Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
1  
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

1 Answer 1

up vote 1 down vote accepted

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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.