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

The Gershgorin circle theorem states that every eigenvalue of $A=(a_{ij})\in {\mathbb{R}}^{n\times n}$ belongs to at least one of the closed Gershgorin discsThe eigenvalue may be complex even though the matrix is real. To prove the theorem, choose a nonzero complex eigenvector $v$ with $Av=\lambda v$ and an index $i$ at which $|v_i|$ is maximal. Then $|v_i|>0$, and the $i$th row givesThe triangle inequality yieldsThus $\lambda \in D_i$, proving the assertion without any symmetry assumption.