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Linear Algebra problem

L|amaL|ama Registered User regular
edited October 2009 in Help / Advice Forum
Studying for an exam coming up and this sort of question has been in there every time for the past few years:


It doesn't really seem to be covered in the notes though. I can do the first part easily, but I've got no idea where to go with B.

L|ama on


  • FuzzywhaleFuzzywhale Registered User
    edited October 2009
    I think part (b) wants you to diagonalize the matrix A. That should not be so bad, because (a) gets you to find the eigenvectors and values.

    Since this doesn't sound familiar to you, here is a handy wikipedia link:

    Fuzzywhale on
  • L|amaL|ama Registered User regular
    edited October 2009
    Hmm, diagonalization wasn't even mentioned in our lectures, and I don't see how that would help. I think it just wants it rewritten some other way in terms of the eigenvectors, not explicitly calculated. Trying to follow the method on wikipedia gets me with rows 1 and 2 both being (1,0), so unless I'm doing something wrong (very possible) I don't think it's even diagonalizable.
    Oh, there's some basic answers to the exams up:
    Q9 (a) Characteristic equation is L(L − 1) = 0, e’values are L = 0 with corresponding e’vector (2, 3),
    and L = 1 with corresponding e’vector (1, 1).
    (b) A^4(v1 + v2) = v1.

    uh... huh. It works out and makes sense, but seems like an almost trivial question.
    And for the similar questions for the other two sets of answers given, one of them is 256v1+v2, the other is v1+243v2, so there must be some other method than just multiplying it out and doing it by inspection.

    L|ama on
  • FuzzywhaleFuzzywhale Registered User
    edited October 2009
    Isn't the characteristic equation det(A-lambda*I)=0?
    If it is, then for this matrix the equation is x^2-6x+5=0, which has roots 1 and 5. Plugging those back into the matrix and then finding the null space should give you the eigenvectors.

    I obviously don't want to contradict your exam solutions, so I guess i'm wrong. At any rate I kind of want to know what else it could be...

    edit: yeah i missed the minus on the bottom right 2. thanks!

    Fuzzywhale on
  • ClipseClipse Registered User regular
    edited October 2009
    For part (b):

    A^4 * (v1 + v2) = A^4 * v1 + A^4 * v2, of course. Let e1 and e2 be the corresponding eigenvalues, then we have:

    A^4 * v1 = A^3 * A * v1 = A^3 * e1 * v1.

    Continuing to reduce the exponent on A in that fashion gives the final result

    A^4 * v1 = e1^4 * v1, and likewise with A^4 * v2. So

    A^4 * (v1 + v2) = e1^4 * v1 + e2^4 * v2.

    The eigenvalues of this matrix are 0 and 1, so declaring that e1 = 1 and e2 = 0 gives

    A^4 * (v1 + v2) = v1.

    (bolded line is the important one, should explain the question with the more complicated answer as well.)

    Fuzzywhale: Det(A - x*I) = x^2 - x. I think you lost a minus sign or something to get that result.

    Clipse on
  • L|amaL|ama Registered User regular
    edited October 2009
    Aha, that makes sense (and I can follow it based on previous knowledge) so thanks.

    L|ama on
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