Proving that a linear isometry on $\mathbb{R}^{n}$ is an orthogonal matrix

Solution 1:

For every $x, y \in \mathbb{R}^n$ we have $$ \langle Ax,Ay\rangle=\frac{1}{4}\left[|A(x+y)|^2-|A(x-y)|^2\right]=\frac{1}{4}\left[|x+y|^2-|x-y|^2\right]=\langle x,y\rangle. $$ Hence, for every $x,y \in \mathbb{R}^n$ we have $$ \langle A^TAx,y\rangle=\langle x,y\rangle, $$ i.e. $A^TA=I_n$.

Solution 2:

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

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