How to show $x\mapsto Ax$ is an isometry

Question

Given that $A$ is an $n\times n$ real orthogonal matrix . $x$ is any column vector of $\mathbb R^{n}$ .Define $$x\mapsto Ax$$ This is an isometry on $\mathbb R^{n}$ .

Then I have to show $$d(Ax_1,Ax_2)=d(x_1,x_2)$$ i.e. $$||Ax_1-Ax_2||=||x_1-x_2||(||\ \ ||\ \ is\ \ the\ \ \ Euclidean\ \ norm\ \ on\ \ \mathbb R^n)$$ i.e. $$||A(x_1-x_2)||=||(x_1-x_2)||$$

$A$ is an element of $\mathbb R^{n\times n}$ and $(x_1-x_2)$ , of $\mathbb R^n$.

What should I do next $?$

Answer 1

We have $ \|Ax\|^2 = \langle Ax, Ax \rangle = \langle A^TAx, x \rangle = \langle x, x \rangle = \|x\|^2 $ and so $ \|Ax\| = \|x\| $.

Now take $x=x_1-x_2$.

Answer 2

Suppose $ x_i = A^Ty_i$, we can get anything in $\mathbb{R}^n$ via this relation because $A$ is a bijection (because it is orthogonal, so the inverse is $A^T$).

So just plug $x_i = A^Ty_i$ into your norm, and you're done.