Suppose the linear transformation $T:R^n \rightarrow R^n$ where T(x)=Ax. Prove that T is an isometry if and only if ${T(e_1),T(e_2),\cdots T(e_n)}$ forms an orthonormal basis of R^n.
I started working on my proof. Please let me know if I am on the right track. I began trying to prove the statement if $T$ is an isometry then ${T(e_1),T(e_2),\cdots T(e_n)}$ forms an orthonormal basis of $R^n$. Since $T$ is an isometry, $\forall e_i\in R^n$ where $1 \leq i \leq n: ||T(x)||=||x||$ which makes this space normal. Then, $T(x)=Ax$ as defined in the problem. $\forall {T(e_i), T(e_j)}$ where $1 \leq i,j \leq n$ and $i \neq j$. $$\langle T(e_i),T(e_j) \rangle = 0$$ $$\langle Ae_i,Ae_j \rangle =0$$ $$A\langle e_i,Ae_j \rangle =0$$ $$A\overline{A}\langle e_i,e_j \rangle =0$$ $$AA^t(0)=0$$ Which shows the transformation of all of the basis vectors to be orthogonal from each other and linearly independent. I still have to work on the other direction but is this a logically sound start.
Update: Dealing with real numbers, $$\langle x,y \rangle =x_1y_1 + x_2y_2 + \cdots + x_ny_n$$ So, $$\langle T(e_i),T(e_j) \rangle =0$$ $$\langle Ae_i,Ae_j\rangle = 0$$ $$Ae_i\bullet Ae_j=0$$ $A_i{th}$ column $\bullet A_j{th}$ column
I don't know how to show these as linearly independent from here.
The orthonormal basis $\implies$ isometry direction:
Suppose that the vectors $T(e_i)$ form an orthonormal basis. Let $v=a_1e_1+...+a_ne_n$.
Then $\|T(v)\|=\|a_1T(e_1)+...+a_nT(e_n)\|=\sqrt{a_1^2+...+a_n^2}=\|v\|$.
The isometry $\implies$ orthonormal basis direction:
WLOG, suppose that $\langle T(e_1),T(e_2)\rangle=d\ne 0$.
Then $\|T(e_1+e_2)\|^2=\|e_1+e_2\|^2=2$.
But it is also true that $\|T(e_1+e_2)\|^2=\|T(e_1)+T(e_2)\|^2=2+d$.