Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 4 / 39C / b / Solution

The given vectors satisfyand $y_k,y_{k+1}$ are independent, so their span is invariant under $A$. On that plane the characteristic polynomial is ${\lambda }^2-2\lambda +2$. ThereforeIndeed $A{v}_{\pm }=y_{k+2}-(1\mp i)y_{k+1}=(1\pm i)v_{\pm }$. No third eigenvalue is determined by these data.

The displayed $x_k=(1,1,1)$ has norm $\sqrt{3}$, although a large-$k$ normalized iterate should have norm one. This harmless source normalization is resolved by dividing all three displayed vectors by $\sqrt{3}$; the recurrence, eigenvalues and eigenvector directions are unchanged.