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Homework Help: Basis of the eigenspace

  1. Sep 12, 2009 #1
    1. The problem statement, all variables and given/known data

    The matrix A=
    2 0 4
    -2 0 -4
    -1 0 -2

    has one real eigenvalue. Find this eigenvalue and a basis of the eigenspace.

    Eigenvalue =

    Basis ( , , )T , ( , , )T

    2. Relevant equations

    3. The attempt at a solution

    I have found the eigenvalue to be 0, but I can't seem to figure out how to come up with 2 bases. I found the first one:

    2x + 4z = 0
    -2x - 4z = 0
    -1x - 1z = 0

    All three equations reduce to :
    x + 2z = 0

    So then,

    x = t , y = 0, and z = -(1/2)t
    which gives a basis of (1, 0, -1/2)T

    I'm not completely sure that I have calculated this correctly, and I have no idea how to come up with the second basis. The eigenvalue I calculated was marked correct, though. If anyone could help me, I would greatly appreciate it. :)
  2. jcsd
  3. Sep 12, 2009 #2
    You only have one eigenvalue so you will have only one basis.

    Your question states "Find this eigenvalue and a basis of the eigenspace".

    Correct me if I am wrong but you should only need to find one basis.

  4. Sep 12, 2009 #3


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    Could you show your work here?
  5. Sep 12, 2009 #4
    I also thought that there should be only one basis, but the question asks for 2, and my prof confirmed this. I still don't understand how this would work though.

    My work:
    2z = -x
    z = - x/2

    So if I set x with the parameter t,

    and z = -t/2, or (-1/2) t

    Sorry I didn't include this before .
  6. Sep 12, 2009 #5


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    y doesn't appear in that equation -- how did you conclude something about it?
  7. Sep 12, 2009 #6
    Because the equation was

    x +2z = 0 or
    1x + 0y +2z = 0

    y is nonexistent in this case. I assumed that would mean it would be equal to 0 in the basis, but like I said, I wasn't sure if I had calculated the basis correctly. How would I find a value for y?
  8. Sep 12, 2009 #7


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    All equations reduce to x+ 2z= 0 so x= -2z. z can be anything at all and y can be anything at all. Your basis contains two vectors.
  9. Sep 12, 2009 #8
    So how would I find the bases in that case?
  10. Sep 13, 2009 #9
    Oh, I just got it.
    x = -2t
    y = s, not 0
    and z = t

    which means that the vector would be:

    (-2t , s, t)T = (-2t, 0, t)T + ( 0, s, 0)T

    which gives the bases

    (-2, 0, 1)T, and (0, 1, 0)T
  11. Sep 13, 2009 #10
    Thank you for your help!
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