kernel matrix is sufficient to prove observability.

\[prop:1\] Let ${k}:{\Omega}\times{\Omega}\to{\mathbb{R}}$ be a positive definite kernel on a domain ${\Omega}$. Let $C = [c_1,  c_2, \cdots , c_{{M}}\}$, $c_j\in{\Omega}$ be the points generating a finite-dimensional covering of the reproducing kernel Hilbert space ${\mathcal{H}}$ associated to ${k}(x,y)$, and consider the discrete linear system on ${\mathcal{H}}$ given by the evolution and measurement equations [(\[eq:k\_measure\])]{} and [(\[eq:k\_measure1\])]{}. Let $A\in{\mathbb{R}}^{{M}\times{M}}$ be a full-rank Jordan decomposition of the form $A = {P}{\Lambda}{P}^{-1}$, where ${\Lambda}= \operatorname{diag}(\begin{bmatrix}{\Lambda}_1 & {\Lambda}_2 &\cdots & {\Lambda}_{{O}}\end{bmatrix})$, and there are no repeated eigenvalues. Given a set of time instances ${\Upsilon}= \{t_1,t_2,\dots,t_{{L}}\}$, and a set of sampling locations ${\mathcal{X}}={\{x_1, \dots, x_{{N}}\}}$, the system [(\[eq:k\_measure\])]{} is observable if the kernel matrix ${\ensuremath{K}}_{ij} := {k}(x_i,c_j)$ is shaded, ${\ensuremath{K}}^D$, the row vector generated by summing the rows of ${\ensuremath{K}}$, has all nonzero entries, ${\Upsilon}$ has distinct values, and $|{\Upsilon}| \geq {M}$.

To begin, consider a system where $A = {\Lambda}$, with Jordan blocks $\{{\Lambda}_1, {\Lambda}_2, \dots, {\Lambda}_{{O}}\}$ along the diagonal. Then $A^{t_i} = \operatorname{diag}(\begin{bmatrix}{\Lambda}_1^{t_i} & {\Lambda}_2^{t_i} & \cdots & {\Lambda}_{{O}}^{t_i}\end{bmatrix})$. We have that $${\mathcal{O}}_{{\Upsilon}} = 
 \begin{bmatrix}
  {\ensuremath{K}}A^{t_1}\\
  \cdots\\
  {\ensuremath{K}}A^{t_{L}}
 \end{bmatrix}
 =
 \underbrace{
 \begin{bmatrix}
  {\ensuremath{K}}& \cdots & {\ensuremath{K}}\end{bmatrix}}_{{\widehat{\mathbf{K}}}\in{\mathbb{R}}^{{N}\times{M}{L}}}
 \underbrace{
 \begin{bmatrix}
  {\Lambda}_1^{t_1} & \cdots & 0\\
   \vdots & \ddots 