Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 1 / 3 / c / Solution

If $Av=\lambda v$, then $A^{2}v={\lambda }^{2}v$. Alternatively, direct matrix multiplication givesThus the normalized eigenvectors $v_{+},v_{-}$ from part (b) have respective eigenvaluesIf $\sin 2\theta =0$, then $B=I$, and the whole two-dimensional space is its eigenspace. This degeneracy includes cases where the two eigenvalues of $A$ were distinct but opposite.