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

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.

For $A+\gamma B$ with $\gamma \ge 0$, the three Gershgorin discs have positive real centres $a+\gamma$, $2c+2\gamma$, $6b+6\gamma$, and radii $b+c$, $a+b+\gamma$, $a+c+\gamma$. Centre minus radius is respectivelyThus none of the discs contains zero. By the Gershgorin circle theorem, zero is not an eigenvalue, so $A+\gamma B$ is nonsingular for every $\gamma \ge 0$. Consequently