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Homework Help: Eigenvalues: Matrix corresponding to projection

  1. Dec 8, 2009 #1
    Let A be a matrix corresponding to projection in 2 dimensions onto the line generated by a vector v.

    A) lambda = −1 is an eigenvalue for A
    B) The vector v is an eigenvector for A corresponding to the eigenvalue lambda = −1.
    C) lambda = 0 is an eigenvalue for A
    D) Any vector w perpendicular to v is an eigenvector for A corresponding to the eigenvalue lambda = −1.
    E) Any vector w perpendicular to v is an eigenvector for A corresponding to the eigenvalue lambda = 0.

    A) A matrix must have a non-zero eigenvalue.
    B) The vector is an eigenvector of corresponding to the eigenvalue.
    C) See A
    D) When a vector is perpendicular to A, it must be zero for eigenvalue (If it was nonzero eigenvalue, then it is not perpendicular)
    E) See D

    So far I believe A, B, E are true? What do you think? Please help me.
     
  2. jcsd
  3. Dec 8, 2009 #2

    Dick

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    Thanks for giving reasons this time, but I don't believe A) is true. And I believe your reason even less. Let's just start from there. A matrix doesn't HAVE to have a nonzero eigenvalue, much less (-1). I will say your answers to D and E seem to be correct, but if you think E is true, then why do you think C is false? That's just crazy.
     
  4. Dec 8, 2009 #3
    When I read the text book, it stated that "a square matrix A is invertible if and only if lambda = 0 is not an eigenvalue of A."

    Therefore, I assume C) was false.

    After I read your message, it appears that B, C, E are correct for this problem?
     
  5. Dec 8, 2009 #4

    Dick

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    Who said A was invertible? I'm not going to tell you what's correct until you convince me you know what's correct. Why do you think B is true?
     
  6. Dec 9, 2009 #5
    I would say that eigenvector corresponding to the eigenvalue λ = -1 is any multiple of the basic vector. So, both make the eigenspace corresponding to the eigenvalue.

    or

    The eigenvector corresponding to the eigenvalue λ = −1 is the solution of the equation Ax = -x
     
  7. Dec 9, 2009 #6

    Dick

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    If you project v onto the line generated by the vector v (which is what A does to v), what do you get? They aren't talking about just any vector that happens to have eigenvalue -1. They are talking about that particular v.
     
    Last edited: Dec 9, 2009
  8. Dec 9, 2009 #7
    I would get linearly indepedent eigenvectors, Dick?
     
  9. Dec 9, 2009 #8

    Dick

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  10. Dec 9, 2009 #9
  11. Dec 9, 2009 #10

    Dick

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    Right. So what is Av if A is the projection onto the line generated by v?
     
  12. Dec 9, 2009 #11
    It will be parallel? so, there are many vectors that could correspond to the given?

    That means C and E are true, right?
     
    Last edited: Dec 9, 2009
  13. Dec 9, 2009 #12

    Dick

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    Av is (v.v)v/|v|^2. (v.v) is the same as |v|^2. Av is v!
     
  14. Dec 9, 2009 #13
    Therefore C and E are true, right?
     
  15. Dec 9, 2009 #14

    Dick

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    Av=v has nothing to do with whether C and E are true or false. It's about a different part of the problem.
     
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