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Another linear algebra proof

  1. Jun 5, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that if s is an eigenvector of matrix A, then it is also an eigenvector
    of matrix A^2.


    2. Relevant equations

    As=Ics

    3. The attempt at a solution
    We know that As=Ics.
    So AAs=AIcs
    A(As)=A(Ics)
    A(As)-A(Ics)=0
    A(As-Ics)=0
    Since As-Ics=0
    A(As-Ics)=A(0)=0
     
  2. jcsd
  3. Jun 5, 2009 #2
    Looks about right, but perhaps a bit cumbersome. Just apply the matrix twice to the vector s. In your notation:
    (AA)s = A(As) = A(Ics) = Ic(As) = (Ic)^2s
    Which is all you need.
     
  4. Jun 5, 2009 #3

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    I think it is correct, although admittely I didn't check all of it.
    There is an easier way.
    From the "relevant equations", you see that you have to show that
    A^2 s = I k s
    for some number k.

    If you start from
    A^2 s = A I c s,
    can you use two properties of matrices to get the A in front of the s and use A s = I c s again?

    [edit]Actually xepma has already given you the complete answer :tongue: [/edit]
     
  5. Jun 5, 2009 #4
    Thanks for the help!
     
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