Transformation of $\mathbb R^2$ Let $T:\mathbb R^2\to \mathbb R^2$ be s.t. for all $x,y\in\mathbb R^2$, $\mid\mid T(x)-T(y)\mid\mid = \mid\mid x-y\mid\mid.$  Then, 2 questions:


*

*If $T(0)=0$, does it follow that $T$ must be linear?

*Show that $T$ is a translation, then a rotation, and then a reflection through the $x$ axis (in which any of those three could be the "empty" one).
I'm rather stumped by this problem.... for 1 think yes, proven perhaps by taking $x=0$? But I haven't actually been able to completely figure it out. 
And I have no idea how to proceed with 2... perhaps looking at the matrices that generate such movements?  
 A: Let $e_1=[1,0]^t$ and $e_2=[0,1]^t.$ Let $M$ be the $2\times 2$ matrix whose first column is $T(e_1)$ and whose second column is $T(e_2)$. Show that if $T(0)=0,$ then $T(v)=Mv$ for all $v\in\Bbb R^2.$
Now, note that the matrix $M$ above must be invertible (why?), and in particular, has determinant $\pm1$ (why?). Show that: a $2\times2$ real matrix with determinant $1$ represents a (possibly "empty") rotation; a $2\times2$ real matrix with determinant $-1$ represents a reflection about a line through the origin, which can be obtained instead by a (possibly "empty") rotation followed by a reflection about the $x$-axis (why?).
For your second part, put $b=T(0)$ and let $S(v)=T(v)-b$ for all $v$. What can you say about $S$ in light of the first exercise? Use this to show that there is a vector $c$ and a matrix $M$ with determinant $\pm 1$ and a vector $c$ such that $$T(v)=M(v+c).$$

Edit: Let me expand on the first two paragraphs (since that's where the proof gets sticky).
Now, we're assuming that $$T(0)=0\tag{1}$$ and that for all $v,w\in\Bbb R^2$ we have $$\lVert T(v)-T(w)\rVert =\lVert v-w\rVert.\tag{2}$$ Putting $w=0$ in $(2)$, we have by $(1)$ that $$\lVert T(v)\rVert=\lVert v\rVert\tag{3}$$ for all $v\in\Bbb R^2$.
Letting $x\cdot y$ indicate the dot product of $x$ and $y$, it is easily proved that $$\lVert v\rVert^2=v\cdot v$$ for all $v\in\Bbb R^2$ (assuming that our norm is the Euclidean norm), so for all $x,y\in\Bbb R^2,$ we have by dot product properties that $$\begin{align}\lVert x-y\rVert^2 &= (x-y)\cdot(x-y)\\ &= x\cdot x-x\cdot y-y\cdot x+y\cdot y\\ &= x\cdot x-2(x\cdot y)+y\cdot y\\ &= \lVert x\rVert^2-2(x\cdot y)+\lVert y\rVert^2.\end{align}\tag{4}$$ Applying $(2),$ $(3)$, and $(4)$, we find that for all $v,w\in\Bbb R^2$, we have $$\begin{align}\lVert v\rVert^2-2\bigl(T(v)\cdot T(w)\bigr)+\lVert w\rVert^2 &= \lVert T(v)\rVert^2-2\bigl(T(v)\cdot T(w)\bigr)+\lVert T(w)\rVert^2\\ &= \lVert T(v)-T(w)\rVert^2\\ &= \lVert v-w\rVert^2\\ &= \lVert v\rVert^2-2(v\cdot x)+\lVert w\rVert^2,\end{align}$$ so in particular $$T(v)\cdot T(w)=v\cdot w.\tag{5}$$
Now, apply $(3),$ $(4),$ $(5),$ and dot product properties, so we see that for any $x,y,z\in\Bbb R^2$ we have $$\begin{align}\lVert z-x-y\rVert^2 &= \lVert z-x\rVert^2-2\bigl((z-x)\cdot y\bigr)+\lVert y\rVert^2\\ &= \lVert z-x\rVert^2-2(z\cdot y)+2(x\cdot y)+\lVert y\rVert^2\\ &= \lVert z\rVert^2-2(z\cdot x)+\lVert x\rVert^2-2(z\cdot y)+2(x\cdot y)+\lVert y\rVert^2\\ &= \lVert T(z)\rVert^2-2\bigl(T(z)\cdot T(x)\bigr)+\lVert T(x)\rVert^2-2(z\cdot y)+2(x\cdot y)+\lVert y\rVert^2\\ &= \lVert T(z)-T(x)\rVert^2-2(z\cdot y)+2(x\cdot y)+\lVert y\rVert^2\\ &= \lVert T(z)-T(x)\rVert^2-2\bigl(T(z)\cdot T(y)\bigr)+2\bigl(T(x)\cdot T(y)\bigr)+\lVert T(y)\rVert^2\\ &= \lVert T(z)-T(x)\rVert^2-2\Bigl(\bigl(T(z)-T(x)\bigr)\cdot T(y)\Bigr)+\lVert T(y)\rVert^2\\ &= \lVert T(z)-T(x)-T(y)\rVert^2.\end{align}\tag{6}$$ In particular, let $x,y\in\Bbb R^2,$ and put $z=x+y,$ so by $(6),$ we find that $$0=\lVert T(x+y)-T(x)-T(y)\rVert^2,$$ so $$0=\lVert T(x+y)-T(x)-T(y)\rVert$$ by nonnegativity, and so $$T(x+y)=T(x)+T(y)\tag{$\star$}$$ for all $x,y\in\Bbb R^2.$ You should be able to prove from here that $T$ is linear, so in particular is given by $$T(v)=Mv\tag{$\heartsuit$}$$ (where $M$ is given in the first paragraph).
Since $T$ is one-to-one (why?), then $M$ is invertible. Moreover, using the fact that $x\cdot y=x^ty$ for all $x,y\in\Bbb R^2,$ we have that $$\begin{align}x\cdot(M^tM-I)y &= x^t(M^tM-I)y\\ &= x^t(M^tMy-y)\\ &= x^tM^tMy-x^ty\\ &= (Mx)^tMy-x^ty\\ &= (Mx)\cdot(My)-x\cdot y,\end{align}$$ so by $(\heartsuit)$ and $(5)$ we have $$x\cdot(M^tM-I)y=0$$ for all $x,y\in\Bbb R^2.$ In particular, then, for any $y\in\Bbb R^2$, we have that $$\lVert(M^tM-I)y\rVert^2=(M^tM-I)y\cdot(M^tM-I)y=0,$$ whence $$(M^tM-I)y=0$$ for all $y\in\Bbb R^2,$ and so $M^tM=I.$ From this, we conclude that $\det(M)=\pm 1$. (Why?)
Can you take it from there?
