Equivalent definitions of isometry

Question

Consider a map $T:\mathbb{R}^2\to\mathbb{R}^2$ such that $\lVert T(x)\rVert=\lVert x\rVert$.

Is this equivalent to stating that $\langle x, y\rangle=\langle T(x), T(y)\rangle$ for all $x,y\in\mathbb{R}^2$, where $\langle\cdot,\cdot\rangle$ denotes the usual inner product?

Answer 1

YES it is equivalent.

1. If $\langle x,y\rangle=\langle Tx,Ty\rangle$, for $x=y$: $$\|x\|^2=\langle x,x\rangle=\langle Tx,Tx\rangle=\|Tx\|^2,$$ and hence $\|x\|=\|Tx\|,$ for all $x$.

2. If $\|x\|=\|Tx\|,$ then $\|x\|^2=\|Tx\|^2,$ for all $x$. Replace $x$ by $x+y$ and obtain: $$ \langle x,x\rangle+2\langle x,y\rangle+\langle y,y\rangle=\langle x+y,x+y\rangle=\|x+y\|^2=\|Tx+Ty\|^2=\langle Tx,Tx\rangle+2\langle Tx,Ty\rangle+\langle Ty,Ty\rangle, $$ and since $\langle x,x\rangle=\langle Tx,Tx\rangle$ and $\langle y,y\rangle=\langle Ty,Ty\rangle$, then $$ \langle x,y\rangle=\langle Tx,Ty\rangle, $$ for all $x,y$.