Results 1 to 2 of 2

Math Help - Euclidean geometry...subgroups

  1. #1
    Senior Member
    Joined
    Jan 2009
    Posts
    296

    Euclidean geometry...subgroups

    Let E*(2) be the set of maps t:R2->R2 x->Ux+a with a in R2 and U an orthogonal matrix of determinant 1. Show that E*(2) is a subgroup of the group of Euclidean transformations and determine which of the isometries are in E*(2).

    I'm not really sure where to even start. Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member HappyJoe's Avatar
    Joined
    Sep 2010
    From
    Denmark
    Posts
    234
    By "Euclidean transformation", do you mean a map T\colon\mathbb{R}^2 \rightarrow \mathbb{R}^2, such that T(x) = Ax + b, where A is an orthogonal matrix, and b is some vector?

    If so, to show that E^*(2) is a subgroup of the group of Euclidean transformations, you need to show that E^*(2) is closed under composition of maps, under inverses and that it contains the neutral element.

    Closed under composition:

    Take t_1,t_2\in E^*(2), and say t_1(x) = O_1x+b_1 and t_2(x) = O_2x+b_2, where O_1 and O_2 are orthogonal matrices of determinant 1. Then:

    (t_1\circ t_2)(x) = t_1(t_2(x)) = t_1(O_2x+b_2) = O_1(O_2x+b_2) + b_1 = (O_1O_2)x + (O_1b_2+b_1),

    and provided that O_1O_2 is an orthogonal matrix of determinant 1, this composition is indeed in E^*(2). But this is true, as you may try to show.

    Closed under inverses.

    Let t\in E^*(2). We need to prove that the inverse t^{-1} is also an element of E^*(2). Suppose t is of the form t(x)=Ox+b, where O is orthogonal of determinant 1. Then the inverse of t is t^{-1}(x) = O^{-1}x - O^{-1}b. Since O^{-1} is again an orthogonal matrix of determinant 1 (which you may check), the inverse is an element of E^*(2) as needed.

    The neutral element is there.

    The neutral element is the map T given by T(x)=Ix = x, where I is the identity matrix. This is certainly an element of S^*(2).

    As for the question "Determine which of the isometries are in E^*(2)", it depends on your definition of isometry. Are you isometries linear or are they just supposed to preserve the distance in \mathbb{R}^2?

    All of the Euclidean transformations preserve the distance in \mathbb{R}^2, and so does E^*(2) in particular. If you are thinking in geometric terms, then E^*(2) consists of the maps, which are rotations followed by a translation.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Euclidean Geometry
    Posted in the Geometry Forum
    Replies: 5
    Last Post: November 19th 2010, 01:00 AM
  2. Non-Euclidean Geometry Homework
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: April 12th 2010, 04:04 PM
  3. euclidean geometry! help
    Posted in the Geometry Forum
    Replies: 3
    Last Post: June 8th 2008, 01:08 PM
  4. geometry -Euclidean
    Posted in the Geometry Forum
    Replies: 2
    Last Post: June 6th 2008, 02:44 PM
  5. Help me with Angles-Euclidean Geometry
    Posted in the Geometry Forum
    Replies: 3
    Last Post: September 3rd 2007, 04:48 PM

Search Tags


/mathhelpforum @mathhelpforum