= Solution
Number the interior grid vertices in any order. The diagonal of $A$ is $10/3$, and off-diagonal weights are $-2/3$ for axial neighbors and $-1/6$ for diagonal neighbors. Each link weight is symmetric, so $A$ is symmetric. Extend the interior grid values by zero to the boundary. Since the eight stencil weights sum to $10/3$, grouping links gives
$$
u^TAu=\sum_{\{i,j\}\text{ interior link}}w_{ij}(u_i-u_j)^2+
\sum_{i\text{ interior},\ j\text{ boundary link}}w_{ij}u_i^2,
$$
where $w_{ij}=2/3$ or $1/6$ and each interior link is counted once. Every term is nonnegative. If the <quadratic form> vanishes, neighboring interior values agree, and every vertex linked to the boundary has value zero. The axial-link grid is connected and reaches the boundary, so all values are zero. Thus $\boxed{A\text{ is symmetric positive definite}}$. Reordering vertices replaces $A$ by $PAP^T$ for a <permutation matrix> $P$, preserving both properties.
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