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Let $A(\theta)=\begin{pmatrix}
\cos \theta& \sin \theta \\
-\sin \theta& \cos \theta 
\end{pmatrix}$, where $\theta \in (0, 2\pi)$. Mark the correct statement below.

  1. $A(\theta)$ has eigenvectors in $\mathbb{R}^2$ for all $θ \in (0, 2\pi)$ 
  2. $A(\theta)$ does not have an eigenvector in $\mathbb{R}^2$ , for any $θ \in (0, 2\pi)$ 
  3. $A(\theta)$ has eigenvectors in $\mathbb{R}^2$ , for exactly one value of $θ \in (0, 2\pi)$ 
  4. $A(\theta)$ has eigenvectors in $\mathbb{R}^2$ , for exactly $2$ values of $θ \in (0, 2\pi)$
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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),

Answer:

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