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I have a function $f:V \rightarrow V$, $f(v) = v − 2 \langle u, v\rangle u$, where $u$ is a fixed unit vector.

I want to show that this $\langle f(v), f(w) \rangle= \langle v,w \rangle$ but I am unsure on how to manipulate the inner products. Can someone show this step by step?

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    $\begingroup$ What is the argument of $f$? Is what you wrote above $f(v)$ and $u$ is a fixed vector, or $f(u)$ and $v$ is a fixed vector? $\endgroup$ – Viktor Vaughn Nov 20 '16 at 16:23
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    $\begingroup$ @SpamIAm corrected! $\endgroup$ – user390960 Nov 20 '16 at 16:25
  • $\begingroup$ And the $v$ in the $f(v)$ of the inner product is the same as $v$ in the definition ? $\endgroup$ – Emilio Novati Nov 20 '16 at 16:26
  • $\begingroup$ @EmilioNovati It should be for all $v,w \in V$. The OP's map is a reflection. $\endgroup$ – Viktor Vaughn Nov 20 '16 at 16:27
  • $\begingroup$ I believe that we must assume that $V$ is a real vector space. $\endgroup$ – bthmas Nov 20 '16 at 17:28
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We know that

$$f(v) = v − 2 \langle u, v\rangle u$$

So we plug that in:

$$\langle f(v), f(w) \rangle = \langle v − 2 \langle u, v\rangle u \rangle, w − 2 \langle u, w\rangle u \rangle $$

And now from the linearity of the dot product:

$$\langle v − 2 \langle u, v\rangle u \rangle, w − 2 \langle u, w\rangle u \rangle =\\ \langle v, w \rangle + \langle - 2\langle u, v \rangle v, w \rangle + \langle v, -2 \langle u, w \rangle w \rangle + \langle -2\langle u, v \rangle v, -2 \langle u, w \rangle w \rangle $$

Now can you take the constants out of the dot products to show that everything but the first dot product cancels?

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Hint:

from the definition show that $\langle f(v),f(v) \rangle=\langle v,v\rangle$, so $f$ is an isometry and conserve angles.

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