Do Commuting Linear Operators A and B Satisfy the Exponential Property?

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Discussion Overview

The discussion revolves around the relationship between two linear operators, A and B, and the conditions under which the exponential property e^A * e^B = e^(A+B) holds true. The scope includes theoretical aspects of linear algebra and operator theory.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant states that e^A * e^B = e^(A+B) is true when A and B commute (AB=BA).
  • Another participant questions whether the condition is necessary or merely sufficient for the exponential property to hold.
  • A different participant raises a point about the requirement for A and B to be of the same size for A+B to exist and suggests that they must be square matrices to have eigenvectors and eigenvalues.
  • There is uncertainty expressed regarding the clarity of the original question and the need for verification of the conditions mentioned.

Areas of Agreement / Disagreement

Participants generally agree that A and B must commute for the exponential property to hold, but there is no consensus on whether this condition is necessary. Additionally, there is uncertainty about the clarity of the question posed.

Contextual Notes

Limitations include the need for A and B to be of the same size and square for eigenvalues and eigenvectors to be defined, as well as the unresolved nature of whether commuting is a necessary condition for the exponential property.

frederick
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If A & B are linear operators, and AY=aY & BY=bY, what is the relationship between A & B such that e^A*e^B=e^(A+B)?? --where e^x=1+x+x^2/2+x^3/3!+...+x^n/n!
 
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e^A e^B = e^(A+B) is true when A and B commute (AB=BA). Or are you asking when e^A e^B Y=e^(A+B) Y, where Y is an eigenvector of A and B? That is always true.
 
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e^A*e^B=e^(A+B)??

A and B commute is sufficient - I don't know if it's necessary.
 
I'm not quite sure what the question is asking here...

I definitely did pick up that A and B must be the same size for A+B to exist (and must be square to have eigenvectors and values). Also, you need AB=BA (I think?). Someone else should verify this
 

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