Matrix Factorization: Spherical & Cartesian Vectors

[psholtz]

The matrix giving the relation between spherical (unit) vectors and cartesian (unit) vectors can be expressed as:

\left( \begin{array}{c} \hat{r} \\ \hat{\phi} \\ \hat{\theta} \end{array} \right) = <br /> \left( \begin{array}{ccc} \sin\theta \cos\phi &amp; \sin\theta \sin\phi &amp; \cos\theta \\ -\sin\phi &amp; \cos \phi &amp; 0 \\ \cos\theta \cos\phi &amp; \cos\theta \sin\phi &amp; -\sin\theta \end{array}\right) \cdot \left( \begin{array}{c} \hat{x} \\ \hat{y} \\ \hat{z} \end{array} \right)

or

T = \left( \begin{array}{ccc} \sin\theta \cos\phi &amp; \sin\theta \sin\phi &amp; \cos\theta \\ -\sin\phi &amp; \cos \phi &amp; 0 \\ \cos\theta \cos\phi &amp; \cos\theta \sin\phi &amp; -\sin\theta \end{array}\right)

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

Can this matrix T be factored into simpler matrices?

 

[I like Serena]

Hi psholtz!

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.

 

[jasonRF]

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

T = \left( \begin{array}{ccc} 0 &amp; 0 &amp; 1 \\ 0 &amp; 1 &amp; 0 \\ 1 &amp; 0 &amp; 0 \end{array}\right) \left( \begin{array}{ccc} \cos \theta &amp; 0 &amp; -\sin \theta \\ 0 &amp; 1 &amp; 0 \\ \sin \theta &amp; 0 &amp; \cos\theta \end{array}\right) \left( \begin{array}{ccc} \cos \phi &amp; \sin \phi &amp; 0 \\ -\sin \phi &amp; \cos \phi &amp; 0 \\ 0 &amp; 0 &amp; 1 \end{array} \right)

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

 

[nceba]

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.

 

[nceba]

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.