# Dividing by matrices

1. Sep 24, 2011

### 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?

Last edited by a moderator: Apr 26, 2017
2. Sep 24, 2011

### 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" [Broken] which has uniqueness.

There is also the http://en.wikipedia.org/wiki/Drazin_inverse" [Broken]. 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.

Last edited by a moderator: May 5, 2017
3. Sep 28, 2011

### spamiam

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.
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