[proofplan] The proof has two parts. First, we establish symmetry by inspecting the explicit entries of $A$: the diagonal entries are $a_{ii} = -4$ and the off-diagonal entries are $a_{ij} \in \{0, 1\}$, with $a_{ij} = a_{ji}$ by the symmetry of the five-point stencil. Since $A$ is real symmetric, all its eigenvalues are real. Second, we show every eigenvalue is strictly negative. We use a Gershgorin-type maximum-modulus argument on the eigenvector to show $|\lambda + 4| \le 4$, which restricts $\lambda$ to $[-8, 0]$. We then rule out $\lambda = 0$ by a propagation argument: if $Ax = 0$, then every component achieving the maximum modulus must have all its stencil neighbours also at maximum modulus, and this propagates to the boundary, where the Dirichlet condition forces $x = 0$. [/proofplan] [step:Verify that $A$ is real and symmetric by inspecting the stencil structure] The five-point stencil assigns to each interior grid point $(i, j)$ the equation \begin{align*} u_{i-1,j} + u_{i+1,j} + u_{i,j-1} + u_{i,j+1} - 4u_{i,j} = h^2 f_{i,j}. \end{align*} Under the natural ordering, each grid point $(i, j)$ receives a global index $k = k(i, j)$, and the equation for point $k$ involves $u_k$ with coefficient $-4$ and the values $u_\ell$ at the (at most four) neighbouring grid points $\ell$ with coefficient $+1$. All other entries in row $k$ are zero. Therefore the matrix entries are: \begin{align*} A_{kk} &= -4, \\ A_{k\ell} &= \begin{cases} 1 & \text{if } \ell \text{ is a grid neighbour of } k, \\ 0 & \text{otherwise.} \end{cases} \end{align*} The neighbour relation is symmetric: $\ell$ is a neighbour of $k$ if and only if $k$ is a neighbour of $\ell$. Therefore $A_{k\ell} = A_{\ell k}$ for all $k, \ell$, so $A = A^\top$. Since all entries of $A$ are real, $A$ is a real symmetric matrix. By the spectral theorem for real symmetric matrices, all eigenvalues of $A$ are real. [/step] [step:Show every eigenvalue satisfies $-8 \le \lambda \le 0$ using Gershgorin's theorem] Each row $k$ of $A$ has diagonal entry $a_{kk} = -4$ and at most four off-diagonal entries equal to $1$ (grid points on the boundary of $\Omega$ have fewer than four interior neighbours). The row sum of absolute values of the off-diagonal entries is \begin{align*} r_k := \sum_{\ell \neq k} |A_{k\ell}| \le 4. \end{align*} By the [Gershgorin Circle Theorem](/theorems/1365), every eigenvalue $\lambda$ satisfies $|\lambda - a_{kk}| \le r_k$ for some $k$, i.e., $|\lambda + 4| \le 4$. Since $\lambda$ is real (established above), this gives \begin{align*} -4 - 4 \le \lambda \le -4 + 4, \quad \text{i.e.,} \quad -8 \le \lambda \le 0. \end{align*} In particular, $\lambda > 0$ is impossible. [guided] The [Gershgorin Circle Theorem](/theorems/1365) states that every eigenvalue of an $n \times n$ matrix lies in at least one of the discs $\Gamma_k = \{z \in \mathbb{C} : |z - a_{kk}| \le r_k\}$, where $r_k = \sum_{\ell \neq k} |a_{k\ell}|$ is the row sum of off-diagonal absolute values. For our matrix $A$, every disc is centred at $a_{kk} = -4$ and has radius $r_k \le 4$. We verify the hypothesis of the Gershgorin theorem: $A \in \mathbb{R}^{N \times N}$ (where $N = m^2$), which is certainly in $\mathbb{C}^{N \times N}$. Since $A$ is real symmetric, all eigenvalues are real, so the Gershgorin discs restrict to real intervals. The disc $|z + 4| \le 4$ intersected with the real line gives $[-8, 0]$. This eliminates all positive eigenvalues but does not yet eliminate $\lambda = 0$. [/guided] [/step] [step:Rule out $\lambda = 0$ by propagating the maximum-modulus condition to the boundary] Suppose for contradiction that $\lambda = 0$ is an eigenvalue with eigenvector $x \neq 0$, so $Ax = 0$. Choose an index $k$ with $|x_k| = \max_\ell |x_\ell| > 0$. The $k$-th row of $Ax = 0$ reads \begin{align*} -4 x_k + \sum_{\ell \sim k} x_\ell = 0, \end{align*} where the sum is over the grid neighbours $\ell$ of $k$ (there are at most four such neighbours). Rearranging: \begin{align*} x_k = \frac{1}{4} \sum_{\ell \sim k} x_\ell. \end{align*} [claim:Every grid neighbour of $k$ satisfies $x_\ell = x_k$] Taking absolute values: \begin{align*} |x_k| = \frac{1}{4}\left|\sum_{\ell \sim k} x_\ell\right| \le \frac{1}{4} \sum_{\ell \sim k} |x_\ell| \le \frac{1}{4} \cdot 4 \cdot |x_k| = |x_k|, \end{align*} where the second inequality uses $|x_\ell| \le |x_k|$ for all $\ell$ (by the maximality of $|x_k|$) and the fact that $k$ has at most $4$ neighbours. Since the first and last expressions are equal, both inequalities must be equalities. Equality in the triangle inequality $\left|\sum_{\ell \sim k} x_\ell\right| = \sum_{\ell \sim k} |x_\ell|$ requires all the $x_\ell$ (for $\ell \sim k$) to have the same complex argument. Equality in $\sum_{\ell \sim k} |x_\ell| = 4|x_k|$ when there are exactly $4$ neighbours requires $|x_\ell| = |x_k|$ for each $\ell \sim k$. Moreover, since $x_k = \frac{1}{4}\sum_{\ell \sim k} x_\ell$ and all summands have the same argument as $x_k$ and the same absolute value $|x_k|$, we conclude $x_\ell = x_k$ for each neighbour $\ell$ of $k$. When $k$ has fewer than $4$ interior neighbours (i.e., $k$ is adjacent to the boundary), the bound $\sum_{\ell \sim k} |x_\ell| \le 4|x_k|$ is strict: there are at most $3$ terms, each bounded by $|x_k|$, giving $\sum_{\ell \sim k} |x_\ell| \le 3|x_k| < 4|x_k|$. But the equation demands $|x_k| = \frac{1}{4}|\sum_{\ell \sim k} x_\ell|$, which requires the sum to have absolute value $4|x_k|$, yet at most $3$ terms each of absolute value $\le |x_k|$ can sum to at most $3|x_k|$. This is a contradiction, so $k$ cannot be a boundary-adjacent grid point. [/claim] [proof] The computation above establishes both assertions: equality forces $x_\ell = x_k$ for all neighbours when there are exactly $4$, and the bound fails when there are fewer than $4$ neighbours. [/proof] We now propagate. Starting from the index $k$ of maximum $|x_k|$, the claim shows $k$ must have $4$ interior neighbours (so $k$ is not adjacent to $\partial\Omega$), and each neighbour $\ell$ satisfies $x_\ell = x_k$, so $|x_\ell| = |x_k| = \max |x_j|$. Apply the same argument to each such $\ell$: it too must have $4$ interior neighbours with value $x_k$. Since the grid is finite and connected, this propagation eventually reaches a grid point $p$ adjacent to the boundary $\partial\Omega$. At such a point, at least one of the four stencil neighbours lies on $\partial\Omega$, where the Dirichlet condition $u = 0$ applies. This means $p$ has fewer than $4$ interior neighbours. But the claim shows this is impossible when $|x_p| = \max |x_j| > 0$. This contradiction shows $x = 0$, contradicting the assumption that $x$ is an eigenvector. Therefore $\lambda = 0$ is not an eigenvalue of $A$. [guided] The Gershgorin bound showed $\lambda \in [-8, 0]$, eliminating positive eigenvalues. To show $A$ is negative definite (not merely negative semidefinite), we must exclude $\lambda = 0$. The strategy is a discrete maximum principle argument. If $Ax = 0$, the equation at each grid point $k$ reads $x_k = \frac{1}{4}\sum_{\ell \sim k} x_\ell$ — that is, each value equals the average of its neighbours. This is the discrete analogue of the mean value property for harmonic functions. At a point $k$ where $|x_k|$ is maximal, the averaging property forces all neighbours to have the same value. Why? Because $x_k$ is a weighted average (with equal weights $\frac{1}{4}$) of values that are each bounded in absolute value by $|x_k|$. An average of quantities bounded by $M$ can equal $M$ only if every quantity equals $M$ (with the same sign/argument). Precisely: \begin{align*} |x_k| = \frac{1}{4}\left|\sum_{\ell \sim k} x_\ell\right| \le \frac{1}{4}\sum_{\ell \sim k} |x_\ell| \le \frac{n_k}{4}|x_k|, \end{align*} where $n_k \le 4$ is the number of interior neighbours of $k$. For $|x_k| > 0$, this requires $n_k = 4$ (the point $k$ must have all four neighbours in the interior) and $|x_\ell| = |x_k|$ for each neighbour $\ell$. This "equality propagation" proceeds like a chain reaction through the grid. Starting from any maximum-modulus point $k$, all its neighbours must also be at maximum modulus, then all *their* neighbours, and so on. The grid graph (interior points connected by the stencil) is connected — one can reach any interior point from any other by a sequence of horizontal and vertical steps. So the propagation reaches every interior point, including those adjacent to the boundary. But a point adjacent to the boundary has a stencil neighbour on $\partial\Omega$ where $u = 0$. Such a point has only $3$ (or $2$ or $1$) interior stencil neighbours, so $n_k < 4$, and the inequality $|x_k| \le \frac{n_k}{4}|x_k|$ with $n_k < 4$ gives $|x_k| \le \frac{3}{4}|x_k|$, which forces $|x_k| = 0$. Since this point was forced to have $|x_k| = \max|x_j|$, we get $\max|x_j| = 0$, i.e., $x = 0$. This contradicts $x$ being an eigenvector, so $\lambda = 0$ is not an eigenvalue. [/guided] [/step] [step:Conclude that $A$ is negative definite] Combining the results of the previous steps: - $A$ is real symmetric (from the stencil structure), so all eigenvalues are real. - Every eigenvalue satisfies $-8 \le \lambda \le 0$ (by Gershgorin). - $\lambda = 0$ is not an eigenvalue (by the propagation argument). Therefore every eigenvalue of $A$ satisfies $\lambda < 0$. Since $A$ is a real symmetric matrix with all eigenvalues strictly negative, $A$ is negative definite: for all $x \in \mathbb{R}^N \setminus \{0\}$, \begin{align*} x^\top A x = \sum_{i=1}^N \lambda_i |\langle x, q_i\rangle|^2 < 0, \end{align*} where $\lambda_1, \ldots, \lambda_N < 0$ are the eigenvalues and $q_1, \ldots, q_N$ the corresponding orthonormal eigenvectors. The inequality is strict because $x \neq 0$ implies at least one coefficient $\langle x, q_i \rangle \neq 0$, and all $\lambda_i$ are strictly negative. [/step]