Thread: help with eigenvalue proof

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.

Originally Posted by chuckienz

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

FYI, its eigenvalues are $\displaystyle e^{i\pi}~,~ e^{-i\pi}$

Originally Posted by Moo

FYI, its eigenvalues are $\displaystyle e^{i\pi}~,~ e^{-i\pi}$

$\displaystyle e^{i\pi}=-1=e^{-i\pi} \in \mathbb{R}$

Originally Posted by Gamma

$\displaystyle e^{i\pi}=-1=e^{-i\pi} \in \mathbb{R}$

I meant theta

$\displaystyle e^{i\theta}~,~ e^{-i\theta}$

ahhh! Okay, I should have realized that is what you meant. Sorry