# Rotation operator is positive.

I want to know if my proof is right. the problem is:

Let V be $$\mathbb{R^2}$$, with the standard inner product. If $$\theta$$ is a real number, let T be the linear operator 'rotation through $$\theta$$, $$T_{\theta}(x_1, x_2) = (x_l \cos \theta - x_2 \sin \theta, x_1 \sin \theta + x_2 \cos \theta).$$ For which values of $$\theta$$ is $$T_{\theta}$$ a positive operator?

The conditions for $$T_\theta$$ to be a positive operator are $$T=T^*$$ and $$\langle T\alpha,\alpha\rangle >0$$.

First of all, I computed the associated matrix $$[T]_b=\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$$.

The only way that $$T=T^*$$ is $$\sin\theta=0$$ or $$\theta=\pi k$$ with $$k\in \mathbb{Z}$$.

Then, $$\langle T\alpha,\alpha\rangle=\cos\theta(x_1^2+x_2^2)$$, to $$\langle T\alpha,\alpha\rangle$$ be positive $$\theta$$ must be in the first or fourth quadrant.

Hence, $$\theta=2\pi k$$ with $$k\in \mathbb{Z}$$.

Is anything wrong?

• That's perfect. In other words, only the identity is positive among the rotations. – Berci Nov 17 '18 at 18:47
• Nice, thank you I didn't realize that!! – Bayesian guy Nov 18 '18 at 3:47