Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Matrix Factorization

  1. Jun 15, 2011 #1
    The matrix giving the relation between spherical (unit) vectors and cartesian (unit) vectors can be expressed as:

    [tex]\left( \begin{array}{c} \hat{r} \\ \hat{\phi} \\ \hat{\theta} \end{array} \right) =
    \left( \begin{array}{ccc} \sin\theta \cos\phi & \sin\theta \sin\phi & \cos\theta \\ -\sin\phi & \cos \phi & 0 \\ \cos\theta \cos\phi & \cos\theta \sin\phi & -\sin\theta \end{array}\right) \cdot \left( \begin{array}{c} \hat{x} \\ \hat{y} \\ \hat{z} \end{array} \right) [/tex]

    or

    [tex]T = \left( \begin{array}{ccc} \sin\theta \cos\phi & \sin\theta \sin\phi & \cos\theta \\ -\sin\phi & \cos \phi & 0 \\ \cos\theta \cos\phi & \cos\theta \sin\phi & -\sin\theta \end{array}\right) [/tex]

    where phi is the polar angle and theta is the azimuthal angle.

    Can this matrix T be factored into simpler matrices?
     
  2. jcsd
  3. Jun 17, 2011 #2

    I like Serena

    User Avatar
    Homework Helper

    Hi psholtz! :smile:

    I'm not really sure where you're headed with your question.
    But the short answer would be: no, you can't simplify this matrix.
     
  4. Jun 19, 2011 #3

    jasonRF

    User Avatar
    Science Advisor
    Gold Member

    There is more than one way to decompose this. Usually for this kind of thing you simply perform rotations to construct the transformation from one set of coordinates to the other. In your case it is a little weird since you didn't order your output coordinates in a right-handed way. By the way, I am assuming here that you mis-typed: phi is azimuthal and theta is polar?

    Anyway, here is one option

    [tex]T = \left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}\right) \left( \begin{array}{ccc} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos\theta \end{array}\right) \left( \begin{array}{ccc} \cos \phi & \sin \phi & 0 \\ -\sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{array} \right)[/tex]

    edit: The left-most matrix is not a rotation; a non-rotation is required here since your output coordinates are not in rigt-handed order. I mapped to (theta,phi,r), which is right-handed, and then used the left matrix to map to your ordering. Given my ordering, the right-most matrix rotates about z to make y and phi coincide, the middle matrix then rotates about the y axis to make z and r coincide and x and theta coincide.

    jason
     
    Last edited: Jun 19, 2011
  5. Jun 24, 2011 #4
    The only way to solve this matrix is to try and use LU factorisation ,but if look @ entries (t11 and t13 are will never be reduced and this is similar to the other entries in the matrix ,HENCE this matrix can not be reduced to row echelon form. 4 an advice try and use trigonometry identites and see if u r goinn to find U. or u can try and use (sin) as the scalar multiple of T.
     
  6. Jun 24, 2011 #5
    The only way to solve this matrix is to try and use LU factorisation ,but if look @ entries (t11 and t13 are will never be reduced and this is similar to the other entries in the matrix ,HENCE this matrix can not be reduced to row echelon form. 4 an advice try and use trigonometry identites and see if u r goinn to find U. or u can try and use (sin) as the scalar multiple of T.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Matrix Factorization
  1. Factoring Method (Replies: 7)

  2. Factoring Help (Replies: 2)

  3. Co factor (Replies: 1)

Loading...