Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 1 / 40A / i / Solution

Take $b$ to be the vector of $f_{ij}$ and absorb $h^{-2}$ into $A$. Its diagonal entry is $-10/(3h^2)$, each axial-neighbour entry is $2/(3h^2)$ and each diagonal-neighbour entry is $1/(6h^2)$. Since these weights agree in both directions, $A$ is a symmetric matrix.

Extend a grid vector $u$ by zero to the whole integer lattice. Let ${\mathcal{E}}_a$ contain unordered axial-neighbour pairs and ${\mathcal{E}}_d$ unordered diagonal-neighbour pairs. Only finitely many differences involving nonzero values contribute. Each lattice vertex has total neighbour weight $4(2/3)+4(1/6)=10/3$, so expanding the squared differences givesIt is nonpositive. If it vanishes, every axial difference vanishes. The axial lattice is connected and $u$ is zero outside the finite interior square, forcing $u=0$. Thus $u^{T}Au<0$ for every nonzero $u$: $\boxed{A\text{ is negative definite}}$.

Reordering the grid replaces $A$ by $P^{T}AP$ for a permutation matrix $P$. This preserves symmetry and the sign of the quadratic form, proving the assertion for any ordering. Multiplying the whole system by $h^2$ merely rescales $A$ by a positive number and has the same conclusions.