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If A is nxn nilpotent matrix, this char(A) = x^n

  1. May 10, 2017 #1
    1. The problem statement, all variables and given/known data
    If ##A## is an ##n \times n## nilpotent matrix, then the characteristic polynomial of ##A## is ##x^n##

    2. Relevant equations


    3. The attempt at a solution
    Suppose that ##A## has an eigenvalue with corresponding eigenvector such that ##A v = \lambda v##. Then ##A^k v = \lambda^k v = 0##, and since ##v \ne \vec{0}##, ##\lambda^k = 0 \implies \lambda = 0##. Since ##0## is the only eigenvalue, and since the characteristic polynomial must be of degree n, the characteristic polynomial must be ##x^n##.
     
  2. jcsd
  3. May 10, 2017 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    You should introduce k. Apart from that: right.
     
  4. May 10, 2017 #3

    fresh_42

    Staff: Mentor

    Yes.
     
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