TIFR-2014-Maths-A-9

Question

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.

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

Answer 1

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),

Comments

Why eigen values are present at only  θ = 0 or π? Please Explain??