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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?

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

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