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Affine fixed points

  1. Dec 5, 2008 #1

    EWW

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    Hello,

    I'm having difficulties with finding fixed points of affine transformations. I understand that given a matrix A of barycentric coefficients, I want to produce a point that is equal to the given point, i.e. Ax = y, where y = x. But all I get is a homogenous linear system whose only solution is zero? Help and thanks, EWW.
     
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  3. Dec 8, 2008 #2

    turin

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    I don't know anything about this, other than what I just read on wikipedia, but is this somehow different from simply finding the eigenvectors of A?
     
  4. Dec 9, 2008 #3

    EWW

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    I understand that for the matrix of an affine transformation, if I take a point x, y and send it through the matrix, I get another point, x', y'. For the fixed point of an affine transformation, the two points are equal. For example, with some 2 X 2 matrix with vector columns -2,3 and 5,-4, I want to send point x, y through it. After multiplying through I get a system of linear equations, like -2x + 5y = x', 3x - 4y = y'. I set x' = x, y'= y, then get 3x - 4y = 0 twice. I'm clearly not getting something- does the solution to finding the fixed point have something to do with how an affine transformation can be composed of a linear part and an affine part? I notice also that in some instances of affine transformations, a 4 x 4 matrix is used for a 3D space . . . with the bottom diagonal entry always a 1. I need to study more of this and maybe the fixed point problem will become clear. Also, I don't know much about eigenvectors yet, so maybe more work on those will help. Thanks for the comment, EWW.
     
  5. Dec 10, 2008 #4

    turin

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    I think that, since this line passes through the origin, then every point on this line would correspond to the same point in the projected space, in this case a 1-D space. So I think this is the right idea. You just have to project the result into the space. However, I think that the matrix cannot be so general. I think that the bottom row must be all zeros except for the last entry that must be a 1. I'm not sure about that.



    Yes, this is what I though it means for a matrix to represent an affine transformation. It's a really cool idea, actually, that I had never heard of, so thanks for posting this question. Are you familiar with a 3x3 matrix representing a rotation? If so, then the 4x4 matrix representing the affine transformation just has this kind of 3x3 matrix in the upper left, and then the right-most column is a translation vector. the bottom diagonal 1 is just to close the group of matrices (so that the combination of two affine transformation matrices is also an affine transformation matrix).

    BTW, did your particular question have something to do with a tetrahedron?



    Well, I suppose you don't need to. You are essentially assuming an eigenvalue of 1 and finding the corresponding eigenvector(s).
     
  6. Dec 17, 2008 #5

    EWW

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    Thanks Turin,

    I went back and carefully worked out how an affine transformation is composed of a linear transformation and a translation f(v) = l(v) + a. For a triangle, your left with a 2 X 2 matrix plus the translation vector. If you imagine this to be embedded in 3-space you can shift the plane of the triangle to z = 1, then you would get a 3 X 3 matrix which has the 2 X 2 matrix of l(v) in the upper left, a next to it, and the bottom row would be 0,0,1. To find the fixed point, you set x = x', and you get a system of linear equations that can be solved.

    So you don't use barycentric coordinates if you're trying to find a fixed point, you use the vector coordinates whose basis and origin refer to the triangle.

    I've been able to find the fixed point of a triangle transformation synthetically, does anyone know of a good way to do this for a tetrahedron? Since I'm using intersections of lines for the 2- space case, I've speculated that I would just use intersections of planes for 3-space . . . . haven't done it yet, but maybe there's another way to do this.

    Yes, this has everything to do with tetrahedrons, I'm hoping to create architectural forms out of these transformations. EWW.
     
  7. Dec 18, 2008 #6

    turin

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    I don't know what you mean by "triangle transformation" nor "synthetically". Please explain.

    I'm supposin' that you just get a system of 3 equations in three variables (x,y,z) with coefficients in terms of the rotation and translation elements, but I might not understand what is your goal.
     
  8. Dec 22, 2008 #7

    EWW

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    By "triangle transformation" I mean the affine transformation of 3 points in 2-space.

    By "synthetically", I mean by ruler and compass- sometimes this method leads to spatial/geometrical insights that I wouldn't have thought of.

    My goal is to apply mathematical transformations to architectural transformations.
     
  9. Dec 22, 2008 #8

    turin

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    That's not what I meant by "goal", but anyway, I think that all you have to do is to solve a linear system of three equations in three variables. There is nothing more to it. From your previous posts I gathered that you were comfortable to do this.

    For concreteness, we can work together through the following example of an affine transformation in 3-D which envolves rotating about the y-axis by [itex]\theta[/itex], then rotating about the z-axis by [itex]\phi[/itex], and finally translating by [itex]\Delta{x}[/itex] in the x-direction. Note that the order DOES matter.

    [tex]
    \left(\begin{array}{cccc}
    \cos\theta\cos\phi&-\sin\phi&\sin\theta\cos\phi&\Delta{x}\\
    \cos\theta\sin\phi&\cos\phi&\sin\theta\sin\phi&0\\
    -\sin\theta&0&\cos\theta&0\\
    0&0&0&1
    \end{array}\right)
    \left(\begin{array}{c}
    x\\y\\z\\1
    \end{array}\right)
    =
    \left(\begin{array}{c}
    x\\y\\z\\1
    \end{array}\right)
    [/tex]

    Your first task is to get three equations out of this, linear in x, y, and z. The equations will be inhomogeneous, in general, due to the translation. I believe that, if the translation is actually zero, then this matrix equation has no (real) solution unless the rotation is also trivial. So, for now, let's assume that all three transformation parameters are nonzero.

    I may not be back on the forum for a while, so anyone else is more than welcome to step in to help out.
     
    Last edited: Dec 22, 2008
  10. Dec 24, 2008 #9

    EWW

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    Thanks a lot Turin, I'll review your post and work through your example . . . happy holidays, EWW
     
  11. Jan 7, 2009 #10

    turin

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    OK, I am back. I actually gave this some thought over the holidays, and as is usually the case when I try to explain stuff (and the reason why I love doing it), I discovered some interesting features of combining rotations and translations. One interesting result was that, in order to have a fixed point, the translation must be in the plane perpendicular to the axis of rotation. Of course, this is quite intuitive in retrospect: since the effect of the rotation must exactly cancel the effect of the translation, any translation parallel to the axis of rotation cannot be canceled by the rotation, so such an affine transformation cannot have a fixed point. In other words, be careful:

    Not all affine transformations have (real) fixed points!!!

    In particular, I'm afraid, the equation that I wrote in my previous post has no (real) solution. I appologize. You can see this if you calculate the axis of rotation and find that it is not perpendicular to the x-axis. That was the other thing that I discovered: how to determine the resultant angle and axis of rotation produced by the rotation about y followed by the rotation about z. Well, actually, I did not quite finish the calculation, but maybe I will do it soon. At any rate, I will generalize the equation to discuss, and again appologize for the unsolvable equation in my previous post.

    [tex]
    \left(\begin{array}{cccc}
    \cos\theta\cos\phi&-\sin\phi&\sin\theta\cos\phi&\Delta{x}\\
    \cos\theta\sin\phi&\cos\phi&\sin\theta\sin\phi&\Delta{y}\\
    -\sin\theta&0&\cos\theta&\Delta{z}\\
    0&0&0&1
    \end{array}\right)
    \left(\begin{array}{c}
    x\\y\\z\\1
    \end{array}\right)
    =
    \left(\begin{array}{c}
    x\\y\\z\\1
    \end{array}\right)
    [/tex]

    There is now actually a restriction on [tex]\Delta{x}[/tex], [tex]\Delta{y}[/tex], and [tex]\Delta{z}[/tex], namely that the vector must be perpendicular to the axis of rotation. But, maybe we can discuss the generic fixed point without this restriction.
     
    Last edited: Jan 7, 2009
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