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Let $(V,\langle\,,\rangle)$ be a vector space with the dot product $\langle\cdot,\cdot\rangle$. Let $f:V\to V$ be a linear map with the property $\lVert f(x)\rVert =\lVert x \rVert$ for all $x\in V$. Show that this already implies $\langle f(x),f(y) \rangle = \langle x,y \rangle$ for all $x,y \in V$

My first idea was to show this with the Cauchy–Schwarz inequality. But it didn't yield anything useful since all i get is $\lVert \langle f(x),f(y) \rangle \rVert \le ||f(x)||^2 ||f(y)||^2 =||x||^2||y||^2 = \langle x,x \rangle \langle y,y \rangle$

I would appreciate a little hint.

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You have to use the fact that $\forall v,w\in V: \left \| v+w \right \|^2=\left \| v \right \|^2+\left \| w \right \|^2+2\langle v,w\rangle$. This implies that: $$\langle v,w\rangle=\frac{1}{2}\left(\left \| v+w \right \|^2-\left \| v\right \|^2-\left \| w \right \|^2\right).$$ So, using the linearity of $f$ and the hypothesis we obtain: $$\begin{align}\langle f(v),f(w)\rangle &=\frac{1}{2}\left(\left \| f(v)+f(w) \right \|^2-\left \| f(v)\right \|^2-\left \| f(w) \right \|^2\right)\\&=\frac{1}{2}\left(\left \| f(v+w) \right \|^2-\left \| f(v)\right \|^2-\left \| f(w) \right \|^2\right)\\&=\frac{1}{2}\left(\left \| v+w \right \|^2-\left \| v\right \|^2-\left \| w \right \|^2\right)\\&=\langle v,w\rangle\end{align}$$

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