Moore penrose inverse expressed as geometric product?

Peeter
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I getting far enough into my geometric algebra books now that I'm at linear transformations, including the result showing how a linear transformation inverse can be expressed directly as a geometric product, using the adjoint and pseudoscalar multiplication, instead of using matrix inversion.

Really I think it amounts to the same thing since to calculate the adjoint of the linear transformation, you'll have to pick a basis and calculate the reciprocal frame vectors to find the components of the adjoint transformation (at least that's the way "Geometric algebra for physicists" outlines it).

Anyhow, the idea is interesting, and makes me wonder if it can be carried further. In particular I observe that the geometric product vector inverse is consistent with the moore penrose "generalized inverse" of a Nx1 matrix. Since that inverse is intrinsically related to projection onto the matrix image, and we can express subspace projection so naturally with the geometric product (ie: dot product of blades), I'd guess that the Moore Penrose inverse could be also be expressed using the geometric product, and that this may highlight some interesting structural features of the generalized inverse hard to see in matrix form.

Since I'm a newbie to the subject I'd also guess that somebody else has already done this (it's not in my books though). Any pointers to where to look (if not would probably be fun to try to calculate)?
 
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Have you tried singular value decomposition of a pseudoinverse? Then you will have rotation, scaling and rotation again... That will give you the intuition.
 
thanks for the hint. I'll look at that (that's mentioned later in the linear algebra chapter but I hadn't worked through an example or details and didn't make the connection).
 
Took a look at some info on SVD. Wow, that's a pretty powerful construct. Gives you rank, an orthonormal basis for both kernel and image, and inverse or pseudoinverse ... all in one shot.
 
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