Proving (AB)^{-1}=B^{-1} A^{-1} in Operator Algebra

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SUMMARY

The discussion centers on proving the identity (AB)^{-1} = B^{-1} A^{-1} in the context of operator algebra. Participants clarify that the proof involves verifying that (AB)(B^{-1} A^{-1}) equals the identity operator I, which is indeed true by the definition of an inverse. The confusion arises from a misinterpretation of Cramer's rule, which does not apply to the properties of operator inverses in this context.

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ian2012
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I hope someone can help me with this:

Let the the inverse A A^{-1}=A^{-1} A=I, where I is the identity operator. Proofing that (AB)^{-1}=B^{-1} A^{-1} :

"First, you want to check whether (AB)(B^{-1} A^{-1})=I. "

However that means the inverse of AB multiplied by AB gives the identity operator, which isn't true, surely, due to Cramer's rule?
 
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ian2012 said:
that means the inverse of AB multiplied by AB gives the identity operator, which isn't true
It is. That's just the definition of "inverse".
 

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