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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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If f preserves orthogonal vectors, what does it do to an orthonormal basis? –  Qiaochu Yuan Jan 21 '11 at 20:09
    
@Qiaochu Yuan: Actually orthogonal vectors stay orthogonal, but I don't know what happens to the length. Therefore I would say the orthonormal basis is turned into an orthogonal basis. But I don't see why the length of every vector changes by a common factor. –  Listing Jan 21 '11 at 20:12
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@user3123: The zero map is a counterexample. –  Jonas Meyer Jan 21 '11 at 20:27
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@user3123: If you want to be careful with this minor point, you could either add the hypothesis that $f$ is nonzero, or change the conclusion to the statement that there exists an isometry $g$ and a real number $C$ such that $f=Cg$. –  Jonas Meyer Jan 21 '11 at 20:34
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@user 3123: "I don't see how the angle would be related to length of vectors in any way." Then you should prove the statement that a norm on a vector space satisfying the parallelogram identity determines a unique inner product which induces it. In characteristic zero, quadratic forms and bilinear forms are the same thing. (There is an annoying technicality in this exercise if you try to do it over C, so do it over R.) –  Qiaochu Yuan Jan 21 '11 at 20:47
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1 Answer

up vote 2 down vote accepted

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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Thank you for your answer, I get the general idea but I have some problems with the proof. How would $C$ not depend on the choice of $x$ if you say that $C = 1/|f(x)|$? Also first you say that $|y|=1$ and afterwards you say "If $y=0$". This seems to be a bit odd, how would the length of the 0 vector be 1? –  Listing Jan 21 '11 at 20:22
    
The lines "To see this, notice..." and "By assumption..." are the arguement that $C$ doesn't depend on $x$ (so long as $|x| = 1$). I'll edit the last two paragraphs/lines to be clearer. –  Jason DeVito Jan 21 '11 at 20:25
    
Thank you I understood it now. –  Listing Jan 21 '11 at 20:32
    
As Jonas Meyer said above, this is false when $f$ is the $0$ map. My proof breaks down when I set $C = 1/f(x)$ - if $f$ is the $0$ map, I just divided by $0$. However, so long as $f$ is not the $0$ map, the proof works ok. –  Jason DeVito Jan 21 '11 at 20:38
    
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. –  Listing Jan 21 '11 at 20:45
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