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Matrix problem

  1. Nov 1, 2008 #1
    1. The problem statement, all variables and given/known data
    Matrix A is
    -4 4 4
    -4 4 4
    4 -4 -4

    It has two real eigenvalues, one of multiplicity 1 and one of multiplicity 2. Find the eigenvalues and a basis of each eigenspace.


    3. The attempt at a solution

    I got the eigenvalues:
    the one of multiplicity 1 is -4
    the one of multiplicity 2 is 0

    I can also get the eigenvectors for both:
    for -4 : [1 1 -1]^T
    for 0: [2 1 1]^T

    But... I don't know where to go from here. At all. Any help would be greatly appreciated.
     
    Last edited: Nov 1, 2008
  2. jcsd
  3. Nov 1, 2008 #2

    gabbagabbahey

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    Remember, when you diagonalize a matrix you get [itex]D=P^{-1}MP \Rightarrow M=PDP^{-1}[/itex] not just [itex]M = D[/itex]. This means that

    [tex]M^n=(PDP^{-1})^n=(PDP^{-1}) \cdot (PDP^{-1}) \cdot (PDP^{-1}) \ldots (PDP^{-1})=PD(P^{-1}P)D(P^{-1}P)\ldots DP^{-1}=PD^nP^{-1}[/tex]

    So, while you have correctly calculated [itex]D^n[/itex], you forgot to multiply it by [itex]P[/itex] and [itex]P^{-1}[/itex] to get [itex]M^n[/itex].

    Edit-this was in reply to the original question ;0)
     
  4. Nov 1, 2008 #3

    gabbagabbahey

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    Your second eigenvector is incorrect. You should also be looking for two (generalized) eigenvectors for [itex]\lambda=0[/itex] since that eigenvalue has multiplicity of 2.

    Your first eigenvector [itex]x[/itex] will be given by [itex](A-\lambda I)x=0[/itex] but(!) your second one will not. Your matrix A is degenerate, and so you need to look for generalized eigenvectors. In this case, since [itex]\lambda=0[/itex] has multiplicity two, its eigenvectors are given by [itex](A-\lambda I)x=Ax=0[/itex] and [itex](A-\lambda I)^2x=A^2x=0[/itex].
     
  5. Nov 1, 2008 #4
    edit: oh okay, so the eigenvectors for 0 are:
    (1, 1, 0)
    (1, 0, 1) ?
     
  6. Nov 1, 2008 #5

    gabbagabbahey

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    Yup!:smile:
     
  7. Nov 1, 2008 #6
    Sweet, how do I get eigenspaces though? To be honest, I'm not even sure what they are...
     
  8. Nov 1, 2008 #7

    gabbagabbahey

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    So, what is the eigenspace of the [itex]\lambda=-4[/itex] eigenvalue? How about the [itex]\lambda=0[/itex] eigenvalue.
     
  9. Nov 1, 2008 #8

    HallsofIvy

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    You already have two eigenvectors for the space of eigenvectors corresponding to eigenvalue 0, <1,1,0> and <1, 0,1>. Since those are not multiples of one another, they are independent. Two independent vectors in a space of dimension 2? What does that tell you about a basis?
     
  10. Nov 1, 2008 #9
    So... For eigenvalue -4, the eigenspace would just be the eigenvector? Like [-1, 1, 1]?

    And for the 0, it'd be:
    [1 1]
    [1 0]
    [0 1]?
     
  11. Nov 1, 2008 #10

    gabbagabbahey

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    For the -4, the eigenspace is just <1,1,-1>.

    And for the eigenvalue zero, the eigenspace is {<1,1,0>,<1,0,1>} (that is the set of the two vectors)

    Clearly, the eigenspace of the -4 eigenvalue is 1D, while the eigenspace of the eigenvalue zero is 2D
     
  12. Nov 1, 2008 #11

    HallsofIvy

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    No, no, no! An eigenspace is never a single vector. The eigenspace corresponding to eigenvalue -4 is the subspace spanned by {<1, 1, -1>} or having that set as basis.

    Similarly, the eigenspace corresponding to eigenvalue 0 is NOT "{<1, 1, 0>, <1, 0, 1>}, it is the subspace spanned by that set or having that set as basis.
     
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