det\Bigl{(}(sI_{n}-A_{0})\,\big{(}I_{n}-(sI_{n}-A_{0})^ {-1}\widetilde{A}\big{)}\Bigr{)}\\ &=\det(sI_{n}-A_{0})\,\det\Bigl{(}I_{n}-(sI_{n}-A_{0})^{-1} \widetilde{A}\Bigr{)};\forall s\in\mathbf{C}\end{split}  \tag{1}\]  
and then \(\det(sI_{n}-A)=\det(sI_{n}-A_{0})\,\det\Bigl{(}I_{n}-(sI_{n}-A_{0})^{-1} \widetilde{A}\Bigr{)}\
eq 0\) Then, for all \(s\
otin sp(A)\), and also for all \(s\in sp(A_{0})\) if the \(H_{\infty}\)-norm of \((sI_{n}-A_{0})^{-1}\widetilde{A}\), which exists since \(A_{0}\) is a stability matrix, satisfies \(\left\|(sI_{n}-A_{0})^{-1}\widetilde{A}\right\|_{\infty}<1\), which is guaranteed if \(\left\|\widetilde{A}\right\|_{2}<1/\sup\limits_{\omega\in\mathbf{R}_{0+}}\left\|( i\omega I_{n}-A_{0})^{-1}\right\|_{2}\). Then, \(A\) is a stability matrix if Conditions 1 or 2 holds. On the other hand, if \(A_{0}\) and \(A\) are negative (implying that \(A<-A_{0}\)), or if they are both Metzler-stable (implying for all off-diagonal entries that \(\widetilde{A}_{ij}\geq-A_{0_{ij}}\); \(\forall i,j(\
eq i)\in\overline{n}\)), then their dominant abscissa (perhaps multiple) eigenvalue is real and negative since \(A_{0}\) being a stability matrix is claimed to guarantee that \(A\) is stable. Since \(A_{0}\) is a stability matrix, it is non-singular with eigenvalues with negative real parts. Then, by the continuity of the eigenvalues with respect to the matrix entries, \(A=A_{0}\bigl{(}I_{n}+A_{0}^{-1}\widetilde{A}\bigr{)}\) is a stability matrix if \(\rho\big{(}A_{0}^{-1}\widetilde{A}\big{)}\leq\left\|A_{0}^{-1}\widetilde{A} \right\|_{2}\leq\left\|A_{0}^{-1}\right\|_{2}\left\|\widetilde{A}\right\|_{2}<1\) leading to the sufficiency of Conditions 3-5 for the stability of \(A\) if is \(A_{0}\) stable. The last, sufficient condition comes directly by upper-bounding Condition 4 by norm product and it is equivalent to \(\left\|\widetilde{A}\right\|_{2}=\lambda_{\max}^{1/2}\big{(}\widetilde{A}^{T} \widetilde{A}\big{)}<1/\left\|A_{0}^{-1}\right\|_{2}=1/\lambda_{\max}^{1/2} \big{(}A_{0}^{-1}A_{0}^{-T}\big{)}=\lambda_{\min}^{1/2}\Big{(}A_{0}^{T}A_{0} \big{)}\). \(\Box\)  
**Theorem A2. **_Assume that \(A_{0}\in\mathbf{R}^{n\times n}\) is a stability matrix and that \(A(t)=A_{0}+\widetilde{A}(t)\), where \(\widetilde{A}:\mathbf{R}_{0+}\to\mathbf{R}^{n\times n}\) is piecewise-continuous and bounded. Then, \(A:\mathbf{R}_{0+}\to\mathbf{R}^{n\times n}\) is stable if \(\frac{k_{0}}{\rho_{0}}\sup\limits_{0\leq\tau\leq t}\left\|\widetilde{A}(\tau) \right\|<1\); \(\forall t\in\mathbf{R}_{0+}\), where \((-\rho_{0})<0\) is the stability abscissa of \(A_{0}\) and \(k_{0}=k_{0}(\left\|A_{0}\right\|)\geq 1\) is a real constant satisfying \(\left\|e^{A_{0}t}\right\|\leq k_{0}e^{-\rho_{0}t}\forall t\in\mathbf{R}_{0+}\)._  
**Proof. ** Consider the linear time-varying system:  
\[\dot{x}(t)=\big{(}A_{0}+\widetilde{A}(t)\big{)}x(t),\ x(0)=x_{0}\ ;\ \forall t\in\mathbf{R}_{0+} \tag{2}\]where \(\widetilde{A}:\mathbf{R}_{0+}\to\mathbf{R}^{n\times n}\) is piecewise-continuous and bounded [1]. Such a system is globally asymptotically stable if and only if \(A:\mathbf{R}_{0+}\to\mathbf{R}^{n\times n}\) is a stability matrix. The state-trajectory solution of Equation (A2) satisfies:  
\[ \begin{split}&\|x(t)\|\leq\|e^{A_{0}t}\|\|x_{0}\|+\int_{0}^{t}\! \!\!\left\|e^{A_{0}(t-t)}\right\|\|\|\overline{A}(\tau)\|\|x(\tau)\|\|x(\tau)\| \,d\tau\leq\,k_{0}e^{-\rho_{0}t}\|x_{0}\|+\frac{k_{0}}{\rho_{0}}\sup_{0\leq t \leq t}\|x(\tau)\|\\ &=K_{0}e^{-\rho_{0}t}\|x_{0}\|+\frac{k_{0}}{\rho_{0}}\sup_{0\leq t \leq t}\!\!\left(\|\overline{A}(\tau)\|\right)\|x(t)\|;\forall t\in\mathbf{R}_{0+} \end{split} \] (A3)  
Let \(t\prime=t\prime(t)\) be defined for each \(t\in\mathbf{R}_{0+}\) as \(t\prime=\left\{z=\max_{0\leq t\leq t}\tau\,:\,\|x(z)\|\geq\|x(t)\|\right\}\).