Divide by Matrix: Is it Possible?

[spamiam]

In post #7 of https://www.physicsforums.com/showthread.php?t=532666" thread, the OP asked whether one could meaningfully divide by a matrix. Certainly this is possible for invertible matrices, but I'm wondering if it's possible to define something similar even for singular matrices.

For instance, suppose I have a singular matrix A. If B = \lambda A, it seems natural to define \frac{B}{A} := \lambda I. However, I don't think this operation is well-defined. Since A is singular, left multiplication by A has a nontrivial kernel, so there is some nonzero vector v such that Av = 0. Letting V be the matrix with columns v, then B = A \cdot \lambda I = A \cdot (\lambda I + V), so \frac{B}{A} could just as well be equal to \lambda I + V.

My question is, is there a way to make this division well-defined? Would working over a ring with specific properties help?

 

[pwsnafu]

Firstly, if you can't do it over a field, you probably can't do it over a ring. The field is the best case for scalars.

What you are looking for is what we call http://en.wikipedia.org/wiki/Von_Neumann_regular_ring" which has uniqueness.

There is also the http://en.wikipedia.org/wiki/Drazin_inverse" . The Drazin inverse has a further generalization called the g-Drazin inverse aka Drazin-Koliha inverse, but this is more C*-algebra stuff, rather than matrix theory proper.

 

[spamiam]

Thanks for the reply--the links were very interesting.

I guess what makes \lambda I the "natural" choice for \frac{B}{A} if B = \lambda A is that \lambda I is a scalar matrix and lies in the center of the ring. I guess these various pseudoinverses all have some nice properties and I'll have to work out some examples.