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Little question about eigenvalues

  1. Dec 16, 2014 #1
    Hey there, I'm thinking about if one of the eigenvalues is zero (means determinant is 0. right?) So, is there any possibility to non-zero eigenvalue also exists?
  2. jcsd
  3. Dec 16, 2014 #2


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    for a diagonal matrix, the eigenvalues are the numbers on the diagonal. so what do you think can happen?
  4. Dec 17, 2014 #3
    Well, now that I'm more inside of these things, I realized that I've asked a stupid question :D Of course there is.
  5. Dec 17, 2014 #4


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    another point of view is that eigenvalues are roots of the characteristic polynomial. so if one root is zero can other roots be non zero?
  6. Dec 19, 2014 #5
  7. Dec 19, 2014 #6
    Well, you could have something like:
    ## \lambda^3 + 18\lambda^2 + 81\lambda##
    Where you have eigenvalues ##\lambda_1 = 0## and ##\lambda_2 = -9##.
    So, I certainly think you can have an eigenvalue of zero and other nonzero eigenvalues. And also, I'm not sure what you mean by the determinant being zero; to get a characteristic polynomial, you have to take the determinant and set it equal to zero to solve for the eigenvalues in the first place.
    In other words, you find eigenvalues using:
    ##\det(A - \lambda I_A) = 0## (Given a matrix A and its identity matrix, IA).
  8. Dec 23, 2014 #7
    I misspoke above; I listed two eigenvalues, whereas a ##3 \times 3## matrix (with a cubic characteristic polynomial) would have three eigenvalues. Of course, in the case I listed above, two of the eigenvalues, ##\lambda_2## and ##\lambda_3## would be the same (λ2 = λ3 = -9). I was treating them as roots of a cubic, where (with three solutions) there are only two roots, given two of them are the same. In the context of matrix algebra, though, all eigenvalues should be accounted for.
  9. Dec 23, 2014 #8


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    Well, if there is an eigenvalue ## \lambda=0 ##, then Det(A-0I_A)=Det(A)=0.
  10. Dec 24, 2014 #9
    Why is that a problem? Having zero determinant just means that it's not invertible.
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