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Remmert page 15 chapter 0 it says that angle preserving mapping is R-linear and injective.

We want to prove:

Given $$T:\mathbb{C}\rightarrow \mathbb{C}$$ a $\mathbb{R}$ linear map which satisfies $|z||w|\langle T(z),T(w)\rangle = |T(z)||T(w)|\langle z,w\rangle$ that isn't a 0 map. Then it is injective.

There exist equivalent definitions, which is not yet introduced but we don't want cheat.

For injectivity we need: $T(z)=T(w) \Rightarrow z=w $ for all $z,w \in \mathbb{C} $

But then, how to continue? Please, do tell.

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Assume that $T$ isn't injective. Then you can find $z'\neq w'$ such that $T(z')=T(w')$. If $T(z')\neq 0$ then set $z=z'$ and $w=w'$. Else, since $T$ isn't identically 0, there exists $x$ such that $T(x)\neq 0$. Set $z=x+z'$ and $w=x+w'$. Then $T(z)=T(w)\neq 0$.
Now, using the equality: $$|z||w||T(w)|^2=|z||w|\langle T(z),T(w)\rangle = |T(z)||T(w)|\langle z,w\rangle=|T(w)|^2\langle z,w\rangle$$ It follows that $|z||w|=\langle z,w\rangle$. Hence, the angle between $z$ and $w$ is 0. This implies that there exists a non-zero scalar $\alpha$ such that $z=\alpha w$. But then: $T(z)=\alpha T(w)=T(w)$ $\Rightarrow$ $\alpha=1$ $\Rightarrow$ $z=w$.

[BTW, It's equivalent to show that $ker (T)=\{0\}$, since if $T(z)=T(w)$ then $T(z-w)=0$ and hence $z-w\in ker(T)$. (And clearly, if $ker (T)\neq\{0\}$ then there exists $0\neq x\in ker (T)$ such that $T(x)=T(0)=0$. Hence $T$ is not injective).]

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Thank you, Dennis! –  VVV Feb 10 '12 at 18:50
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