Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 1 / 38C / a / 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 giveswhere $w_{ij}=2/3$ 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 $PA{P}^T$ for a permutation matrix $P$, preserving both properties.