Show that if 0 < θ < π then the matrix $\displaystyle B=\left(\begin{array}{cc}cos\theta&-sin\theta\\sin\theta&cos\theta\end{array}\right)$
has no real eigenvalues and consequently no eigenvectors.
$\displaystyle \left(\begin{array}{cc}cos\theta -\lambda &-sin\theta\\sin\theta&cos\theta -\lambda \end{array}\right)$
So we take the determinate of the above matrix to get
$\displaystyle (\cos(\theta)-\lambda)^2+\sin^2(\theta)=0$
$\displaystyle \cos^2(\theta)-2\lambda \cos(\theta)+\lambda^2+\sin^2(\theta)=0$
$\displaystyle \lambda^2-2\cos(\theta) \lambda+1=0$
This will only have real solutions if the discriminate is posative
So the discriminate of this quadratic is
$\displaystyle (-2\cos(\theta))^2-4(1)(1)=4\cos^{2}(\theta)-4=4(\cos^2(\theta)-1)$
This is negative for $\displaystyle 0< \theta < \pi$
So the equation have no Real solutions