are nonzero and real, we say the matrix has a *full-rank Jordan decomposition*.

\[def:shaded\] **(Shaded Kernel Matrix)** Let ${k}:{\Omega}\times{\Omega}\to{\mathbb{R}}$ be a positive-definite kernel on a compact 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 let ${\mathcal{X}}= {\{x_1, \dots, x_{{N}}\}}$, $x_i\in{\Omega}$ Let ${\ensuremath{K}}\in{\mathbb{R}}^{{N}\times{M}}$ be the kernel matrix, where $ {\ensuremath{K}}_{ij} := {k}(x_i,c_j)$. For each row ${\ensuremath{K}}_{(i)} := [k(x_i,c_1), k(x_i,c_2),\dots, k(x_i,c_{{M}})]$, define the set ${\mathcal{I}}_{(i)} := \{\iota_1^{(i)},\iota_2^{(i)},\dots, \iota_{{M}_i}^{(i)}\}$ to be the indices in the kernel matrix row $i$ which are nonzero. Then if $$\begin{aligned}
{\label{eq:shaded_cond}}
    \bigcup_{1 \leq i \leq {N}} {\mathcal{I}}^{(i)} = \{1,2,\dots, {M}\},
    \end{aligned}$$ we denote ${\ensuremath{K}}$ as a *shaded kernel matrix* (see figure \[shaded\_matrix\]).

This condition implies that the null space of the adjoint of ${\ensuremath{K}}$ as a linear operator between Euclidean spaces, i.e. ${\ensuremath{K}}^T:{\mathbb{R}}^{{N}}\to{\mathbb{R}}^{{M}}$ is trivial. Note that, in principle, for the Gaussian kernel, a single row generates a shaded kernel matrix, although this matrix can have many entries that are extremely close to zero. With this definition in place, we can prove the following proposition, which shows that if $A$ has a full-rank Jordan decomposition, a shaded 