Q:

Showing an Isometry between normed space.

Question: 
  Show that a linear map $A:V\rightarrow W$ between Normed spaces is an Isometry IFF $\left \| A\left ( \vec{v} \right ) \right \|=\left \| \vec{v} \right \|$.

Let A be a linear map and an Isometry.
Because A is an Isometry,
we have that $\forall \vec{v_{1}},\vec{v_{2}}$:
$d_{v}\left ( \vec{v_{1},\vec{v_{2}}} \right )=d_{w}\left ( A\left ( \vec{v_{1}} \right ),A\left ( \vec{v_{2}} \right ) \right )=\left | A\left ( v_{1} \right )-A\left ( \vec{v_{2}} \right ) \right |=\left | \vec{v_{1}}-\vec{v_{2}} \right |$
But V is a normed space so
$\left | \left \| A\left ( \vec{v_{1}} \right )-A\left ( \vec{v_{2}} \right ) \right \|  \right |=\left | \left \| \vec{v_{1}} \right \| -\left \| \vec{v_{2}} \right \|\right |$.
Hence, $\left \| A\left ( \vec{v_{1}} \right ) \right \|=\left \| \vec{v_{1}} \right \|$. 
Does this look correct thus far? I'm not exactly too sure.
Any input will be appreciated.
Thanks in advance.

A:

I don't understand the last two sentences of the proof. It seems that you made a leap at that point. But you are already half there:
As you noticed, $A$ is an isometry if and only if $\|Ax-Ay\|=\|x-y\|$, for every $x,y\in V$. By the linearity of $A$, this is the case if and only if $\|A(x- y)\|=\|x-y\|$, for every $x,y\in V$. Now you need to set $w=x-y$ and you are done.

