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Diagonalizable endomorphism has trivial null space

  1. Dec 14, 2013 #1
    Is it correct to state that a diagonalizable endomorphism has always kernel = {0}?
     
  2. jcsd
  3. Dec 14, 2013 #2
    Is the zero operator a diagonalizable endomorphism?
     
  4. Dec 14, 2013 #3
    I still don't understand.
     
  5. Dec 14, 2013 #4

    jgens

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    R13's point was that the zero map is diagonalizable and has kernel the whole space.
     
  6. Dec 14, 2013 #5
    Right, that's the case of the zero operator. But what if my eigenvalues are all non-zero (hence my eigenvectors are all linear indipendent)?
     
  7. Dec 14, 2013 #6

    jgens

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    In that case your operator has trivial kernel.
     
  8. Dec 14, 2013 #7
    Perfect. Thank you.
     
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