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My math book gives this theorem:

Let $f$ be a rigid motion in a real inner product space $X$, and let $T (x) := f (x) - f (0)$. Then T is an orthogonal transformation.

From this theorem, we are asked to prove a lemma.

Part of this lemma is: $(T(x),T(y))=(x,y)$

My thought was $(T(x),T(y))=(f(x)-f(0),f(y)-f(0))=(x-0,y-0)=(x,y)$, as desired.

Is this enough of a proof? My book gives a different proof using $||T (x)-T(y)||^2$, but I do not really follow that proof and to me what I did seems to make sense. Am I correct in that assumption?

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  • $\begingroup$ What is your definition of rigid motion? In not $|f(x)-f(y)|=|x-y|?$ If that is the definition, how can you show $(f(x)-f(0),f(y)-f(0))=(x,y)?$ $\endgroup$
    – mfl
    Dec 5, 2016 at 19:39
  • $\begingroup$ Ah! That makes sense. So can I say: ||Tx-Ty||^2= ||x-y||^2, and ||Tx-Ty||^2= ||Tx||^2+||Ty||^2 -2(Tx,Ty), and ||x-y||^2 = ||x||^2 + ||y||^2 2(x,y), and this cancels out from the first equality to be -2(Tx,Ty)=-2(x,y) so (Tx,Ty)=(x,y) ? $\endgroup$
    – readyforthebettys
    Dec 5, 2016 at 19:42
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    $\begingroup$ Yes. You got it. Note hat $|Tx-Ty|^2=|f(x)-f(y)|^2=|f(x)-f(0)+f(0)-f(y)|^2|x-y|^2.$ $\endgroup$
    – mfl
    Dec 5, 2016 at 19:50

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