How to Derive the Inverse of the Sum of Two Operators?

[poonintoon]

Hi,
I am trying to show that for two operators S and P:
(S+P)^{-1}=S^{-1}-S^{-1}P(S+P)^{-1}
I can't get anywhere and searching on google I am not even sure if it is possible
to solve the general case but the question gives no more hints.
Any help appreciated. Thanks.
J.

 

[micromass]

You'll need to show that (S^{-1}-S^{-1}P(S+P)^{-1})(S+P)=I. Id begin by multiplying both sides by S...

 

[HallsofIvy]

Hi,
I am trying to show that for two operators S and P:
(S+P)^{-1}=S^{-1}-S^{-1}P(S+P)^{-1}
I can't get anywhere and searching on google I am not even sure if it is possible
to solve the general case but the question gives no more hints.
Any help appreciated. Thanks.
J.

Are you sure about this? The standard way to show that A is the multiplicative inverse of B is to multiply them together to show that you get the identity. But if we multiply both sides, on the right, by S+ P we get
(S+ P)^{-1}(S+ P)= S^{-1}(S+ P)+ S^{-1}P(S+P)^{-1}(S+P)
I= S^{-1}S+ S^{-1}P+ S^{-1}= I+ S^{-1}(P+ S)
S^{-1}(P+S)= 0
which is certainly not always true!

 

[D H]

Your approach was correct but you made an error, Halls.

poonintoon: Follow Hall's approach and you will have your proof.

 

[HallsofIvy]

Actually two errors! I lost a negative sign and a "P"!

 

[Anonymous217]

Just a quick curiosity, but how would you find that inverse without having known it in the first place? Surely it wasn't found by simply trying different functions was it?

 

[D H]

Just a quick curiosity, but how would you find that inverse without having known it in the first place?

It exists. That's all one needs to know. Well, that and the fact that operators form a ring.

 

[HallsofIvy]

Just a quick curiosity, but how would you find that inverse without having known it in the first place? Surely it wasn't found by simply trying different functions was it?

Why not? That's a well respected method!

 

[Anonymous217]

Why not? That's a well respected method!

Well, I was hopeful of some type of actual procedure so that it could be applied to more complicated operations, but I guess not.

 

[Office_Shredder]

The more you guess and check the better you get at it