Let $T:\Bbb R^3\to\Bbb R^3$ be a linear transformation that has: $$1, \frac{\sqrt{3}+i}{2},\frac{\sqrt{3}-i}{2}$$ as the roots of its characteristic polynomial. Knowing that $T$ is an orthogonal transformation and that the sub-space associated to $1$ is $(1,0,1)$ find $T(0,1,0)$.
I feel like I am missing something here, is there a relation between the sub-spaces associated to $\frac{\sqrt{3}+i}{2},\frac{\sqrt{3}-i}{2}$? Also, are not orthogonal matrixes/transformations supposed to have only eigengalues of module $1$? The solution says the answer is $\frac12\big(\frac{-1}{\sqrt{2}} , \sqrt{3} , \frac{1}{\sqrt{2}}\big)$
Since the eigenvalues are $\exp\left(\pm\frac{\pi i}6\right)$ and since $T$ is a rotation of $\pm\frac\pi6$ radians around the line $l$ passing through the origin and $(1,0,1)$, and since $(0,1,0)$ and $(1,0,1)$ are orthogonal, there are only two possible answers, which are the two vectors that you obtain if you ratate $(0,1,0)$ by $\pm\frac\pi6$ radians around $l$. These vectors are$$\pm\left(-\frac1{2\sqrt2},\frac{\sqrt3}2,\frac1{2\sqrt2}\right).$$