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A weird eigenvalue problem

  1. May 25, 2009 #1
    Hi,

    Is there any solution for the following problem:

    [tex]Ax = \lambda x + b[/tex]

    Here [tex]x[/tex] seems to be an eigenvector of [tex]A[/tex] but with an extra translation vector [tex]b[/tex].
    I cannot say whether [tex]b[/tex] is parallel to [tex]x[/tex] ([tex]b = cx[/tex]).

    Thank you in advance for your help...

    Birkan
     
    Last edited: May 26, 2009
  2. jcsd
  3. May 26, 2009 #2

    jbunniii

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    You didn't say what vector space you are working with, so I will assume a complex finite-dimensional vector space.

    We can rewrite your problem as

    [tex](A - \lambda I)x = b[/tex]

    If [tex]b = 0[/tex], this has a solution if and only if lambda is an eigenvalue of A. Every map in a complex finite-dimensional vector space has an eigenvalue, so a solution exists in this case.

    If [tex]b \neq 0[/tex], then this is equivalent to

    [tex]b \in image(A - \lambda I)[/tex]

    For this to happen, it suffices that [tex]A - \lambda I[/tex] be surjective. This is true if and only if [tex]A - \lambda I[/tex] is invertible, which is true if and only if lambda is NOT an eigenvalue of A. Thus there are plenty of solutions in this case!
     
  4. May 26, 2009 #3

    I got it. By the way, vector space is actually finite-dimensional (d=9000) Euclidean Space.

    Since I do not know the [tex]\lambda[/tex], (only [tex]A[/tex] and [tex]b[/tex] are known) how can I find an numeric solution for that? Is there any way using eigenvalue logic here?
    Such as
    -- find eigenvalues of [tex]A[/tex],
    -- check if [tex]b[/tex] is parallel to any
    -- select the appropriate eigenvector etc.
     
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