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Characteristic polynomial for nilpotent matrix.

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data

    How do I show that the cp is p(x)=x^n, dimA=n?

    2. Relevant equations

    A^k=0 for some k (obviously need to show k=n); p(A)=0

    3. The attempt at a solution

    p(A)=0 <=> A^n + ... + det(A)=0
  2. jcsd
  3. Oct 6, 2009 #2
    How do I show that the cp of A is p(x)=x^n, dimA=n?
  4. Oct 6, 2009 #3
    This is a trivial consequence of Cayley-Hamilton.
  5. Oct 6, 2009 #4
    i see that p(A)=0 and that p(A)=A^n +...+det(A) BUT I'm still a bit confused by some details. A is nilpotent <=> A^k=0 ... must show k=n for one thing.
  6. Oct 6, 2009 #5
    Suppose p(A) is not xn. Then it is some nontrivial (not xn) polynomial of degree n, which implies A satisfies said polynomial. Show that this leads to a contradiction.
  7. Oct 7, 2009 #6


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    Suppose [itex]\lambda[/itex] is an eigenvalue of A: that is, [itex]Av= \lambda v[/itex] for some non-zero vector v. Then [itex]A^2v= \lamba Av= \lambda^2 v[/itex] and, continuing like that [itex]A^n v= \lambda^n v[/itex]= 0[/itex]. Actually, that proves that all eigenvalues are 0 but what you need is that all eigenvalues satisfy [itex]\lambda^n= 0[/itex].
  8. Oct 7, 2009 #7


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    This was also posted under "Mathematics- Abstract and Linear Algebra". I have merged the threads here.
  9. Oct 8, 2009 #8
    I am almost always tempted to correct the theorem statement as

    "... satisfies the matrix version of the characteristic polynomial"

    since the right hand side zeros of the the two cases are different objects. But it is a technical puristic bla bla. I know!
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