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Contraction Transform

  1. Aug 3, 2010 #1
    Hi folks,

    This is my first post here. I hope this is the right forum for this question.

    I am trying to come up with a linear transform to that will take as input a vector (x, y, z) and output a vector that is scaled in the direction of another vector.

    For example:
    Suppose I have the corners of a square defined by the four vectors
    (4, 4, 0)
    (-4, 4, 0)
    (-4, -4, 0)
    (4, -4, 0)

    I want to scale those vectors by 50% in the direction specified by the vector < 1, 1, 0 >

    I want to end up with the four vectors
    (2, 2, 0)
    (-4, 4, 0)
    (-2, -2, 0)
    (4, -4, 0)

    The initial square has been “squashed” by 50% in the northeast/southwest direction.
    Can anybody come up with a transform for that?
     
  2. jcsd
  3. Aug 4, 2010 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    So you want a matrix of the form
    [tex]\begin{pmatrix}a & b & c \\ d & e & f \\ g & h & i\end{pmatrix}[/tex]
    such that
    [tex]\begin{pmatrix}a & b & c \\ d & e & f \\ g & h & i\end{pmatrix}\begin{pmatrix}4 \\ 4 \\ 0\end{pmatrix}= \begin{pmatrix}2 \\ 2 \\ 0\end{pmatrix}[/tex]
    That gives the three equations 4a+ 4b= 2, 4d+ 4e= 2, and 4g+ 4h= 0.

    You also want
    [tex]\begin{pmatrix}a & b & c \\ d & e & f \\ g & h & i\end{pmatrix}\begin{pmatrix}-4 \\ 4 \\ 0\end{pmatrix}= \begin{pmatrix}-4 \\ 4 \\ 0\end{pmatrix}[/tex]
    That gives the three equations -4a+ 4b= -4, -4d+ 4e= 4, -4g+ 4h= 0.

    Doing the same with the other two points will give you a total of twelve equations for the 9 values, a, b, c, d, e, f, g, h, and i. But I think you will find the last three redundant. Once you have the first three points, the requirement that the figure be a parallelogram fixes the fourth.
     
    Last edited: Aug 5, 2010
  4. Aug 4, 2010 #3
    Thank you HallsofIvy.

    But I’m looking for a formula that will convert any vector. I will be using this formula in a computer program that will contract any shape by any amount in any direction.
    This is my current strategy:

    For any given point (x, y, z), direction vector (vx, vy, vz) and scale g;
    Rotate the point so that its x-axis lines up with the direction vector.
    That means two separate rotations for a point in 3d.
    Then multiply the x value by the scale.
    Then back out the two rotations in reverse order.

    I built the following:
    Matrix A rotates the point about the Z-axis
    Matrix B rotates the point about the X-axis
    Matrix C scales only the x dimension.
    Matrix D is the inverse of B
    Matrix E is the inverse of A

    So if P (x, y, z) is the input and P’ (x’, y’, z’) is the output I have
    P’ = [E][D][C][A]P
    Unfortunately, the strategy is not working.
     
  5. Aug 4, 2010 #4
    So you want a matrix T that scales by a factor of g along the vector v, while leaving everything orthogonal to v fixed. That is, Tv = gv, and Tu = u for all u orthogonal to v. You now have the eigenvalues and eigenspaces of T. Since the eigenspaces are orthogonal, to find T all you need is an orthogonal matrix Q such that Qv is on the first coordinate (x) axis; then you'll have T = QTDQ, where D = diag(g, 1, ..., 1) is the diagonal matrix of eigenvalues.

    Observe that since your eigenvalues are real and eigenspaces are orthogonal, T will be a symmetric matrix.
     
  6. Aug 5, 2010 #5
    This sounds great, adriank, but what does Q look like? I'm thinking it has to be a transform that rotates v into the x-axis. And the inverse of Q would rotate it back. I got something that works when u and v are two dimensional vectors.

    But my three dimensional attempt fails.
     
  7. Aug 5, 2010 #6
    I think I got it!

    My problem was more to do with the syntax of the programming language.

    Thank you all.
     
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