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

Put $c=\cos \theta$, $s=\sin \theta$. The characteristic polynomial is $(c-\lambda {)}^2-s^2$, givingWhen $s=0$, these eigenvalues coincide and $A=cI=\pm I$.

Adding and subtracting the two coordinates shows that $A(1,1{)}^T=(c+s)(1,1{)}^T$ and $A(1,-1{)}^T=(c-s)(1,-1{)}^T$. The normalized eigenvectors can therefore be chosen asTheir inner product is $(1-1)/2=0$, and their squared norms are one, proving orthonormality. If $s=0$, every nonzero vector is an eigenvector; this chosen orthonormal basis still works.

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.