Answer C
p(λ) = (cos θ − λ) 2 + sin2 θ
λ2 − 2λ cos θ +1=0
λ = cos θ ± √ (cos2 θ − 1)
λ = cos θ ±√ (− sin2 θ) = cos θ ± isin θ
the operator A(θ) only has eigenvalues when θ = 0 or θ = π
i.e. λ =1,-1 when θ = 0 and π
As θ = 0 is not included in (0,2 π) range
we have A(θ) has eigenvector in R2 for exactly one value of θ = π where θ = (0,2π),