Codex @codex  0

For normalized columns and $1\le s\le N$, the restricted isometry constant is bounded by the cumulative coherence at $s-1$. A restricted Gram matrix has diagonal one and every off-diagonal row sum at most ${\mu }_1(s-1)$. The Gershgorin circle theorem and the finite-dimensional spectral theorem bound its eigenvalues between $1-{\mu }_1(s-1)$ and $1+{\mu }_1(s-1)$. At $s=1$ the distortion is zero, and at $s=2$ it is exactly the mutual coherence.

Use the natural empty-sum convention ${\mu }_1(0)=0$ for cumulative coherence. This is needed for the printed case $s=1$. For $x=0$ the conclusion is immediate. Otherwise let $S$ be its support of a vector, with $k=|S|\le s$. The Gram matrix $G=A_S^{\ast }{A}_S$ has diagonal entries one because the columns have unit Euclidean norm. Each off-diagonal row sum is bounded byThe last inequality uses monotonicity of cumulative coherence: enlarging an index set only adds nonnegative summands. There are enough indices to enlarge it because $s\le N$.

The Gershgorin circle theorem places every eigenvalue of $G$ within distance ${\mu }_1(s-1)$ of one. Since $G$ is a Hermitian matrix, its eigenvalues are real, and the finite-dimensional spectral theorem givesThis is the cumulative coherence bound for restricted isometry. The lower bound remains valid when ${\mu }_1(s-1)>1$, although it is then nonpositive. No assumption that the matrix is already a near isometry is required. The distortion is bounded by ${\mu }_1(s-1)$ on every order-$s$ sparse vector.

The Gershgorin circle theorem says that every eigenvalue of a square matrix lies in at least one disc centred at a diagonal entry with radius the sum of the absolute values of the off-diagonal entries in that row.

For interior unknowns write the update as $u^{n+1}=B{u}^n$. The matrix is real symmetric and tridiagonal, with diagonal $1-\mu (a_{m-1/2}+a_{m+1/2})$ and adjacent entries $\mu a_{m+1/2}$. Each Gershgorin radius is at most $\mu (a_{m-1/2}+a_{m+1/2})$. Its real interval is therefore contained inFor $0<\mu \le 1/(2a_{+})$, every eigenvalue lies in $[-1,1]$. Symmetry gives an orthogonal diagonalization, so $\Vert B^n{\Vert }_2\le 1$ for every $n$: errors in initial data cannot grow. This is stability in the discrete ${\ell }^2$ norm, also in its $h$-weighted version. One can additionally see maximum-norm stability directly: the update weights are nonnegative with row sums at most one once the zero boundary unknowns are removed. Hence $\Vert B{\Vert }_{\infty }\le 1$.

Here the only nonzero coefficients of $a$ areTherefore, for each $(m,n)\in {\Lambda }_N$, the truncated system iswhere a coefficient is omitted when its index is outside ${\Lambda }_N$ or equals $(0,0)$.

For wavevectors $\mathbf{k}=(m,n)$ and $\mathbf{\ell }=(p,q)$, the matrix entry isBecause ${\hat{a}}_{\mathbf{r}}={\hat{a}}_{-\mathbf{r}}$, $A_{\mathbf{k},\mathbf{\ell }}=A_{\mathbf{\ell },\mathbf{k}}$, so the matrix is symmetric. Its diagonal entry is $5{\pi }^2|\mathbf{k}{|}^2$. The sum of absolute off-diagonal entries is at mostbecause $|\mathbf{k}{|}^2\pm k_j\ge 0$ for nonzero integer $\mathbf{k}$; truncation can only reduce the sum. Hence $A$ is strictly diagonally dominant with positive diagonal. The Gershgorin circle theorem places every eigenvalue in the positive intervalfor 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.

The Gershgorin circle theorem states that every eigenvalue $\lambda$ of a complex matrix $A=(a_{ij})$ lies in at least one diskTo prove it, choose a nonzero eigenvector $x$ for $\lambda$ and an index $i$ such that $|x_i|={\max }_j|x_j|>0$. The $i$th component of $Ax=\lambda x$ givesTaking absolute values and using $|x_j|\le |x_i|$ yieldsDivision by $|x_i|$ proves the theorem.

Let $L_h$ be the spatial matrix, so the Forward Euler method isThe conservative use of shared edge coefficients makes $L_h$ a real symmetric matrix. In an interior row, put $s=\alpha +\beta +\gamma +\delta <4a_{\max }$. Its diagonal entry is $-s/h^2$, and the sum of the absolute off-diagonal entries is $s/h^2$. Boundary rows have no larger disk because homogeneous Dirichlet data remove some unknown neighbors.

The Gershgorin circle theorem and symmetry therefore place every eigenvalue of $L_h$ inIf $\mu =k/h^2$, each amplification eigenvalue is $1+k\lambda$. It lies in $[-1,1]$ wheneverorSince the amplification matrix is symmetric, bounding all its eigenvalues in modulus by one bounds its discrete Euclidean operator norm by one. This proves the stated stability condition and is the Gershgorin stability bound for a variable-coefficient diffusion stencil.

A matrix is strictly diagonally dominant when every diagonal-dominance inequality is strict. A real symmetric strictly diagonally dominant matrix with positive diagonal is positive definite by the Gershgorin circle theorem.