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

In summary, the definition of inverse in operator algebra states that an operator is considered the inverse of another operator if their product results in the identity operator. Proving the property (AB)^{-1}=B^{-1}A^{-1} is important in simplifying expressions and solving equations in operator algebra. The proof for this property involves using the definition of inverse and basic algebraic manipulations. This property can be extended to any number of operators and is applicable to all types of operators that satisfy the definition of inverse, such as linear, unitary, and Hermitian operators.
  • #1
ian2012
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I hope someone can help me with this:

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

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

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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  • #2
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".
 

1. What is the definition of inverse in operator algebra?

In operator algebra, the inverse of an operator is defined as another operator that, when multiplied together, results in the identity operator. In other words, an operator A is considered the inverse of another operator B if A*B = B*A = I, where I is the identity operator.

2. Why is it important to prove that (AB)^{-1}=B^{-1}A^{-1} in operator algebra?

Proving this property is important because it allows us to manipulate and simplify algebraic expressions involving operators, making it easier to solve equations and perform calculations in operator algebra.

3. How is the proof for (AB)^{-1}=B^{-1}A^{-1} derived?

The proof for (AB)^{-1}=B^{-1}A^{-1} is derived using the definition of inverse in operator algebra and basic algebraic manipulations. It involves showing that (AB)(B^{-1}A^{-1}) = (BA)(B^{-1}A^{-1}) = I, which satisfies the definition of inverse.

4. Can this property be extended to more than two operators?

Yes, this property can be extended to any number of operators. For example, for three operators A, B, and C, we can show that (ABC)^{-1} = C^{-1}B^{-1}A^{-1}.

5. Is the property (AB)^{-1}=B^{-1}A^{-1} applicable to all types of operators?

Yes, this property is applicable to all types of operators as long as they satisfy the definition of inverse. This includes linear operators, unitary operators, and Hermitian operators.

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