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Assume $n\ge 3$, so the corner entry is separate from the superdiagonal. For $A(\delta )$, the row radii areThus every eigenvalue lies in the union ${\bigcup }_{i}D(i,r_i)$. The discs with centers $2,\dots ,n$ remain mutually disjoint. The expanding disc centered at one meets the disc centered at $j$ exactly when $j-1\le \delta +ϵ+r_j$.

Let $I_{\delta }=\{1\}\cup \{j\ge 2:j-1\le \delta +ϵ+r_j\}$. Then the union of discs indexed by $I_{\delta }$ contains exactly $|I_{\delta }|$ eigenvalues; every other disc contains exactly one. Every isolated eigenvalue is real by conjugation and obeys $|\lambda -j|\le r_j$. Once discs merge, the second Gershgorin circle theorem gives a bound and count for their whole component, rather than one eigenvalue per original disc. Eigenvalue branches coming from that component stay in its union, but may no longer be distinguished by individual discs. These statements describe the bounds through all merger thresholds as $\delta$ increases.

For each row, a Gershgorin disc has centre $a_{jj}=-2$ and radius $r_j={\sum }_{k\ne j}|a_{jk}|$. Here $r_1=r_3=1$ and $r_2=2$, so the Gershgorin circle theorem givesSince $A$ is a real symmetric matrix, its eigenvalues are real. Intersecting the union of these Gershgorin discs with the real axis yieldsThis bound permits zero but does not assert a zero eigenvalue. The fixed-end Dirichlet boundary conditions exclude it: for any nonzero real vector $v=(v_1,v_2,v_3{)}^T$, extend by $v_0=v_4=0$ and calculateIndeed, equality would force every adjacent value to agree with the zero endpoint. Thus $-A$ is a symmetric positive-definite matrix, $A$ is a negative-definite matrix, and the eigenvalues satisfy $-4\le \lambda <0$. The same argument works for any number of interior grid points, with prefactor $c^2/h^2>0$ in the sum. The fixed string has no zero-frequency rigid translation.

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.