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Solving Ax=b, when b is an eigenvector

  1. Apr 5, 2009 #1
    Let A be a nonsingular matrix. What can you say about the convergence of GMRES to the solution of Ax=b when b is an eigenvector of A?

    I know that if A is a nonsingular matrix with minimal polynomial of degree m, then GMRES solves Ax=b in at most m iterations.

    Because b is an eigenvector of A then: (A-[tex]\lambda[/tex]I)b=0.

    The only thing I've sort of come up with is that because b is an eigenvector, it cannot equal 0. At first I though that that implies that A-[tex]\lambda[/tex]I=0, but I'm thinking that that is not necessarily true. However, IF it is true that A-[tex]\lambda[/tex]I=0, I completely understand, because this would imply that the minimal polynomial is q(x)=x-[tex]\lambda[/tex]. This would be a minimal polynomial of degree 1, so GMRES would converge in 1 iteration. Can anyone help me out because I don't think this reasoning is correct, but I'm thinking that A-[tex]\lambda[/tex] is not necessarily nonsingular...or is it? Thank you!!
     
  2. jcsd
  3. Apr 6, 2009 #2
    Can anyone offer any insight?
     
  4. Apr 7, 2009 #3
    I just read the Wikipedia page on GMRES. Perhaps I misinterpret, but it seems like GMRES sould solve [tex] Ax = b [/tex] in just one step if [tex] b [/tex] is an eigenvector, since the exact solution [tex] x = b/\lambda [/tex] is in the span of [tex] b [/tex] (i.e., the first Krylov subspace).
     
  5. Apr 8, 2009 #4
    Oh wow! That makes so much sense and seems so obvious. Thank you!
     
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