= Solution
If $Av=\lambda v$, then $A^2v=\lambda^2v$. Alternatively, direct <matrix multiplication> gives
$$
B=\begin{pmatrix}1&\sin2\theta\\\sin2\theta&1\end{pmatrix}.
$$
Thus the normalized <eigenvectors> $v_+,v_-$ from part (b) have respective <eigenvalues>
$$
\boxed{(c+s)^2=1+\sin2\theta,\qquad(c-s)^2=1-\sin2\theta.}
$$
If $\sin2\theta=0$, 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.
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