Proving that rotations are linear

Question

Problem

Let $T$ rotates every point through the same angle $\phi$ about the origin. That is, $T$ maps a point with polar coordinates $(r,\theta)$ onto the point with polar coordinates $(r,\theta+\phi)$, where $\phi$ is fixed. Also, $T$ maps $0$ onto itself. Is $T$ linear?

$T$ is linear but I am not able to prove it.

Attempt

\begin{align*} &\, T(r_1,\theta_1) + T(r_2,\theta_2) \\ =&\, (r_1,\theta_1+\phi) + (r_2,\theta_2+\phi) \\ =&\, (r_1+r_2, \theta_1+\theta_2+2\phi) \\ \neq&\, T((r_1,\theta_1),(r_2,\theta_2)) \end{align*} What I am missing ?

Answer 1

You must show linearity of $T$ in regard of the vectors: E.g. $$ T(\lambda u + \mu v) = \lambda T(u) + \mu T(v) $$ where $u$ and $v$ are arbitrary vectors of your vector space and $\lambda$ and $\mu$ arbitrary scalars of the field.

Answer 2

Note that\begin{align}T(r\cos\theta,r\sin\theta)&=(r\cos(\theta+\phi),r\sin(\theta+\phi))\\&=(r\cos(\theta)\cos(\phi)-r\sin(\theta)\sin(\phi),r\sin(\theta)\cos(\phi)+r\cos(\theta)\sin(\phi))\end{align}and that therefore$$T(a,b)=(a\cos(\phi)-b\sin(\phi),a\sin(\phi)+b\cos(\phi)).$$Now, it is easy to check that $T$ is linear.

Answer 3

You can prove linearity using just polar coordinates by using two formulas proved by Mark Viola here. Let $\vec r_1$ and $\vec r_2$ denote vectors with magnitudes $r_1$ and $r_2$, respectively, and with angles $\phi_1$ and $\phi_2$, respectively.

Let $\vec r$ denote the vector with angle $\phi$ and magnitude $r$ that denotes the sum of $\vec r_1$ and $\vec r_2$. Thus $$\vec r=\vec r_1+\vec r_2$$

Then, as proved by Mark Viola, $$r=\sqrt{r_1^2+r_2^2+2r_1r_2\cos (\phi_2-\phi_1)} \tag 1$$ and $$\phi =\phi_1+\operatorname{arctan2}\left(r_2\sin(\phi_2-\phi_1),r_1+r_2\cos(\phi_2-\phi_1)\right) \tag 2$$ where the function arctan2(y,x) is described in this article.

Let $T$ be the rotation transformation. We need to prove that given two vectors $\vec r_1$ and $\vec r_2$ and a scalar $\lambda$ $$\bbox[5px,border:2px solid #C0A000]{T(\lambda \vec r_1 + \vec r_2) = \lambda T(\vec r_1) + T(\vec r_2)}$$ Notice that for a given vector $\vec r$ with polar coordinates $(r, \phi)$ we have that $$T(\vec r) = (r, \phi + \theta) \tag 3$$ where $\theta$ is the rotation angle. So the magnitud of the vector is not modified by this transformation.

Also note that $\lambda (r, \phi) = (\lambda r, \phi) \tag 4$ which means that multiplying a vector by a scalar changes only the magnitude of the vector but not the angle.

Lets check the polar coordinates of $\lambda \vec r_1 + \vec r_2$ and use equations $(1)$, $(2)$ and $(3)$ $$\eqalign{ \lambda \vec r_1 + \vec r_2 &= \lambda (r_1, \phi_1) + (r_2, \phi_2) \\ &= (\lambda r_1, \phi_1) + (r_2, \phi_2) \\ &= \left( \sqrt{(\lambda r_1)^2 + r_2^2+2r_1r_2\cos (\phi_2-\phi_1)}, \phi_1+\operatorname{arctan2}\left(r_2\sin(\phi_2-\phi_1),\lambda r_1+r_2\cos(\phi_2-\phi_1)\right) \right ) }$$ So, by $(3)$ $$\eqalign{ T(\lambda \vec r_1 + \vec r_2) &= \left( \sqrt{(\lambda r_1)^2 + r_2^2+2\lambda r_1r_2\cos (\phi_2-\phi_1)}, \phi_1+\operatorname{arctan2}\left(r_2\sin(\phi_2-\phi_1),\lambda r_1+r_2\cos(\phi_2-\phi_1)\right) + \theta \right ) }$$

Now lets check $\lambda T(\vec r_1) + T(\vec r_2)$ and use equations (1) and (2) $$\eqalign{ \lambda T(\vec r_1) + T(\vec r_2) &= \lambda (r_1, \phi_1 + \theta) + (r_2, \phi_2 + \theta) \\ &= (\lambda r_1, \phi_1 + \theta) + (r_2, \phi_2 + \theta) \\ &= \left( \sqrt{(\lambda r_1)^2 + r_2^2+2\lambda r_1r_2\cos ((\phi_2 + \theta) -(\phi_1 + \theta))}, ( \phi_1 + \theta) + \operatorname{arctan2} \left(r_2\sin((\phi_2 + \theta) - (\phi_1 + \theta)),\lambda r_1+r_2\cos((\phi_2 + \theta) - (\phi_1 + \theta))\right) \right ) \\ &= \left( \sqrt{(\lambda r_1)^2 + r_2^2+2\lambda r_1r_2\cos (\phi_2 - \phi_1)}, \phi_1 + \operatorname{arctan2}\left(r_2\sin(\phi_2-\phi_1),\lambda r_1+r_2\cos(\phi_2 - \phi_1)\right) + \theta \right ) \\ }$$

So $$T(\lambda \vec r_1 + \vec r_2) = \lambda T(\vec r_1) + T(\vec r_2) \tag*{$\blacksquare$}$$

Answer 4

Here is a proof using words. A parallelogram with a vertex at origin after rotation around origin still remains a parellelogram (not a general quadrilateral). A line through origin is mapped by a rotation to another line through origin.

This is the meaning behind the condition $T(u+v)= T(u) + T(v)$ and $T(cv)=cT(v)$ expected of a linear transformation. (assuming of course that in the plane the sum of vectors $u,v$ is the vector that is the the fourth vertex of the parallelogram formed by $0,u$ and $v$. )

Answer 5

Such a rotation satisfies $|Tx|=|x|$ and $T(\alpha x)=\alpha Tx$. And $Tx\cdot Ty=x\cdot y$, where "$\cdot$" is the dot product on $\mathbb{R}^2$. That is, rotation preserves length and preserves the angle between two vectors. Therefore,

\begin{align} &|T(x+y)-Tx-Ty|^2 \\ =& |T(x+y)|^2+|Tx|^2+|Ty|^2 \\ &-2T(x+y)\cdot Tx-2T(x+y)\cdot Ty-2Tx\cdot Ty \\ = &|x+y|^2+|x|^2+|y|^2 \\ &-2(x+y)\cdot x - 2(x+y)\cdot y - 2x\cdot y \\ =&|(x+y)-x-y|^2 =0. \end{align}

Therefore $T(x+y)=Tx+Ty$. We knew $T(\alpha x)=\alpha Tx$. So $T$ is linear.