Problem 5. A function f R Rn (not necessarily linear) is called an isometry if i
ID: 1721419 • Letter: P
Question
Problem 5. A function f R Rn (not necessarily linear) is called an isometry if it preserves distances, ie if for all T, y E ri. For each TE R", ne the map Tr R" RT by Ta(T) i. The map T is called translation by i (a) Prove that Tif is an i for each UERT For which i is Te a linear transformation (b) Prove that if f and g are isometries of R then g of is also an isometry of IR (c) Suppose T R" Rn is a linear transformation. Prove that if T is an isometry then T is an orthogonal transformation. (d) Prove that if f is an isometry of R then f is a linear transformation if and only if f 0 Hint: compute the lengths of ir and f(ii) f(i in terms of the dot product in order to show that if f is an isometry with f(0) 0, then f(i) f(i) ii for all ii, i.) (e) Prove that for every isometry f of Rn, there is an orthogonal n x n matrix A and a vector be R" such that f(T AE b for a FER"Explanation / Answer
a) ||Tv(x) -Tv(y)|| = ||x+v-y-v|| = ||x-y||.
So Tv preserves distances...hence an isometry.
Tv is linear iff Tv(0) =v =0. So v must be the zero vector.
b) ||g*f(x)-g*f(y)||
= ||g(f(x))-g(f(y))||
=||f(x)-f(y)|| (because g is an isometry)
= ||x-y|| (because g is an isometry)
So g*f is an isometry.
(c) ||Tx-Ty||2 = <Tx-Ty,Tx-Ty>
= <Tx,Tx> -2<Tx,Ty> +<Ty,Ty>
= ||Tx||2 -2<Tx,Ty> + ||Ty||2
= ||x||2 -2<Tx,Ty> + ||y||2 (||Tx||= ||Tx-T0|| = ||x-0||=||x||)...................(1)
But LHS is ||x-y||2 = ||x||2 -2<x,y> + ||y||2 ...............................................................(2)
From (1) and (2) , <Tx,Ty> = <x,y>, implying T is an orthogonal transformation
(d) Follows the same argument as in (c)
(e) If f is an isometry and f(0) = b,
let g(v) = f(v)-b.
Then g is an isometry and g(0) =0, So g is a linear transformation and an isometry,,hence there exists an Orthogonal matrix A such that g(v) =Av
This means f(v) = g(v)+b = Av+b, as desired.
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