Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 2 / 11 / a / Solution

Solving $(A-4I)\mathbf{v}=0$ gives $v_2=0$ and $v_1=v_3$. A convenient normalized eigenvector isThe displayed nonzero vector proves that four is an eigenvalue. Every nonzero multiple represents the same eigendirection.