Prove Ux$\cdot$Uy = x$\cdot$y $\forall$ x,y $\in$ $\mathbb{R}^n$ where U is Isometric transformation

Question

Suppose that U is a linear transformation from $\mathbb{R}^n$ into $\mathbb{R}^m$ that is isometric, i.e., $\lVert Ux\rVert$ =$\lVert x\rVert$ for all x $\in$ $\mathbb{R}^n$

a) Prove that Ux$\cdot$Uy = x$\cdot$y $\qquad$ $\forall$x,y$\in \mathbb{R}^n$

I have an attempt that basically plays on using $\lVert U(x+y)\rVert^2$ as my starting point:

Near the end through some manipulation I get: $\lVert x\rVert^2$ + 2($x\cdot$y) + $\lVert y\rVert^2$

I feel like I've gotten somewhere as clearly $\lVert x\rVert^2$ = $\lVert Ux\rVert^2$ by how isometric is defined. Similarly, $\lVert y\rVert^2$ = $\lVert Uy\rVert^2$

But I don't know if this actually lets me show: Ux$\cdot$Uy = x$\cdot$y

And even if it does, I'm not sure how to make the last argument to show that.

Also, sorry for leaving out some of the intermediate details. I'm still very new to using this MathJax thing (I need to get around to learning Latex!) and it took very long to type that little bit out. Let me know if I need to clarify anything. Thanks!

Answer 1

You're right to look at $||U(x+y)||^2$, but I suggest trying to use the polarization identity $$ x\cdot y=\frac{1}{4}\Big(||x+y||^2-||x-y||^2\Big)$$

Answer 2

You're basically there. Expanding that in two different ways will give you what you want: \begin{align*} \|U(x + y)\|^2 &= \|x + y\|^2 &\text{$U$ preserves the length of $x + y$}\\ \|Ux + Uy\|^2 &= \|x + y\|^2 \\ \|Ux\|^2 + 2(Ux \cdot Uy) + \|Uy\|^2 &= \|x\|^2 + 2(x \cdot y) + \|y\|^2 \\ 2(Ux \cdot Uy) &= 2(x \cdot y) &\text{$U$ preserves the length of $x$ and $y$}\\ Ux \cdot Uy &= x \cdot y \end{align*}