Let $V$ be a finite-dimensional inner product space, and $f:V\to V$ be a map such that $f(0) = 0$ and $\forall x,y\in V$, $||f(x) – f(y)|| = ||x-y||$. Show that $f$ is an orthogonal linear map.
I've been able to show only a part of what's required, and my work is as follows:
Since $||f(x) – f(y)|| = ||x-y||$ holds for all $x,y$, put $y=0$. This gives $||f(x)|| = ||x||$ for all $x\in V$. $f:V\to V$ is a norm-preserving map, and so it is orthogonal if it is linear (proof of this is given at the end).
Now, how do I show that $f$ is linear? Is there a nice way of doing this? I'd appreciate any hints.
Proof of statement above:
I know that a map $T:V\to W$ is orthogonal if $\forall x,y\in V$, we have $\langle Tx,Ty\rangle = \langle x,y\rangle$.
$f$ is norm-preserving, so for arbitrary $x,y\in V$, we have $||f(x+y)|| = ||x+y||$ and hence $||f(x+y)||^2 = ||f(x)+f(y)||^2 = ||x+y||^2$. Expanding, we get $||f(x)||^2 + ||f(y)||^2 + 2\langle f(x),f(y)\rangle = ||x||^2 + ||y||^2 + 2\langle x,y\rangle$ and using $||f(x)||=||x||$ and $||f(y)|| = ||y||$ we get $\langle f(x),f(y)\rangle = \langle x,y\rangle$ as desired.
Best Answer
Hint
All isometry can be written as $$f(v)=\ell(v)+b,$$ where $\ell:V\to V$ is a linear map.