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In an attempt to show that if $f(v)=Av$ is an isometry implies that $A$ is orthogonal I wish to show that $\forall x,y : \langle x-y,x-y \rangle=\langle A(x-y),A(x-y) \rangle \implies A$ is orthogonal .

I thougt of of showing $\forall x,y : \langle x,y \rangle=\langle Ax,Ay \rangle $ but I didn't manage to do it.

Can someone please help with this ?

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    @N.I - I don't know this, and I can't see why this is true...if $A$ preserves distance from any point to $0$ why does is preserves the distance between any two points ?2012-05-31

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Let be $A:V\to V$ a linear map and $\langle\cdot,\cdot\rangle$ a scalar product on $V.$

We say that $A$ is orthogonal when it preserves the scalar product, i.e.: $\langle Au,Av\rangle=\langle u,v\rangle,\ \forall u,v\in V.$

For the orthogonality of $A$ it is not only necessary but even sufficient that $A$ preserves the associated quadratic form, i.e.: $\langle Au,Au\rangle=\langle u,u\rangle,\ \forall u\in V.$


In order to prove the sufficiency, we can use the Polarization Identity: $\langle u,v\rangle=\frac{1}{4}\left(\langle u+v,u+v\rangle-\langle u-v,u-v\rangle\right)\tag{P.I.},\ \forall u,v\in V.$ Infact, let us assume that $A$ preserves the quadratic form associated to $\langle\cdot,\cdot\rangle,$ i.e.: $\langle Au,Au\rangle=\langle u,u\rangle,\ \forall u\in V,\tag{*}$ then, for any $u,v\in V,$ by the polarization identity we get $\langle Au,Av\rangle\stackrel{P.I.}{=}\frac{1}{4}\left(\langle A(u+v),A(u+v)\rangle-\langle A(u-v),A(u-v)\rangle\right)\\\stackrel{*}=\frac{1}{4}\left(\langle u+v,u+v\rangle-\langle u-v,u-v\rangle\right)\stackrel{P.I.}{=}\langle u,v\rangle,$ that is $A$ preserves the scalar product $\langle\cdot,\cdot\rangle,$ and we have done.

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    Did you fix $v$ before P.I. ? (why $\forall u$ and not $\forall u,v$ ?)2012-05-31