I am doing some Gaussian process modeling for generating a random vector field in $\mathbb{R}^D$. As explained in https://arxiv.org/abs/1106.6251 (section 3.3), if you are given some inputs $\mathbf{X}=\{\mathbf{x}_1,\dots ,\mathbf{x}_N\}$, we can generate our field at those points as a random vector $\mathbf{f}(\mathbf{X})$ of size $ND$: $$ \mathbf{f}(\mathbf{X}) \sim \mathcal{N} (m(\mathbf{X}), \mathbf{K}(\mathbf{X}, \mathbf{X}))\,, $$ where $m(\mathbf{X})$ is a vector of size $ND$ that concatenates the mean vectors associated to the outputs and the covariance matrix $\mathbf{K}(\mathbf{X}, \mathbf{X})$ is the $ND \times ND$ block partitioned matrix $(\mathbf{K}(\mathbf{x}_i, \mathbf{x}_j ))_{d,d'}$ for $i,j\in \{1 \ldots N \}$ and $d, d' \in \{1 .. D \}$. Explicitly: $$ \mathbf{K}(\mathbf{X},\mathbf{X})= \begin{bmatrix} \mathbf{K}({\mathbf{x}_1},{\mathbf{x}_1}) & \cdots & \mathbf{K}({\mathbf{x}_1},{\mathbf{x}_N})\\ \mathbf{K}({\mathbf{x}_2},{\mathbf{x}_1}) & \cdots & \mathbf{K}({\mathbf{x}_2},{\mathbf{x}_N})\\ \vdots \cdots \vdots \\ \mathbf{K}({\mathbf{x}_N},{\mathbf{x}_1})& \cdots & \mathbf{K}({\mathbf{x}_N},{\mathbf{x}_N})\\ \end{bmatrix} $$ where each block $\mathbf{K}(\mathbf{x}_i,\mathbf{x}_j)$ is a $D$ by $D$ covariance matrix.(such that $\mathbf{K}(\mathbf{x}_i,\mathbf{x}_j)=\mathbf{K}.(\mathbf{x}_j,\mathbf{x}_i)$, so that we end up with a symmetric matrix).
My question is: if the matrices $ \mathbf{K}({\mathbf{x}_i},{\mathbf{x}_j})$ are positive definite, is it guaranteed that $\mathbf{K}(\mathbf{X},\mathbf{X})$ is positive definite?
I don't think the claim is true in general, so I'm wondering if I'm misunderstanding something?
No, a simple example would be to have all $K(x_i, x_j)$ be equal. If $N>1$, this makes $K(X, X)$ singular and hence not positive definite.