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

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.

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..

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 $T(\alpha+\beta)=T(\alpha)+T(\beta)$.