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I wish to prove that if $T:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is defined by $T(v)=Av$ (where $A\in M_{n}(\mathbb{R})$) is an isometry then $A$ is an orthogonal matrix.

I am familiar with many equivalent definition for $A\in M_{n}(\mathbb{R})$ to be orthogonal, and it doesn't matter to me which one to show. What I tried to do is the following: $||x-y||=||Ax-Ay||\implies\langle x-y,x-y\rangle=\langle Ax-Ay,Ax-Ay\rangle\implies\langle x-y,x-y\rangle=\langle x-y,A^{t}A(x-y)\rangle$,

from here I thought that I will be able to deduce $A^{t}A=I$ and complete the proof, but I was unable to do so.

How can I complete the proof, or prove this in another fashion ? Help is appreciated!

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So you now that $\langle z, A^tAz \rangle = \langle z,z\rangle$ for every $z$. Try to let $z = x \pm y$ and expand ... –  martini Jul 12 '12 at 13:10

2 Answers 2

up vote 3 down vote accepted

For every $x, y \in \mathbb{R}^n$ we have $$ \langle Ax,Ay\rangle=\frac{1}{2}\left[|A(x+y)|^2-|A(x-y)|^2\right]=\frac{1}{2}\left[|x+y|^2-|x-y|^2\right]=\langle x,y\rangle. $$ Hence, for every $x,y \in \mathbb{R}^n$ we have $$ \langle A^TAx,y\rangle=\langle x,y\rangle, $$ i.e. $A^TA=I_n$.

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We know that $\,\langle\, x,y\,\rangle =0\,\,\,\forall\,y\Longleftrightarrow x=0\,$ , so $$\forall x,y\,\,:\,\langle\,x,y\,\rangle=\langle\,Ax,Ay\,\rangle=\langle\,x,A^tAy\,\rangle\Longrightarrow \langle\,x,(A^tA-I)y\,\rangle=0$$

$$\Longrightarrow (A^tA-I)y=0\,\,\,\,\forall y\,\,\Longrightarrow A^tA-I=0\Longrightarrow A^tA=I$$

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Why $\langle x,y\rangle=\langle Ax,Ay\rangle$ ? –  Belgi Jul 12 '12 at 13:41
1  
@Belgi By definition of being an isometry. –  M Turgeon Jul 12 '12 at 13:42
    
@MTurgeon, how's that ? I am given the definition that $f$ is an isometry if $d(x,y)=d(f(x),f(y))$ where in this case $d(x,y):=||x-y||$ –  Belgi Jul 12 '12 at 13:45
3  
@Belgi , it's the same: $$||x-y||=||Ax-Ay||\Longleftrightarrow \langle x-y,x-y\rangle=\langle Ax-Ay,Ax-Ay\rangle$$ Now just use linearity of inner product –  DonAntonio Jul 12 '12 at 13:50

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