 Solution

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.