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Linear Operators, Eigenvalues

  1. Oct 29, 2006 #1
    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!
     
  2. jcsd
  3. Oct 30, 2006 #2
    Hint: write it out. And remember to watch out for which side you're multiplying on with A or B since, in general, [itex]AB \neq BA[/itex].
     
  4. Nov 1, 2006 #3

    dextercioby

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    Since Y is a common eigenvector for A and B (assuming of course they don't have a continuous spectrum), then A and B commute. Then using Baker-Campbell-Hausdorff formula one sees that "A and B commute" is enough to have [itex] e^{A}e^{B}=e^{A+B} [/itex].

    Daniel.
     
  5. Nov 1, 2006 #4

    Hurkyl

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    What am I missing? Consider the matrices:

    [tex]
    \left(\begin{array}{cc}
    3 & 0 \\
    1 & 1 \\
    \end{array}\right)[/tex] and [tex]\left(
    \begin{array}{cc}
    2 & 0 \\
    0 & 1 \\
    \end{array}\right)
    [/tex]

    These share a common eigenvector [itex][0 1]^T[/itex], but don't commute.
     
  6. Nov 1, 2006 #5

    dextercioby

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    Yes, sloppy me, the theorem goes: 2 linear operators commute iff they share a COMPLETE system of common eigen vectors.

    Daniel.
     
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