Computing on the observable abundance of the linear continuous-time systems with the real and differential eigenvalues

     In my blog article “observable abundance of the linear continuous-time systems”(http://blog.sciencenet.cn/blog-3343777-1071513.html), the observable abundance of the linear continuous-time systems(LCTS) is defined and the corresponding computing equation is got as follows.

$v_{o,T}=\mathrm{Vol}(R_{o,T})=\left|W_{o,T}\right|^{-1}\mathrm{Vol}(\widetilde{R}_{o,T})$

     For the SISO LCTS $\Sigma(A,C)$ , if the system matrix $A$ is diagonal matrix and its eigenvalues $\lambda_{i}(i=1,2,\cdots,n)$ are real and differential, then we have

$W_{o,T}=\left[\begin{array}{cccc} c_{1}^{2}\frac{e^{2\lambda_{1}T}-1}{2\lambda_{1}} & c_{1}c_{2}\frac{e^{(\lambda_{1}+\lambda_{2})T}-1}{\lambda_{1}+\lambda_{2}} & \cdots & c_{1}c_{n}\frac{e^{(\lambda_{1}+\lambda_{n})T}-1}{\lambda_{1}+\lambda_{n}}\\ c_{1}c_{2}\frac{e^{(\lambda_{1}+\lambda_{2})T}-1}{\lambda_{1}+\lambda_{2}} & c_{2}^{2}\frac{e^{2\lambda_{2}T}-1}{2\lambda_{2}} & \cdots & c_{2}c_{n}\frac{e^{(\lambda_{2}+\lambda_{n})T}-1}{\lambda_{2}+\lambda_{n}}\\ \vdots & \vdots & \ddots & \vdots\\ c_{1}c_{n}\frac{e^{(\lambda_{1}+\lambda_{n})T}-1}{\lambda_{1}+\lambda_{n}} & c_{2}c_{n}\frac{e^{(\lambda_{2}+\lambda_{n})T}-1}{\lambda_{2}+\lambda_{n}} & \cdots & c_{n}^{2}\frac{e^{2\lambda_{n}T}-1}{2\lambda_{n}} \end{array}\right]$

where the output matrix $C=[c_{1},c_{2},\cdots,c_{n}]$ . When $\lambda_{i}\in(-\infty,0](i=1,2,\cdots,n)$ , we have

$\widehat{W} =\lim_{T\rightarrow\infty}W_{o,T}=-\left[\begin{array}{cccc} \frac{c_{1}^{2}}{2\lambda_{1}} & \frac{c_{1}c_{2}}{\lambda_{1}+\lambda_{2}} & \cdots & \frac{c_{1}c_{n}}{\lambda_{1}+\lambda_{n}}\\ \frac{c_{1}c_{2}}{\lambda_{1}+\lambda_{2}} & \frac{c_{2}^{2}}{2\lambda_{2}} & \cdots & \frac{c_{2}c_{n}}{\lambda_{2}+\lambda_{n}}\\ \vdots & \vdots & \ddots & \vdots\\ \frac{c_{1}c_{n}}{\lambda_{1}+\lambda_{n}} & \frac{c_{2}c_{n}}{\lambda_{2}+\lambda_{n}} & \cdots & \frac{c_{n}^{2}}{2\lambda_{n}} \end{array}\right]$

     For the determinant of the matrix $\widehat{W}$ , it can be proven as follows.

     For any $\lambda_{i}$ and $\lambda_{j}$ , we have

$\left.\det\left(\widehat{W}\right)\right|_{\lambda_{i}=\lambda_{j}}=0$

that is, $\lambda_{i}-\lambda_{j}$ is a factor of $\det\left(\widehat{W}\right)$ . In addition, for $\lambda_{1}$ and $\lambda_{2}$ , we have

$\left.\frac{\mathrm{d}}{\mathrm{d}\lambda_{1}}\det\left(\widehat{W}\right)\right|_{\lambda_{2}=\lambda_{1}}=0$

that is, $$ is a factor of $\frac{\mathrm{d}}{\mathrm{d}\lambda_{1}}\det\left(\widehat{W}\right)$ , and then, we can conclude that $\left(\lambda_{1}-\lambda_{2}\right)^{2}$ is a factor of $\det\left(\widehat{W}\right)$ 。Similarly, for any   $\lambda_{i}$ and $\lambda_{j}$ , $\left(\lambda_{i}-\lambda_{j}\right)^{2}$ is a factor of   $\det\left(\widehat{W}\right)$ . Therefore, we have

$\det\left(\widehat{W}\right)=G\left[\prod_{1\leq j_{1}

where $G$ is an undetermined function on the eigenvalues $\lambda_{i}(i=1,2,\cdots,n)$ , and then,  we have

$\lim_{T\rightarrow\infty}v_{o,T} =\frac{1}{G}\left|\left(\prod_{1\leq j_{1}