Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a map where $\|f(a)-f(b)\|=\|a-b\|$ and $f(0)=0$, prove $a · b = f(a) · f(b)$ Let $f: \mathbb{R}^n \rightarrow  \mathbb{R}^n$ be a map where $\|f(a)-f(b)\|=\|a-b\|$ and $f(0)=0$, prove $a · b = f(a) · f(b)$.
Trying to solve this problem from a textbook and I'm rather confused. Obviously this is a linear transformation, but how do we go about proving this statement? I really have no idea here.
Thank you.
 A: First, it is obvious that $\lVert f(x) \rVert = \lVert x \rVert$ for all $x\in\mathbb{R}^n$ since $f(0)=0$.
Now, for all $a$ and $b\in\mathbb{R}^n$,
$$\lVert f(a)-f(b) \rVert^2=\lVert a-b \rVert^2.$$
It follows that
$$\lVert f(a) \rVert^2+\lVert f(b) \rVert^2-2f(a)\boldsymbol{\cdot}f(b)=\lVert a \rVert^2+\lVert b \rVert^2-2a\boldsymbol{\cdot}b.$$
Since $\lVert f(a) \rVert = \lVert a \rVert$ and $\lVert f(b) \rVert = \lVert b \rVert$, we can deduce that
$$a\boldsymbol{\cdot}b = f(a)\boldsymbol{\cdot}f(b).$$
A: You're given $f: \mathbb{R}^n \rightarrow  \mathbb{R}^n$ is linear ( I believe you've proved the linearity) a map with $\|f(a)-f(b)\|=\|a-b\|$ and $f(0)=0.$ Since there was a notion of a norm in the vector space, you can define inner product using the polarization identity, which was suggested by Kavi Rama Murthy.
Now, $\|f(a)-f(b)\|^2=\|a-b\|^2$ gives us $ < f(a), f(b)> = <a, b>$ for any $a, b$ $\in \mathbb{R^n}.$
A: Analogous Statement: A linear transformation is norm preserving iff inner product preserving.
Notation. $\langle \boldsymbol{a},\boldsymbol{b} \rangle = \boldsymbol{a}\cdot\boldsymbol{b}$ (the dot product or standard inner product).
Note that $f(\boldsymbol{y}) = \|\boldsymbol{y}\|$ for any $\boldsymbol{y}\in \mathbb{R}^n$ (as $f(\boldsymbol{0}) = \boldsymbol{0}$).
Recall that $\langle \boldsymbol{x} , \boldsymbol{x}\rangle= \|\boldsymbol{x}\|^2$ for any $\boldsymbol{x} \in \mathbb{R}^n$. Let's consider $\boldsymbol{x}, \boldsymbol{y} \in \mathbb{R}^n$. Then we have that
\begin{align}
\|f(\boldsymbol{x})-f(\boldsymbol{y})\|^2 &= \|f(\boldsymbol{x})\|^2+\|f(\boldsymbol{y})\|^2 -2\langle f(\boldsymbol{x}),f(\boldsymbol{y})\rangle
\\&= \|\boldsymbol{x}\|^2 + \|\boldsymbol{y}\|^2 - 2 \langle f(\boldsymbol{x}), f(\boldsymbol{y})\rangle
\end{align}
Observe that
$$ \|f(\boldsymbol{x}) - f(\boldsymbol{y})\|^2 = \|\boldsymbol{x}-\boldsymbol{y}\|^2 = \|\boldsymbol{x}\|^2 + \|\boldsymbol{y}\|^2 - 2 \langle \boldsymbol{x}, \boldsymbol{y}\rangle $$
Combining the two equations together, we obtain
$$ \langle f(\boldsymbol{x}), f(\boldsymbol{y}) \rangle = \langle\boldsymbol{x}, \boldsymbol{y}\rangle $$
