Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 1 / Section B / 8 / c

Computing the characteristic polynomial with the convention $\det (\lambda I-H)$ givesThus the three eigenvalues areThey can also be checked directly on the respective eigenvectors $(1,0,-1{)}^T$, $(1,-1,1{)}^T$, and $(1,2,1{)}^T$.

For the lowest eigenvalue, $-1$, solving $(H+I)\mathbf{v}=0$ gives $v_2=0$ and $v_3=-v_1$. A unit eigenvector is thereforeAny nonzero scalar multiple is an equally valid eigenvector.