\end{aligned}$$ We need to show that one of the columns above can be written in terms of the others. This is equivalent to solving the linear system $$\begin{aligned}
  \begin{bmatrix}
   {k}_{1j_1}\\
   {k}_{2j_1}\\
   \vdots\\
   {k}_{({l}-1)j_1}
  \end{bmatrix}
  &=
  \begin{bmatrix}
   {k}_{1j_2} & \cdots & {k}_{1j_{{l}}}\\
   {k}_{2j_2} & \cdots & {k}_{2j_{{l}}}\\
   \vdots & \ddots & \vdots\\
   {k}_{({l}-1)j_2} & \cdots & {k}_{({l}-1)j_{{l}}}\\
  \end{bmatrix} 
  \begin{bmatrix}
   c_1\\
   c_2\\
   \vdots\\   
   c_{({l}-1)}
  \end{bmatrix}. 
 \end{aligned}$$ Suppose the kernel matrix on the RHS is generated from the Gaussian kernel. From [@micchelli1984interpolation], it’s known that every principal minor of a Gaussian kernel matrix is invertible, which implies that ${\mathcal{O}}_{{\Upsilon}}$ cannot be observable.

We now prove a sufficient condition for the observability of a system with repeated eigenvalues, but with the condition that the Jordan blocks are trivial.

\[prop:3\] Suppose that the conditions in Proposition \[prop:1\] hold, with the relaxation that the Jordan blocks $\begin{bmatrix}{\Lambda}_1 & {\Lambda}_2 &\cdots & {\Lambda}_{{O}}\end{bmatrix}$ may have repeated eigenvalues, and where ${\Lambda}_i$ are single-dimensional. Let ${l}$ be the cyclic index of $A$. We define $$\begin{aligned}
{\label{eq:empKShadFull}}
  {\mathbf{{\ensuremath{K}}}}= \begin{bmatrix}
                    {\ensuremath{K}}^{(1)}\\
                    \vdots\\
                    {\ensuremath{K}}^{({l})}
                  \end{bmatrix}
 \end{aligned}$$ as the *${l}$-shaded matrix* which consists of ${l}$ shaded matrices with the property that any subset of ${l}$ columns in the matrix are linearly independent from 