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Homework Help: Is this normalised eigenvector undefined?

  1. Feb 16, 2012 #1
    1. The problem statement, all variables and given/known data

    I've started off with a 2x2 matrix of

    (0) (i)
    (-i) (0)

    and I found the eigenvalues to be +1, -1

    Then I found the resulting 2x1 eigenvectors to be




    I now need the normalised eigenvectors.

    2. Relevant equations

    3. The attempt at a solution

    I'm getting

    {1/[sqrt(0)]} x eigenvector1
    {1/[sqrt(0)]} x eigenvector2

    Does this mean the normalised eigenvectors in this case are undefined?
  2. jcsd
  3. Feb 16, 2012 #2
    You need to take the dot product with the complex conjugate of the vector if it's over a complex field.

    [itex]|a| = \sqrt{a*\ .\ a}[/itex]
  4. Feb 16, 2012 #3
    Thank you.

    Do you mean then that the normalised eigenvectors would be

    [1/(sqrt(a* . a))] x eigenvector1


    [1/(sqrt(a* . a))] x eigenvector2
  5. Feb 16, 2012 #4
    yes, if a is the eigenvector in question.
    This still holds if the eigenvectors are all real, the magnitude of a real vector a is still [itex]\sqrt{a*\ .\ a}[/itex] but since a is real a*=a and we get the familiar [itex]\sqrt{a\ .\ a}[/itex]
  6. Feb 16, 2012 #5
    I'm getting values for the normalised eigenvectors of

    {1/[sqrt(2)]} x eigenvector1
    {1/[sqrt(2)]} x eigenvector2

    What do you think?
  7. Feb 16, 2012 #6
    That is correct!
  8. Feb 16, 2012 #7

    I noticed there, with the two 2x1 eigenvectors which I in my first post being




    For the number that's going to be under the radical, in this case we discovered it to be 2, can I simply take the [magnitude of (1)]^2 + the [magnitude of (-i)]^2 which equals 1+1=2 rather than take the dot product? Or am I effectively taking the dot product by doing this?
  9. Feb 16, 2012 #8


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    Science Advisor

    Yes, [itex](a+ bi)(a- bi)= a^2+ b^2[/itex].
  10. Feb 16, 2012 #9
    You are taking the dot product when you do this, yes :p
  11. Feb 16, 2012 #10
    That's great guys! Thanks very much for your help. It's much appreciated!
  12. Feb 16, 2012 #11
    The final part of this question states that the original matrix is Hermitian, and that I should check that the eigenvalues and eigenvectors have the required properties for those of a Hermitian matrix.

    The eigenvalues are both real, so that is satisfied.

    Therefore for the eigenvectors to satisfy the required properties of a Hermitian matrix, (eigenvector1)^T x eigenvector2 = (eigenvector2)^T x eigenvector1 = 0

    ^T indicates Transpose

    When I do this I'm getting -2i for both. So they are both equal to each other. Though 2i is not = 0 i.e. a null vector
  13. Feb 16, 2012 #12
    Remember that when you take the dot product of two complex vectors you need to take the complex conjugate of one of them.
    You'll find that [itex]e_1*.e_2 = e_2*.e_1 = 0[/itex], where e1 and e2 are the two eigen vectors and * is complex conjugation.
  14. Feb 16, 2012 #13
    That's great. Thanks very much again genericusrnme!
  15. Feb 16, 2012 #14
    No problem buddy :biggrin:
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