How do I compute the eigenvector of $\begin{bmatrix}\cos\theta&\sin\theta\\\sin\theta&-\cos\theta\end{bmatrix}$?
$\begin{bmatrix}\cos\theta&\sin\theta\\\sin\theta&-\cos\theta\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x\cos\theta+y\sin\theta\\x\sin\theta -y\cos\theta\end{bmatrix}$
The I got this system of equations:
${x\cos\theta+y\sin\theta=\lambda x\\x\sin\theta -y\cos\theta=\lambda y}\implies$
$\det\begin{bmatrix}\lambda-\cos\theta&-\sin\theta\\\sin\theta&-\lambda-\cos\theta\end{bmatrix}=0\implies -\lambda^2+\cos^2\theta+\sin^2\theta=0\implies\lambda=1 $
Questions:
1) How do I compute $\lambda$?
2) Is it possible to get a simple value for the given matrix $A$?
Best Answer
\begin{align} & \det\left(\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} - \lambda\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right) = (\cos\theta-\lambda)^2 + \sin^2\theta \\[10pt] = {} & \lambda^2 - 2\cos\theta + 1 = 0 \\[10pt] &\text{if and only if } \lambda = \cos\theta \pm i\sin\theta. \end{align}