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Homework Help: Linear Transformation question

  1. Feb 5, 2005 #1
    i was trying to figure out something that i didn't understand and the book doesn't have much examples of it either. My question is how do u know whether a transformation is a projection on a line, reflection on a line, or rotation through an angel? With T given. The questions i did from the book, i was able to find the line and reflection, since the same question was previously stated and i was able to do it, n got the answer from there, but didn't understand it.
    Take T[x y]=1/2[x-y y-x] and by solving these i get [1 -1]/[-1 1] (this is not division, just goes at the bottom, its a 2x2 matrix, and the vectors r transposed) so by looking at this
    [ 1 -1]
    [-1 1]
    how can u tell whether its a reflection, projection or rotation?
    Similarly another result i got for another part was
    [1/2 root3/2]
    [root3/2 1/2 ]
    i think this one is rotation but how can u tell?
     
  2. jcsd
  3. Feb 6, 2005 #2

    Galileo

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    A rotation about the origin is a transformation which preserves the lengths and angles of vectors. It's called an orthogonal transformation. They are given by an orthogonal matrix, possesing the following property:
    [tex]A^TA=I[/tex]
    where the T denotes the transpose operation.
    Furthermore: if det(A)=1, then it's a rotation, if det(A)=-1, then it's a reflection followed by a rotation.
    Also, a rotation over an angle [itex]\theta[/itex] can always be given by the following matrix:

    [tex]\left( \begin{array}{cc}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array} \right)[/tex]
    Since [itex]\cos \frac{\pi}{3}=1/2[/itex] and [itex]\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}[/itex] the matrix:
    [tex]\left( \begin{array}{cc}1/2 & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & 1/2 \end{array} \right)[/tex]
    is matrix for the rotation about an angle of [itex]\pi/3[/itex]. (Did you forgot a minus sign in your matrix?)

    Projections always have the following properties:
    [tex]P^T=P[/tex]
    [tex]P^2=P[/tex].
     
  4. Feb 6, 2005 #3

    ehild

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    You have omitted 1/2.

    [tex] T=1/2 \left(\begin {array}{cc} 1&-1\\ -1& 1\end{array}\right)[/tex]

    Find out what does this transformation do with the base vectors. You will see that they are projected onto the [1,-1] direction.

    To be a rotation, the matrix should be unitary, and yours is not (the determinant should be 1).
    A matrix that represents an anti-clockwise rotation by angle alpha in the (xy) plane is

    [tex]T=\left(\begin {array}{cc}\cos(\alpha)&-\sin(\alpha)\\ \sin(\alpha)&\cos(\alpha)\end{array}\right )[/tex]

    ehild
     
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