= Solution
Here the only nonzero coefficients of $a$ are
$$
\widehat a_{0,0}=5,
\qquad
\widehat a_{1,0}=\widehat a_{-1,0}
=\widehat a_{0,1}=\widehat a_{0,-1}=1.
$$
Therefore, for each $(m,n)\in\Lambda_N$, the truncated system is
$$
\begin{aligned}
\pi^2\{&5(m^2+n^2)\widehat u_{m,n}
+[m(m-1)+n^2]\widehat u_{m-1,n}\\
&+[m(m+1)+n^2]\widehat u_{m+1,n}
+[m^2+n(n-1)]\widehat u_{m,n-1}\\
&+[m^2+n(n+1)]\widehat u_{m,n+1}\}
=-\widehat f_{m,n},
\end{aligned}
$$
where a coefficient is omitted when its index is outside $\Lambda_N$ or equals $(0,0)$.
For wavevectors $\mathbf k=(m,n)$ and $\boldsymbol\ell=(p,q)$, the matrix entry is
$$
A_{\mathbf k,\boldsymbol\ell}
=\pi^2(\mathbf k\mathbin\cdot\boldsymbol\ell)\widehat a_{\mathbf k-\boldsymbol\ell}.
$$
Because $\widehat a_{\mathbf r}=\widehat a_{-\mathbf r}$, $A_{\mathbf k,\boldsymbol\ell}=A_{\boldsymbol\ell,\mathbf k}$, so the matrix is <symmetric matrix>[symmetric]. Its diagonal entry is $5\pi^2|\mathbf k|^2$. The sum of absolute off-diagonal entries is at most
$$
\pi^2\sum_{j=1}^2\left(
|\mathbf k|^2+k_j+|\mathbf k|^2-k_j\right)
=4\pi^2|\mathbf k|^2,
$$
because $|\mathbf k|^2\pm k_j\geq0$ for nonzero integer $\mathbf k$; truncation can only reduce the sum. Hence $A$ is <strict diagonal dominance>[strictly diagonally dominant] with positive diagonal. The <Gershgorin circle theorem> places every eigenvalue in the positive interval
$$
[\pi^2|\mathbf k|^2,9\pi^2|\mathbf k|^2]
$$
for at least one row. Since the symmetric matrix has only real eigenvalues, all are positive, and $A$ is a <positive-definite matrix>. Any one-dimensional ordering merely conjugates $A$ by a <permutation matrix>, preserving symmetry and positive definiteness.
Back to article page
