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Eigenvalue Problem

  1. Jul 16, 2009 #1
    1. The problem statement, all variables and given/known data
    Let A be an nxn matrix and let I be the nxn identity matrix. Compare the eigenvectors and eigenvalues of A with those of A+rI for a scalar r.


    2. Relevant equations



    3. The attempt at a solution
    I think I should be doing something like this:
    det(A-[tex]\lambda[/tex]I), and

    det((A+rI)-[tex]\lambda[/tex]I)=det(A-([tex]\lambda[/tex]-r)I).

    The eigenvalues would be [tex]\lambda[/tex] where the det(A-[tex]\lambda[/tex]I)=0
    and det(A-([tex]\lambda[/tex]-r)I)=0.

    So does that mean the eigen values for the first matrix are [tex]\lambda[/tex] =n and the second will be n+r?
     
    Last edited by a moderator: Jul 16, 2009
  2. jcsd
  3. Jul 16, 2009 #2

    Mark44

    Staff: Mentor

    An eigenvector of A is a nonzero vector x such that (A - [itex]\lambda[/itex])x = 0, and [itex]\lambda[/itex] is the eigenvalue for that eigenvector.

    Can you say something in this vein for the eigenvectors of A + rI?
     
  4. Jul 17, 2009 #3
    [/B][/B]
    The eigenvectors are the solution to:
    (A-([tex]\lambda[/tex]-r)I)x=0
    What do you suppose is meant by "compare the eigenvectors and eigenvalues?"
     
  5. Jul 17, 2009 #4

    Mark44

    Staff: Mentor

    So now put together what I wrote and what you wrote.
    An eigenvalue of A is a number [itex]\lambda[/itex] such that (A - [itex]\lambda[/itex]I)x = 0.
    An eigenvalue of A + rI is a number ? such that (A - ([itex]\lambda[/itex] - r)I)x = 0.

    (Fill in the question mark.)
    What can you say about the values of x in either case?
     
  6. Jul 18, 2009 #5
    I don't see why I should be assuming that [tex]\lambda[/tex] in the first equation should be equal to [tex]\lambda[/tex] in the second equation. The same goes for x. It's really giving me trouble as I work with this.
     
    Last edited: Jul 18, 2009
  7. Jul 18, 2009 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    lambda is an eigenvalue corresponding to eigenvector x if Ax=lambda*x, right? Then (A+rI)x=Ax+rx=(lambda+r)x, also right? I haven't changed any lambda's or x's. What does (A+rI)x=(lambda+r)x tell you about eigenstuff of A+rI?
     
  8. Jul 18, 2009 #7
    (A+rI)x=([tex]\lambda[/tex]+r)x
    (A+rI)x-([tex]\lambda[/tex]+r)x=0
    (A+rI-[tex]\lambda[/tex]I-rI)x=0
    (A-[tex]\lambda[/tex]I)x=0

    So the eigenvectors of A and A+rI are the same, right?
     
  9. Jul 19, 2009 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Right.
     
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