Inverse of the difference of two matrices A and B

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Homework Help Overview

The discussion revolves around the expression for the inverse of the difference of two matrices, A and B, where it is proposed that 1/(A-B) can be expressed as a series involving the inverses of A and B. Participants are exploring the conditions under which this expression holds true and the implications of matrix non-commutativity.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss starting from the left-hand side of the equation and manipulating it to derive the right-hand side. There is mention of using the geometric series sum formula, but concerns are raised about the non-commutativity of matrices affecting the validity of the approach. Questions arise regarding the assumptions necessary for the geometric series to apply in the context of matrices.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and raising questions about the assumptions involved in applying the geometric series to matrices. Some guidance has been offered regarding the need for caution in the application of the geometric series formula, particularly in relation to matrix norms and commutativity.

Contextual Notes

There is an acknowledgment that the problem may be poorly worded, and participants are considering the implications of the norms of the matrices involved. The discussion highlights the importance of clarifying the conditions under which the proposed formula is valid.

yukawa
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Homework Statement


Show that
1/(A-B) = (1/A) + (1/A)B(1/A) + (1/A)B(1/A)B(1/A) + (1/A)B(1/A)B(1/A)B(1/A)+...

where A and B are matrices whose inverse exist.


The Attempt at a Solution


I tried to start from the LHS by pulling out (1/A):
LHS = (1/A)[1 + B(1/A) + B(1/A)B(1/A) + B(1/A)B(1/A)B(1/A)+...]
= (1/A)*{1/[1-B(1/A)]} where I have used equation of GP sum to infinity
= 1/(A - AB(1/A))
However, i can't get the expression in RHS since A and B are not commute.

Are there any other possible approach to this problem?

Any help would be great~thanks~
 
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yukawa said:

Homework Statement


Show that
1/(A-B) = (1/A) + (1/A)B(1/A) + (1/A)B(1/A)B(1/A) + (1/A)B(1/A)B(1/A)B(1/A)+...
So, more simply, (1/A)(1/1-(B/A))= (1/A)[(B/A)+ (B/A)2+ (B/A)3+ ...)

Do you know the proof of the sum of a geometric series?>

where A and B are matrices whose inverse exist.


The Attempt at a Solution


I tried to start from the LHS by pulling out (1/A):
LHS = (1/A)[1 + B(1/A) + B(1/A)B(1/A) + B(1/A)B(1/A)B(1/A)+...]
= (1/A)*{1/[1-B(1/A)]} where I have used equation of GP sum to infinity
= 1/(A - AB(1/A))
However, i can't get the expression in RHS since A and B are not commute.

Are there any other possible approach to this problem?

Any help would be great~thanks~
 


The geometric series approach came to mind as well, but since these are matrices some care needs to be taken.
If this is simply a formal problem the suggestion makes sense: usually, however, it is not enough to go through the mechanics - some conditions on the norms of the matrices are required.
This seems to be a rather poorly worded problem
(BUT, I agree with the suggestion of HallsofIvy - that seems to be the intent here)
 


You cannot assume that the formula for the sum of a geometric series of real numbers, but you can copy the proof- being careful that you don't assume commutativity.
 

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