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Nilpotency of Matrix with one eigenvalue

  1. Mar 11, 2009 #1
    Hello everybody,

    I have a question for which I cannot find the answer around,
    any help would be really appreciated.

    Suppose we have a matrix A of a linear transformation of a vector space,
    with only one eigenvalue, say 's'.

    My question is: Is the operator (A-sI) nilpotent? ('I' is the identity matrix).

    I am trying to understand the proof of the Jordan Canonical Form Theorem and there is
    a fuzzy point in my notes about that...

    Thanks a lot in advance..

    EDIT: I am becoming pretty sure about that, but some confirmation would be great..
    Last edited: Mar 11, 2009
  2. jcsd
  3. Mar 11, 2009 #2


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    Science Advisor

    I assume you are talking about finite dimensional vector spaces. A linear transformation from a vector space of dimension n to itself can be represented by an n by n matrix. If an n by n matrix, over the complex numbers, has only the single eigenvalue, [itex]\lambda[/itex], then its characteristic equation must be [itex](x- \lambda)^n= 0[/itex]. Since every matrix satisfies its own characteristic equation, [itex](A- \lambda I)^n= 0[/itex].
  4. Mar 11, 2009 #3
    Oh yes, I see it now! Thanks so much!
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