Real eigenvalues of a rotation matrix

Question

Suppose the rotation matrix $$\pmatrix{ \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta}$$ has real eigenvalues. Then what would the value of $\theta$ be?

I know that the eigenvalues of this matrix are $\{e^{\pm i\theta}\}$. To derive the condition under which the eigenvalues are real, I found the characteristic function and set the discriminant $\Delta=b^2-4ac$ $>0$. From this, I got $\sin^{2}\theta <0$. Since sine is a square term, how can it be negative?! Am I right?

Answer 1

The only thing wrong with your approach is that the eigenvalues can be real if the discriminant is zero, too. $\sin^2 \theta < 0$ has no solution, but $\sin^2 \theta = 0$ gives $\theta = 0$ and $\theta = \pi$. The first gives you the identity matrix, the second its opposite.

Answer 2

Using Euler's formula, the eigenvalues, $e^{\pm i\theta}$ will be real iff $\sin\theta=0$. Thus $\theta=k\pi\,,k\in \Bbb Z$.

Answer 3

The characteristic polynomial of $$ A=\left[\begin{array}{rr} \cos\left(\theta\right) & \sin\left(\theta\right) \\ -\sin\left(\theta\right) & \cos\left(\theta\right) \end{array}\right] $$ is given by $$ \chi_A(t) = \left\lvert\begin{array}{rr} t - \cos\left(\theta\right) & -\sin\left(\theta\right) \\ \sin\left(\theta\right) & t - \cos\left(\theta\right) \end{array}\right\rvert = t^{2} - 2 \, \cos\left(\theta\right) t + 1 $$ To ensure that this polynomial has real roots, we need $$ (-2\,\cos(\theta))^2-4\cdot(1)\cdot(1)\geq0 $$ which is equivalent to $$ \cos^2(\theta)\geq 1 $$ Of course, this is only possible if $\cos(\theta)=\pm1$. Thus our desired values of $\theta$ are $\theta=k\cdot\pi$ for $k\in\Bbb Z$.

Answer 4

Using the characteristic equation seems like a crutch - better to think about what it all means, in my opinion.

Say that matrix is $A_\theta$ and $x$ is an eigenvector. Then $x\ne0$ and $A_\theta x=\lambda x$. This says that rotating $x$ through an angle $\theta$ results in a vector parallel to $x$ (where I'm taking $y$ and $-y$ to be "parallel".) That's clearly impossible unless $\theta $ is $0$ or $\pi$ mod $2\pi$. (Which says in turn that $A_\theta=\pm I$, so $\lambda=\pm1$.)