Inverse of the difference of two matrices A and B

[yukawa]

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~

 

[HallsofIvy]

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~

 

[statdad]

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)

 

[HallsofIvy]

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.