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

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Discussion Overview

The discussion revolves around the conditions under which the inverse of the product of two matrices exists, specifically addressing whether the individual matrices need to be invertible for the inverse of their product to exist. The scope includes theoretical aspects of linear algebra and matrix properties.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that neither A nor B needs to be invertible for (AB)-1 to exist.
  • Others argue that both matrices must be invertible, citing the formula Inv(AB) = inv(B)inv(A) as a basis for their claim.
  • A participant provides a counterexample using specific matrices to illustrate that the product can be invertible even if the individual matrices are not square or invertible.
  • There is a discussion about the conditions under which Inv(AB) = inv(B)inv(A) holds true, suggesting that the existence of the inverses of A and B is not necessary for (AB)-1 to exist.
  • A question is raised regarding the existence of left and right inverses for a non-square matrix C, with acknowledgment that A and B can be different.
  • Another participant questions the applicability of the left inverse in solving linear systems, indicating that it may not always be a viable method.

Areas of Agreement / Disagreement

Participants express disagreement regarding the necessity of invertibility for the individual matrices A and B. Some maintain that invertibility is required, while others provide examples and reasoning to suggest that it is not. The discussion remains unresolved with multiple competing views.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about matrix dimensions and the definitions of inverses. The mathematical steps leading to conclusions about the existence of inverses are not fully resolved.

Gear300
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For any two elements A and B that form AB, neither A nor B have to be invertible for (AB)-1 to exist, right?
 
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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.
 


defunc said:
Both have to be invertible. Inv(AB) = inv(B)inv(A).
See my counterexample.
 


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?
 


Yes. See the wikipedia article here, under the section titled Matrices.
 


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)?
 

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