Positive definiteness is exactly the inverse condition for the observed GP covariance

For every $z\in {\mathbb{R}}^n$, $z^{T}Az=z^{T}Kz+z^{T}Rz\ge 0$ by [F1], and $A$ is symmetric. Hence $A$ is a nonnegative endomorphism of the finite-dimensional real inner-product space ${\mathbb{R}}^n$.

By [F2], choose an orthonormal eigenbasis $e_1,\dots ,e_n$ of $A$ with eigenvalues ${\lambda }_j\ge 0$. In this basis, $A$ is invertible exactly when no ${\lambda }_j$ is zero, while $z^{T}Az={\sum }_j{\lambda }_j{c}_j^2$ for $z={\sum }_j{c}_j{e}_j$ is positive for every nonzero $z$ exactly when all ${\lambda }_j>0$. Thus $A$ is invertible if and only if it is positive definite.

In the same basis, $A^{\ast }A=A^2$ has eigenvalues ${\lambda }_j^2$. By [F4], its unique nonnegative square root has eigenvalues ${\lambda }_j$, so [F3] says the singular values of $A$ are precisely these ${\lambda }_j$. By [F5], the SVD has full rank exactly when all $n$ singular values are positive, which is exactly the invertibility condition in step 2.1.