We are given the following problem:

Suppose that $\theta\in\Bbb R$ is not an integer multiple of π. Show that the matrix

$$ A=\left[\begin{array}{rrr} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\\ \end{array}\right] $$ does not have an eigenvector in $\mathbb{R^2}$.

I start by calculating the difference of the original matrix and the product of lambda and the identity:

$$A-\lambda I = \left[\begin{array}{rrr} \cos\theta - \lambda & -\sin\theta\\ \sin\theta & \cos\theta -\lambda\\ \end{array}\right]$$

We can see that the determinant is now:

$\det(A-\lambda I) = 1-2\lambda \cos\theta +\lambda^2$

because $$\cos\theta^2+\sin\theta^2 =1$$

So $\theta=2\pi n - \cos^{-1}(\frac{\lambda^2 + 1}{2\lambda})$

Is this the right approach? What do I do next?

  • 1
    $\begingroup$ you are looking for roots of $\text{det}(A-\lambda I)$ (you may want to review any text you're using). $\endgroup$
    – yoyo
    May 17 '15 at 0:12
  • $\begingroup$ solve the quadratic equation. $\endgroup$
    – abel
    May 17 '15 at 0:18
  • $\begingroup$ Please use \sin \cos and \det as opposed to sin cos and det. It will look better that way. $\endgroup$ May 17 '15 at 0:19
  • 1
    $\begingroup$ The question is now, "when does the equation $$ 1-2\lambda \cos\theta +\lambda^2 = 0 $$ have real solutions? (for $\lambda$)". $\endgroup$ May 17 '15 at 0:20
  • 1
    $\begingroup$ Solve for $\lambda$, not for $\theta$ $\endgroup$ May 17 '15 at 0:24

The equation you have can be solved using just the quadratic formula.

Solve $\lambda$ in $1-2\lambda\cos(\theta) + \lambda^2 = 0$ with:

$$\frac{2\cos(\theta) \pm 2\sqrt{\cos^2(\theta)-1}}{2} = \cos(\theta) \pm i\sin(\theta)$$

This is Euler's formula, which is equal to $e^{\pm i\theta}$, and always has a modulus of $1$.

  • $\begingroup$ This is a fine answer. I just hope OP also realizes that the result is geometrically obvious, since the linear transformation is just rotation by $\theta$. $\endgroup$ May 17 '15 at 0:35
  • $\begingroup$ I agree. Even though these are the eigenvalues, it is more important to note that they both have a modulus of $1$, implying that a rotation will never have any scaling effect. $\endgroup$
    – Rellek
    May 17 '15 at 0:40

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