I have a Matrix symmetric, non-negative Matrix $\pmb{A} \in {\mathbb{N}_0}^{N\times N}$.
It is defined as follows: $$\pmb{A}=\left[\begin{matrix} \min(x_1,x_1) & \dots &\min(x_1,x_N)\\ \vdots & \ddots & \vdots \\ \min(x_N, x_1) & \dots & \min(x_N, x_N) \end{matrix}\right]$$ where $x_n\in\mathbb{N}_0$.
How can I prove that the matrix is positive-semidefinite?
My approach so far starts at $$\pmb{A}\succcurlyeq0 \iff \vec{x}^T\pmb{A}\vec{x} \ge 0$$ $$\vec{x}^T\pmb{A}\vec{x} = \sum_i^N\sum_j^N A_{ij}x_ix_j\ge0$$ Which I have decomposed using the properties of the matrix into $\sum_k^NA_{kk}x_k^2 + 2\sum_{i=2}^N\sum_{j=1}^{i-1}A_{ij}x_ix_j\ge0$, but I am having trouble getting any further.
Let $M := \max_i x_i$. If we define the map $\phi: \{0, \ldots, M\} \to \{0, 1\}^{M+1}$ by $$\phi(x) = (\mathbf{1}\{x \le 0\}, \mathbf{1}\{x \le 1\}, \mathbf{1}\{x \le 2\}, \ldots, \mathbf{1}\{x \le M\}),$$ we have $\min(x_i, x_j) = \langle \phi(x_i), \phi(x_j) \rangle$.
So for any vector $v$, we have $$v^\top A v = \sum_i \sum_j v_i v_j \langle \phi(x_i), \phi(x_j)\rangle = \left\langle \sum_i v_i \phi(x_i), \sum_j v_j \phi(x_j)\right\rangle = \left\|\sum_i v_i \phi(x_i)\right\|^2 \ge 0.$$