Past exam of the options II course of the University of Oxford / 2007 / P / P4 / e / Solution

Put $R=ϵ+\delta$. Only the first row disc changes: it now has radius $R$. The other discs keep radii $r_i=1/4+ϵ$ for $2\le i\le n-1$ and $r_n=1/4$. They remain mutually disjoint. As $R$ grows, the first disc joins an initial segment of these discs. Define $m$ as the largest index joined to it, or one if no other disc is joined. The joining condition for index $i$ isFor an interior index this starts at $\delta =i-5/4-2ϵ$, and for the final index at $\delta =n-5/4-ϵ$. Until its joining threshold, disc $i$ contains a unique real eigenvalue and retains the bound $|{\lambda }_i-i|\le r_i$.

After the first $m$ discs have joined, the second theorem places exactly $m$ eigenvalues in their union and exactly one in each remaining disc. The cluster is bounded, for instance, byand by $|z-1|\le R$ when $m=1$. The union of the actual discs is a sharper enclosure. Once discs overlap, one cannot assign one eigenvalue to each individual disc merely by its centre; the theorem counts the whole separated cluster.