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Show that if $V$ is a finite-dimensional vector space with a dot product $\langle-,-\rangle$, and $f: V \rightarrow V$ linear with $\forall v,w \in V: \langle v,w \rangle=0 \Rightarrow \langle f(v),f(w) \rangle=0$ then $\exists C \in \mathbb{R}$ such that $(C\cdot f)$ is a linear isometry.

Notes & Thoughts: $g$ is a linear isometry means $\forall v \in V: \lVert g(v)\rVert=v$

Visually the theorem makes sense, if orthogonal vectors remain orthogonal under $f$ then the angles remain, so the original vector just changes its length. (If I understand this correctly)

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    $N$ice hint, I didn't think about the parallelogramm identit$y$.2011-01-21

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Let $x,y\in V$ with $|x| = |y|=1$. I claim that $|f(x)| = |f(y)|$.

To see this, notice that $0 = |x|^2 - |y|^2 = \langle x+y,x-y\rangle$.

By assumption, this is equal to $\langle f(x+y), f(x-y)\rangle = |f(x)|^2 - |f(y)|^2$.

Thus, if we define $C = 1/|f(x)|$ with $|x| = 1$, then $C$ does not depend on the choice of $x$. I claim this $C$ solves the problem.

So, let $z\in V$ be arbitrary. I want to show that $|z| = |f(z)|$.

If $z = 0$, then $Cf(z) = 0$, so $|z| = |f(z)|$.

If $z\neq 0$, then $z/|z|$ is a unit vector, and so $1 = |Cf(z/|z|)|$. Multiplying both sides by $|z|$, w get $|z| = |Cf(z)|$, showing $Cf$ is an isometry.

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    So if $f$ is not the $0$ map you can always be sure that the unit vector is not mapped to $0$? Edit: I see yes, otherwise according to the proof all unit vectors would be mapped to $0$ and therefore all other vectors too, thanks.2011-01-21