The second theorem, or <Gershgorin eigenvalue counting>, states: \b[if a union of $k$ indexed Gershgorin discs is disjoint from all remaining discs, it contains exactly $k$ eigenvalues, counted with algebraic multiplicity.] In particular, an isolated <Gershgorin disc> contains exactly one simple <eigenvalue>. The separation condition is essential; merely drawing overlapping <discs> does not assign one <eigenvalue> to each.

Solution

Past exam of the numerical analysis course of the University of Oxford / 2014 / 3 / b / Solution

 Solution (source code)