Past exam of the probability, statistics and options I course of the University of Oxford / 2011 / P / 1 / a / Solution

The Gershgorin circle theorem states that every eigenvalue of $A$ lies in at least one discFor a nonzero complex eigenvector $v$, choose an index $i$ with $|v_i|={\max }_j|v_j|>0$. The $i$th row of $Av=\lambda v$ givesDividing by $|v_i|$ proves membership in $D_i$. This is valid for real matrices even when the eigenvalues are complex.