Question

Prove that the rotation of the circle is not mixing the Lebesgue
measure for any angle.

Answer 1

To prove that the distance from the origin to a point P is invariant under rotation of the coordinate system, we use a rotation matrix and trigonometric identities to show that the squared distance before and after the rotation remains the same.

Step-by-step explanation:

To prove that the distance of point P to the origin is invariant under rotations of the coordinate system, consider the initial coordinates of point P as (x, y). After a rotation by any angle, the new coordinates of P, say (x', y'), can be obtained by the rotation matrix:

`R(\theta) = \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}`

The new coordinates (x', y') are given by:

`x' = x\cos(\theta) - y\sin(\theta)y' = x\sin(\theta) + y\cos(\theta)`

The square of the distance from the origin to P is initially
`x^2 + y^2.`rotation, the square of the distance to P becomes
`x'^2 + y'^2:`

`x'^2 + y'^2 = (x\cos(\theta) - y\sin(\theta))^2 + (x\sin(\theta) + y\cos(\theta))^2`

Expanding, we use trigonometric identities to simplify and find that:

`x'^2 + y'^2 = x^2\cos^2(\theta) + y^2\sin^2(\theta) + 2xy\sin(\theta)\cos(\theta) - 2xy\sin(\theta)\cos(\theta) + x^2\sin^2(\theta) + y^2\cos^2(\theta)`

Which simplifies to:

`x'^2 + y'^2 = x^2(\cos^2(\theta) + \sin^2(\theta)) + y^2(\sin^2(\theta) + \cos^2(\theta))`

And since
`\(\cos^2(\theta) + \sin^2(\theta) = 1\)`

`x'^2 + y'^2 = x^2 + y^2`

Thus, the distance from the origin to point P remains the same after the rotation, proving the invariance under rotations of the coordinate system. This principle is applied in the context of Lebesgue measure to show that the measure is not mixing for rotations since distances are preserved.