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B Zero eigenvalue and null space

  1. Mar 8, 2016 #1
    Suppose ##T## is an operator in a finite dimensional complex vector space and it has a zero eigenvalue. If ##v## is the corresponding eigenvector, then
    $$
    Tv=0v=0
    $$
    Does it mean then that ##\textrm{null }T## consists of all eigenvectors with the zero eigenvalue?
    What if ##T## does not have zero eigenvalue? Does it mean that its null space is just the zero vector?

    Thanks
     
  2. jcsd
  3. Mar 8, 2016 #2

    Samy_A

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    Yes.
    (Except that ##\textrm{null }T## consists of all eigenvectors with the zero eigenvalue and the 0 vector.)
     
  4. Mar 8, 2016 #3
    So, the answer to all of my questions is affirmative?
    Then, if ##T## does not have a zero eigenvalue, it's equivalent of being injective.
     
  5. Mar 8, 2016 #4

    Samy_A

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    For a linear operator, yes.
    (I should have mentioned that in my first answer too, I just assumed you meant that T is a linear operator.)
     
  6. Mar 8, 2016 #5
    Yes, it's linear.
     
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