I have seen in wikipedia that the orthogonal matrices on $\Bbb{R^n}$, denoted by $O(n)=\{x\in GL_n(\Bbb{R}): x^tx=I\} $ is a subgroup of $E(n)$- the group of isometries.
I am a little bit confused because untill now I used the definition isometry in $C^*$ algebras. There, if we have an isometry acting on $\Bbb{C^n}$ it's autumatically invertible (because $\Bbb{C^n}$ is finite dimensional) thus unitary.
So, first- why the same argument with $\Bbb{R^n}$ is finite dimensional and thus invertible from the right implies invertible, does not imply just $O(n)=\{x\in M_n(\Bbb{R}): x^tx=I\}$?
Actually I know this is true but the explanations I saw are complicated and the argument "$\Bbb{R^n}$ finite dimensional" is not used.
Second, I don't understand why $O(n)$ does not equal $E(n)$. Can you give an example of isometry which is not in $O(n)$? Why in $C^*$ algebras the definitions coincide?
Thank you!
I don't know a lot about $C^*-$ algebras, but $O(n)$ does not consider translation of $\Bbb R^n$, since these are not linear transformation.
For instance: $ v\longmapsto v+w$ where $w$ is a fixed vector in $\Bbb R^n$ is an isometry.