I am following the paper (A Study of Synchronization and Group Cooperation Using Partial Contraction Theory) to understand how contraction theory used to analyze coupled oscillators.
On page 211, the matrix $J_s$ is defined as follows
$$J_s-I^n_{J_{is}}=\sum_{i,j\in \mathcal N}T^n_{K_{ijs}}-U^n_{K_0}$$
Why $s$ is used in the formula? I think since Jacobian matrix is symmetric, its symmetric part is just itself, thus we just need to write $J$ instead.
Since I don't understand this point, my confusion starts again on the condition of the theorem 2, in which one conclusion reads as
$$\lambda_{m+1}(\sum_{i,j\in\mathcal N}T^n_{k_{ijs}})>max_i(\lambda_{max}(J_{is}))$$
I don't understand the right hand side of this inequality. I think $J_{is}$, i.e. $J_i$ defined as $\frac{\partial f(x_i,t)}{\partial x_i}$ (am I right?), which is not a matrix, how can it has eigenvalues?
Thanks in advance!
The text you mention is in the paywall. Please provide more details to help someone who want to answer the question without access to the referred text.
From the free access first two pages of the text we can see that $$\dot{x}=f(x,t)$$ in which $x$ and $f(x,t)$ are a $m\times 1$ vectors, and $t$ is a real number. Hence $$x=\begin{bmatrix}x_1\\x_2\\\vdots\\x_m\end{bmatrix},\qquad f(x,t)=\begin{bmatrix}f_1(x,t)\\f_2(x,t)\\\vdots\\f_m(x,t)\end{bmatrix},$$ and the Jacobian matrix $$Jf(x,t)=\begin{bmatrix}\nabla f_1(x,t)\\\nabla f_2(x,t)\\\vdots\\\nabla f_m(x,t)\end{bmatrix}$$ is a $m\times m$ matrix, with ith-row givem by $$\nabla f_i(x,t)=\begin{bmatrix}\frac{\partial f_i(x,t)}{\partial x_1}& \frac{\partial f_i(x,t)}{\partial x_2}&\ldots & \frac{\partial f_i(x,t)}{\partial x_m}\end{bmatrix}.$$
Perhaps you can find related results seraching for "\( \dot{x}=f(x,t)\) contraction theory" on SearchOnMath, for instance.