Isometry $\phi: \mathbb{R}^2 \rightarrow \mathbb{R}^2$

Question

I'm stuck doing the first part of an exercise on groups in an abstract algebra course.

Let $\phi:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ be an isometry s.t. $\phi((0,0)) = (0,0)$. I'm given that $\phi$ preserves the euclidian distance i.e. that $|\phi(x) - \phi(y)| = |x-y|$.

i ) Show that $\phi(v_1) \cdot \phi(v_2) = v_1 \cdot v_2 $, where $v_1, v_2 \in \mathbb{R}$ and $\cdot$ denotes the usual inner product on $\mathbb{R}$. (Hint: use $|v_1-v_2|^2 = |v_1|^2 + |v_2|^2 -2v_1\cdot v_2$)

I can't really start the exercise, apart from writing $|\phi(v_1) - \phi(v_2)| = |v_1-v_2| \Rightarrow |\phi(v_1) - \phi(v_2)|^2 =|v_1|^2 + |v_2|^2 -2v_1\cdot v_2$.

How do I continue from here?

Answer 1

Using the hint:

$$\left|\phi v_1-\phi v_2\right|^2=|\phi v_1|^2+|\phi v_2|^2-2\phi v_1\cdot\phi v_2$$

But on the other hand, we're given that

$$|\phi v_1-\phi v_2|^2=|v_1-v_2|^2\stackrel{\text{hint}}=|v_1|^2+|v_2|^2-2v_1\cdot v_2$$

Well, now compare both expressiona after you prove that also $\;|\phi x|=|x|\;,\;\;\forall x\in\Bbb R^2$

Answer 2

Hint: use also that $$ |\phi(v_1)-\phi(v_2)|^2 = |\phi(v_1)|^2 + |\phi(v_2)|^2 - 2\phi(v_1)\cdot\phi(v_2). $$

Answer 3

Hint: for all $x$, $|\phi(x)| = |\phi(x) - (0,0)| = |\phi(x) - \phi((0,0))| = |x - (0,0)| = |x|$.

Answer 4

This is what I would do: re-write your hint as:

$v_1\cdot v_2 = \dfrac{1}{2}\left(|v_1|^2 + |v_2|^2 - |v_1-v_2|^2\right)$

and note that $|v_k| = |v_k - (0,0)|$.

So.....

$v_1\cdot v_2 = \dfrac{1}{2}\left(|v_1 - (0,0)|^2 + |v_2 - (0,0)|^2 - |v_1 - v_2|^2\right)$

$= \dfrac{1}{2}\left(|\phi(v_1) - \phi((0,0))|^2 + |\phi(v_2) - \phi((0,0))|^2 - |\phi(v_1) - \phi(v_2)|^2\right)$

and since $\phi((0,0)) = (0,0)$,

$=\dfrac{1}{2}\left(|\phi(v_1)|^2 + |\phi(v_2)|^2 - |\phi(v_1) - \phi(v_2)|^2\right)$

$= \dots ?$