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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
up vote 4 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. $A^TA=I_n$.

share|cite|improve this answer

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

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

share|cite|improve this answer
Why $\langle x,y\rangle=\langle Ax,Ay\rangle$ ? – Belgi Jul 12 '12 at 13:41
@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
@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

Your Answer


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.