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

[frederick]

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!

 

[StatusX]

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.

 

[mathman]

e^A*e^B=e^(A+B)??

A and B commute is sufficient - I don't know if it's necessary.

 

[Office_Shredder]

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