Rigid motion on $\mathbb{R}^2$ which fixes the origin is linear Let $V=\mathbb{R}^2$ be an inner product space with the standard inner product, and let $T$ be a rigid motion of $V$. Suppose $T(0)=0$, prove that $T$ is linear.
(A rigid motion of an inner product space is a function (not necessarily linear) $f:V \to V$, such that $||f\alpha-f\beta||=||\alpha-\beta||$.)
It is obvious that $T$ preserves the inner product (T is unitary), but I don't know how to prove that $T$ is linear. Thanks in advance 
 A: Since $T$ is rigid,
$$\|T(x)\| = \|T(x)-T(0)\| = \|x-0\| = \|x\|.$$
So indeed $T$ is an isometry. However we also know that $\|x-y\| = \|T(x) - T(y)\|$ or equivalently noting that $\|z\|^2 = \langle z,z\rangle$
$$\langle x-y,x-y\rangle = \langle T(x)-T(y),T(x)-T(y)\rangle.$$
Expanding both sides gives
$$\langle x,x\rangle - 2\langle x,y\rangle + \langle y,y\rangle = \langle T(x),T(x)\rangle - 2 \langle T(x),T(y)\rangle + \langle T(y),T(y)\rangle.$$
Making use of the isometric nature of $T$, we can cancel some terms to give
$$\langle x,y\rangle = \langle T(x),T(y)\rangle.$$
So $T$ also preserves inner products. Let's let $y \to y+z$ then our previous expression gives
$$\langle x,y+z\rangle = \langle T(x),T(y+z)\rangle.$$
Expanding the left and side gives
$$\langle x,y\rangle + \langle x,z\rangle = \langle T(x),T(y+z)\rangle.$$
However we know that $T$ preserves inner products so we can actually view the left hand side as being nothing more than $\langle T(x),T(y)\rangle + \langle T(x),T(z)\rangle.$ Rewriting..
$$\langle T(x),T(y)+T(z)\rangle = \langle T(x),T(y+z)\rangle.$$
Moving terms to one side gives
$$\langle T(x),T(y+z)-T(y)-T(z)\rangle = 0.$$
Since this holds for all choices of $x$, $y$ and $z$ we must conclude that
$$T(y+z) - T(y) - T(z) = 0.$$
Showing scalar multiplication is not too different in nature.
A: Hint: Prove that any rigid motion $T$ is a composition of a linear translation and an orthogonal transformation. Then $T(0) = 0$ implies that the translation is the  $0$ operator so $T$ must be an orthogonal transformation hence linear..
A: This is a rather geometrical way of doing this, not really linear algebra, but anyway.
Notice this interesting thing in $\mathbb{R}^{2}$: if $A,B$ and length $r<AB$ then there exist an unique $C$ such that $AC=r$ and $BC=AB-r$.
Fix an $v\not=0$ then $T(v)\not=0$. Consider any $0<\alpha<1$. Apply $A=0$ and $B=T(v)$ and $r=||\alpha v||$ with the above result to show that there is an unique place $T(\alpha v)$ could be. Show that $\alpha T(v)$ is a possible value for $T(\alpha v)$, hence with uniqueness prove it is the only value.
Now consider $\alpha=-1$. Prove similarly, though now $A=T(\alpha)$ and $B=-T(\alpha)$ and $r=||\alpha||$, and now you are trying to show that $0$ must be the midpoint of $AB$.
Combine the first result prove scalar property for all $-1\leq\alpha\leq 1$. Now for $\alpha>1$, simply apply this very result to $\alpha v$, with the scalar being $\frac{1}{\alpha}$. This show that scalar multiplication work for any $\alpha$.
To handle sum, use the very basic property: $\alpha+\beta$ is the midpoint of $2\alpha$ and $2\beta$. Apply $A=T(2\alpha)$ and $B=T(2\beta)$ and $r=||2\alpha-2\beta||$ to show that $T(\alpha+\beta)$ must be the midpoint of $T(2\alpha)$ and $T(2\beta)$. Use the scalar property already proven, it is also the midpoint of $2T(\alpha)$ and $2T(\beta)$ which show that $T(\alpha+\beta)=T(\alpha)+T(\beta)$.
A: Let $\alpha, \beta \in \mathbb{R}^2, c \in \mathbb{R}$.
$$\begin{aligned}
&\Vert T(c \alpha+\beta)-(cT\alpha+T\beta) \Vert^2 \\
&= \Vert (T(c \alpha+\beta)-T\beta)-cT\alpha\Vert^2 \\
&= \Vert T(c \alpha+\beta)-T\beta\Vert^2+c^2\Vert T\alpha\Vert^2 - 2c\langle T(c \alpha+\beta)-T\beta,T\alpha \rangle \\
&= \Vert (c \alpha+\beta)-\beta\Vert^2+c^2\Vert \alpha\Vert^2 + c[(-2\langle T(c \alpha+\beta),T\alpha \rangle)
-(- 2\langle T\beta,T\alpha \rangle) ] \\
&= 2c^2\Vert \alpha\Vert^2 + c[(\Vert T(c \alpha+\beta)-T\alpha \Vert^2-\Vert T(c \alpha+\beta) \Vert^2 - \Vert T\alpha \Vert^2) \\
&- (\Vert T\beta-T\alpha \Vert^2 - \Vert T\beta\Vert^2- \Vert T\alpha \Vert^2)]\\
&= 2c^2\Vert \alpha\Vert^2 + c[(\Vert (c \alpha+\beta)-\alpha \Vert^2-\Vert c\alpha+\beta \Vert^2) 
- (\Vert \beta-\alpha \Vert^2 - \Vert \beta\Vert^2)]\\
&= 2c^2\Vert \alpha\Vert^2 + c[(-2\langle c\alpha+\beta,\alpha \rangle+\Vert \alpha\Vert^2) - (-2\langle \beta,\alpha \rangle+\Vert \alpha\Vert^2)]\\
&= 2c^2\Vert \alpha\Vert^2 - 2c\langle c\alpha,\alpha \rangle\\
&= 0
\end{aligned}$$
I used multiple times that $\Vert a-b \Vert^2=\Vert a \Vert^2 + \Vert b \Vert^2 - 2\langle a,b \rangle$.
This gives as desired $T(c \alpha+\beta)=cT\alpha+T\beta$.
