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Yet another eigenvalue proof

  1. Apr 9, 2009 #1
    Let x not equal to zero be a vector in the nullspace of A. Then x is an eigenvector of A.

    I'm not sure how to start this proof
  2. jcsd
  3. Apr 9, 2009 #2
    If x is a non-zero vector in the null space of A, then you know that A is singular, and you also know that [tex] \lambda = 0 [/tex] is an eigenvalue of A since A is invertible if and only if zero is not an eigenvalue of A. That should start you off.
  4. Apr 9, 2009 #3


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    Saying that x is in the null space of A means that Ax= 0= 0x.
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