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Change of coordinate/Reflection Linear Algebra Problem

  1. Feb 21, 2007 #1
    Change of coordinate/Reflection Linear Algebra Problem!!

    1. The problem statement, all variables and given/known data
    Hi everyone, so here I am yet again :blushing:.

    In R^2 let L be the line y=mx, where m is not zero. Find an expression for T(x,y) where T is the reflection of R^2 about L.
    2. Relevant equations

    A transformation that reflects about the x axis is given by T(x,y) = (x, -y) while one about the y axis is T(x,y)=(-x,y).
    The change of coordinate matrix Q is given by x'_j = (summation) (Q_ij) (x_i)
    (sorry, not sure how to add symbols into these windows), and we probably also need the relationship that the matrix representation of [T]_B' = Q^-1*[T]_B*Q where B' and B are ordered bases for R^2.

    Also, the equation of a line in slope-intercept form is y = mx + b. A line orthogonal to this would be y = - x/m +b.

    3. The attempt at a solution

    I know that if we are talking about reflection about a line L with slope m, that reflection will occur for lines orthogonal to L with slope -1/m. I thought that I could set it up as follows:

    For any ordered pair (x,y) on L, let T(x,y) = (x,y) since the line itself is invariant under transformation. Then consider some (-x,y) on a line L' that is perpendicular to L (it could've been (x, -y), but it there is no loss of generality with what I have). For T(-x,y) = -(-x, y) = (x, -y).

    If we let B be the standard ordered basis for R^2, we can let B' = {(1,1),(-1,1)}. Then we get the change of coordinate matrix (noted here as columns) Q = ((1 1) (1 -1)}, and Q^-1 = {(-1 -1) (-1 1)}.

    For [T]_B we have ((1 0) (0,-1)}. Then it is easy to find [T]_B' = Q^-1*[T]_B*Q.

    From that answer, which should be a 2*2 matrix, it appears that T is left multiplication by [T]_B. Thus for any (x,y) in R^2 I can say

    T(x,y) = [T]_B*((x) (y)) = some set of equations.

    Basically, I am lost at the beginning point where I need to decide what B' and Q are. Once I get those, I think the rest of this will work. I kind of just guessed by saying that B'={(1,1),(-1,1)} -- I'm just not sure how to get this step to work.

    Or should my B' be something more like (x,y), (-x,y) with x and y in it? I am very confused about how to find a B'.

    For reference, the answer in the back of my books is:

    T(x,y) = (1/m^2)*((1-m^2)x + 2my, 2mx + (m^2 -1)y). I have NO idea how this ordered pair is supposed to emerge. :bugeye: Maybe my entire method is wrong. Can anyone help?
  2. jcsd
  3. Feb 23, 2007 #2
    Did you try rotating by an angle -\theta so that the line goes to an axis (say x-axis), followed by reflection, followed by rotating back by \theta? This amounts to a matrix product.
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