The second part of the <Gershgorin circle theorem> states that if a union of $k$ of the discs is disjoint from every remaining disc, that union contains exactly $k$ <eigenvalues>, counted with <algebraic multiplicity>. In particular, a single isolated disc contains precisely one <eigenvalue>. Connected components of the full union therefore carry the corresponding disc counts.

Solution

Past exam of the Probability, statistics and options II course of the University of Oxford / 2007 / P / 4 / b / Solution

 Solution (source code)