1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Homework Helper
    Gold Member

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Linear Transformation question
  1. Linear Transformations (Replies: 2)

  2. Linear Transformation (Replies: 6)

  3. Linear transformation (Replies: 6)

  4. Linear Transformations (Replies: 0)

Loading...