For any two elements A and B that form AB, neither A nor B have to be

  • Level: Graduate 
  • Thread starter Thread starter Gear300
  • Start date Start date
  • Tags Tags
    Elements Form
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
8 replies · 3K views
Gear300
Messages
1,212
Reaction score
10
For any two elements A and B that form AB, neither A nor B have to be invertible for (AB)-1 to exist, right?
 
Physics news on Phys.org


Both have to be invertible. Inv(AB) = inv(B)inv(A).
 


Right. For example..
[tex]\left[ <br /> \begin{array}{c c c}<br /> 1&0&0\\<br /> 0&1&0<br /> \end{array}<br /> \right]<br /> \left[ <br /> \begin{array}{c c}<br /> 1&0\\<br /> 0&1\\<br /> 0&0<br /> \end{array}<br /> \right]~=~\left[ <br /> \begin{array}{c c}<br /> 1&0\\<br /> 0&1\\<br /> \end{array}<br /> \right][/tex]

The matrix on the right is clearly invertible, while the two matrices in the product aren't event square, let alone invertible.
 


Thanks for the reply. So that would mean that Inv(AB) = inv(B)inv(A) iff inv(B) and inv(A) exist (whereas Inv(AB) may exist without a defined inv(A) or inv(B)), right?
 


Looks correct to me. The condition that I am familiar with is:

If A and B are invertible matrices of the same size, then AB is invertible and (AB)-1 = B-1A-1.

This is easy to prove by showing that (AB)(B-1A-1) = A(BB-1)A-1 = AIA-1 = AA-1 = I.

As far as the other way of your "iff", if (AB)-1 = B-1A-1 is given then it seems to me that the existence of B-1 and A-1 would directly follow since they are used in the initial condition.
 


I see. Thanks again for the replies. I have an additional question:

For a matrix C that is not a square matrix, there is no defined inverse; however, it is possible that there is a left inverse A and a right inverse B, in which A =/= B, for the matrix C, right?
 


Thanks for the link.

So in the case of linear systems Ax = b, I suppose it wouldn't always be possible to use the left inverse of A to isolate x as a general method since the left inverse of A does not necessarily exist. How would one isolate x in these linear systems (other than parametrization of x)?