# Proving $A\in M_n(\mathbb{R})$ is orthogonal

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.

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have you tried the Polarization identity $\langle u,v\rangle=\frac{1}{4}\left(\langle u+v,u+v\rangle-\langle u-v,u-v\rangle\right)$? –  Giuseppe Tortorella May 31 '12 at 7:55
@GiuseppeTortorella no, I forgot that (been a couple of years since I studied it), where do you think of using it ? –  Belgi May 31 '12 at 7:57
Recall that $A$ is an isometry iff $\forall x, \langle Ax, Ax \rangle = \langle x, x \rangle$ (and is linear)... –  Najib Idrissi May 31 '12 at 8:07
@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 ? –  Belgi May 31 '12 at 8:19

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.
Did you fix $v$ before P.I. ? (why $\forall u$ and not $\forall u,v$ ?) –  Belgi May 31 '12 at 8:21