 Solution

The second theorem, or Gershgorin eigenvalue counting, states: 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.