Past exam of the options I course of the University of Oxford / 2009 / P / P1 / Solution

The Gershgorin circle theorem states that every complex eigenvalue $\lambda$ of $A=(a_{ij})$ belongs to at least one closed discTo prove it, take an eigenvector $v\ne 0$ and choose $i$ with $|v_i|={\max }_j|v_j|>0$. The $i$th row of $Av=\lambda v$ givesDivision by $|v_i|$ proves the assertion. The proof works for real matrices even when their eigenvalues or eigenvectors are complex.

For $B+\mu I$, the three discs have centres $2+\mu ,3+\mu ,4+\mu$ and radii $2,3,4$. Every point in their union has real part at least $\mu$. Thus for $\mu >0$ no disc contains zero, and zero cannot be an eigenvalue. Hence $B+\mu C$ is nonsingular.