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Eigenvalues of linear operators

  1. Apr 28, 2009 #1
    Let V be the vector space of all real integrable functions on [0,1] with inner product [tex]<f,g>=\int_0^1 f(t)g(t)dt[/tex]

    Three linear operators defined on this space are [tex] A=d/dt [/tex] and [tex]B=t[/tex] and [tex]C=1[/tex] so that [tex]Af=df/dt[/tex] and [tex]Bf=tf[/tex] and [tex]Cf=f[/tex]

    I need to find the eigenvalues of these operators:

    For A:
    [tex]\frac{df}{dt} = \lambdaf[/tex]
    [tex]\frac{d ln(f)}{dt} = \lambda [/tex]
    [tex]d(ln(f))=\lambda dt+c[/tex]
    [tex]f=ce^{\lambda t}[/tex]

    So the eigenvalues for A are continuous and can be any real number.

    For B:
    [tex]tf=\lambda f[/tex]
    [tex]\lambda =t [/tex]

    The eigenvalues are the variable t?

    For B;
    [tex]Cf=f=\lambda f[/tex]
    The eigenvalue is 1.

    Confused about B.
     
  2. jcsd
  3. Apr 29, 2009 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Re: eigenvalues

    I'm not sure I like the statement "the eigenvalues of A are continuous". What does it mean for a number to be "continuous"? But, yes, every real number is an eigenvalue- the set of eigenvalues is a continuous interval rather than discreet.

    No, eigenvalues are NUMBERS, not variables. [itex]tf= \lambda f[/itex] if and only if f(t)= 0 for all t, the "zero vector". This operator has NO eigenvalues.

     
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