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Proving kernel of matrix is isomorphic to 0 eigenvalue's eigenvectors

  1. Sep 25, 2007 #1
    1. The problem statement, all variables and given/known data
    I want to prove that the eigenvectors corresponding to the 0 eigenvalue of hte matrix is the same thing as the kernel of the matrix.


    2. Relevant equations
    A = matrix.
    L = lambda (eigenvalues)

    Ax=Lx


    3. The attempt at a solution

    Ax = 0 is the nullspace.

    Ax = Lx
    Lx = 0.
    L= 0.
    the eigenvectors corresponding to the 0 eigenvalue are the same as the nullspace.

    Is this a sufficient enough proof?
     
  2. jcsd
  3. Sep 25, 2007 #2

    morphism

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    No, it's not. Maybe you have the right idea, but what you've written down doesn't make a lot of sense.

    The nullspace is {x : Ax = 0}. Can you write down what the set of eigenvectors corresponding to zero is?
     
  4. Sep 26, 2007 #3
    Isn't the the set of eigenvectors which correspond to the 0 eigenvalue?
     
  5. Sep 26, 2007 #4

    matt grime

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    What is the definition of the kernel of a matrix? What is the definition of the set of eigenvectors of a matrix with eigenvalue zero? Aren't they trivially the same?
     
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