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Homework Help: Complex numbers + linear algebra = :S

  1. Sep 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Find all complex numbers Z (if any) such that the matrix: (it is 2 by 2)


    multiplied by a vector V = ZV has a nonzero solution V.

    part b)

    for each Z that you found, find all vectors V such that (the same matrix)*V=ZV.

    3. The attempt at a solution

    I'm not sure that I even know what this question is asking... could anyone clarify this a little?
  2. jcsd
  3. Sep 9, 2009 #2
    And please don't mistake my lack of effort for laziness, I genuinely don't understand the question.
  4. Sep 9, 2009 #3


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    The question is "solve Av=zv".

    (A is the matrix you were given)
    (v and z are the indeterminate variables you're solving for)
    (v is a vector variable)
    (z is a complex variable)
  5. Sep 9, 2009 #4


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    In other words, as Hurkyl clarified, it's asking you for eigenvalues and eigenvectors of your matrix. Search on those keywords if you need examples.
  6. Sep 10, 2009 #5
    So I'm looking for complex numbers such that, when multiplied by a vector, it gives the same results as the matrix multiplied by the same vector?
  7. Sep 10, 2009 #6


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    Yes, the number times the vector should be the same as the matrix times the vector.
  8. Sep 10, 2009 #7
    I googled eigenvectors/values but I don't see the relevance to this question :s

    How would I find a set of complex numbers that multiplies the same way as a matrix of real numbers? Any hints please?
    Last edited: Sep 10, 2009
  9. Sep 10, 2009 #8


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    I'm surprised you didn't see the relevance. But look, let v be column vector (x,y). So z*v is column vector (z*x,z*y) What is Av written out as a column vector? Now equate the two column vectors so you have a system of two equations, write them down. What condition does z have to satisfy to get a solution where x and y are not both zero? It's a quadratic. It has two roots. They are complex. See if you can find them. Look back at examples of finding eigenvalues again. Because that is what you are doing.
  10. Sep 10, 2009 #9
    That makes a lot of sense, thanks Dick :)
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