 Solution

For a complex $n\times n$ matrix $B=(b_{ij})$, putThe Gershgorin circle theorem states thatTo prove it, let $Bv=\lambda v$ with $v\ne 0$, and choose $i$ for which $|v_i|={\max }_j|v_j|>0$. The $i$th coordinate of the eigenvalue equation givesThe triangle inequality, followed by division by $|v_i|$, givesThus $\lambda \in D_i$, proving the assertion for every eigenvalue.

The second theorem, Gershgorin eigenvalue counting, says that if a union of $k$ indexed Gershgorin discs is disjoint from the remaining $n-k$ Gershgorin discs, that union contains exactly $k$ eigenvalues, counted with algebraic multiplicity. In particular,Discs that touch are not disjoint; overlapping Gershgorin discs cannot individually be assigned one eigenvalue each.