Let $(V, \langle,\rangle)$ be an euclidean Vector space and let $v_1 \in V$ with $||v_1||=1$. Let $f \in \operatorname{End}(V)$ with the two properties:
$(1) ||f(v_1)||= 1.$
$(2)\space \forall \space v, w \in V$ the following applies: $\langle v, w \rangle = 0 \Rightarrow \langle f(v), f(w) \rangle = 0$
I want to show that $f$ is an orthogonal map.
My thoughts and ideas: In order for $f$ to be an orthogonal map, $\langle f(v), f(w) \rangle = \langle v, w \rangle$ has to be true. Now we can extend $v_1$ to $(v_1,v_2',\dots,v_n')$ with the basis extension theorem and then apply the Gram-Schmid Process so that it is an orthonormal basis $(v_1,v_2,\dots,v_n)$.
Now we can show $||f(v_i)||= 1$ for all $i$ by demonstrating that $\langle v_1+v_i,v_1-v_i \rangle = 0$. Then apply the second condition and conclude the Hypothesis. In the end we can show $\langle f(v), f(w) \rangle = \langle v, w \rangle$ by representing $v,w$ as linear combinations of the basis.
I am not sure about this and maybe I'm not right. If I am I do not know how to write it down properly. Any ideas and thoughts would be very helpful indeed. Thanks!
