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Homework Help: Geometry help

  1. Oct 3, 2005 #1

    Hi there,
    I am have trouble with a proof. I have some steps done, but I am not sure if I am aproaching this correctly.
    the question is:

    Show that the matrix of any reflection in R^n is a symmetric matrix.

    I know that F(x) = Ax + b is an isometry
    where A is an Orthoganol matrix , and vector b is in R^n
    this implies that A^t A = I (identity matrix) iff A^-1 = A^t

    and if f is a reflection then f(f(x)) = x
    then f(f(x)) = A(Ax + b) + b = x
    i need to somehow prove that A = A^t which then means the matrix is symmetric

    thank you for you time and help
    Last edited: Oct 3, 2005
  2. jcsd
  3. Oct 3, 2005 #2


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    Not every reflection in Rn can be represented by a matrix. Now, do you mean that (1) every reflection in an n-1 dimensional hyperplane passing through the origin in Rn is a symmetric matrix, or do you mean (2) the matrix part of every reflection in Rn is symmetric? If f is a reflection in an n-1 dimensional hyperplane L, then f(x) = Ax + b, where A is the matrix corresponding to a reflection in the n-1 dimensional hyperplane passing through the origin, parallel to the hyperplane L, and if d is the shortest vector from 0 to L, then b = 2d. You know that A is orthogonal, so you know that A-1 = AT. It remains to show that A-1 = A. But this is true since:

    A(Ax + b) + b = x
    A²x + Ab + b = x
    (A² - I)x = -(Ab + b)

    Now you could use the fact that Ab = -b, so the right side is:

    -(Ab + b) = -(-b + b) = -0 = 0

    for all x, hence A² - I = 0, hence A² = I, and thus A = A-1. But suppose you didn't know what Ab was. Regardless, you still know that the right side is:

    -(Ab + b) for all x

    but this is a constant, so you have (A² - I)x = y, where y is a constant vector, and this is true for all x. If you plug in x = 0, then you get 0 = y. In other words, the only linear function that is constant is the 0 function. So without knowing explicitly what Ab is, you can deduce still that the right side must be 0 (and so, if you wanted, could then deduce that Ab = -b), and since the right side is 0 for all x, A² - I = 0, and the desired result follows.
  4. Oct 3, 2005 #3
    Oh wow, that really clears things up. thank you very much akg..
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