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Prove - determinant

  1. May 22, 2009 #1
    Hello,
    I don't want how to prove: Matrix A is nilpotent, so A^k=0. Prove that det(A+I)=0.
    Thank you so much :-)
     
  2. jcsd
  3. May 22, 2009 #2

    matt grime

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    What you state is trivially false - take A=0, that is nilpotent, and det(0+I) is not zero.
     
  4. May 22, 2009 #3

    jbunniii

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    In fact, det(A^k) = det(A)^k, so det(A) must be zero. Alternatively, A^k = 0, so A cannot be injective, so det(A) = 0.

    By the way, this shows that 0 is an eigenvalue of A. Something stronger is true: 0 is the ONLY eigenvalue of A.

    I think lukaszh's problem statement should be: prove that det(A+I) = 1.
     
  5. May 26, 2009 #4
    I'm so sorry, because I've made a mistake. Prove that det(A+I)=1. Thank you again for your reactions.
     
  6. May 26, 2009 #5

    jbunniii

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    First, find all the eigenvalues of A+I. Then, use what you know about how to calculate the determinant of a linear map given its eigenvalues.
     
  7. May 26, 2009 #6
    Hello,
    so, I know that
    [tex]A=S\Lambda S^{-1}[/tex]
    Eigenvalues of A are [tex]\{\lambda_1,\lambda_2,\cdots,\lambda_n\}[/tex] for [tex]n\times n[/tex] matrix. If I add to both sides identity, then
    [tex]A+I=S\Lambda S^{-1}+I=S\Lambda S^{-1}+SS^{-1}=S(\Lambda+I)S^{-1}[/tex]
    Its determinant is
    [tex]\mathrm{det}(A+I)=\mathrm{det}S(\Lambda+I)S^{-1}=\mathrm{det}(\Lambda+I)=\prod_{j=1}^{n}(\lambda_j+1)[/tex]
    I don't know, how to utilize fact [tex]A^k=0[/tex] :-(
     
  8. May 26, 2009 #7

    jbunniii

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    [tex]\lambda[/tex] is an eigenvalue of A + I if and only if

    [tex](A + I)x = \lambda x[/tex]

    for some nonzero vector x. This is true if and only if

    [tex]Ax = (\lambda - 1)x[/tex]

    which is true if and only if [tex]\lambda - 1[/tex] is an eigenvalue of A.

    I claim that [tex]A^k = 0[/tex] implies that all of the eigenvalues of A are zero. If you can prove this claim then it implies the result you want.
     
  9. May 26, 2009 #8
    Beautiful, thank you.
     
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