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

Dividing by matrices

  1. Sep 24, 2011 #1
    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 [itex] A [/itex]. If [itex] B = \lambda A [/itex], it seems natural to define [itex] \frac{B}{A} := \lambda I [/itex]. However, I don't think this operation is well-defined. Since [itex]A[/itex] is singular, left multiplication by [itex]A[/itex] has a nontrivial kernel, so there is some nonzero vector [itex] v [/itex] such that [itex] Av = 0[/itex]. Letting [itex] V[/itex] be the matrix with columns [itex] v [/itex], then [itex] B = A \cdot \lambda I = A \cdot (\lambda I + V)[/itex], so [itex] \frac{B}{A}[/itex] could just as well be equal to [itex] \lambda I + V[/itex].

    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. jcsd
  3. Sep 24, 2011 #2


    User Avatar
    Science Advisor

    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
  4. Sep 28, 2011 #3
    Thanks for the reply--the links were very interesting.

    I guess what makes [itex] \lambda I [/itex] the "natural" choice for [itex] \frac{B}{A} [/itex] if [itex] B = \lambda A [/itex] is that [itex] \lambda I [/itex] 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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook