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Quick diagonalize matrix (row reduce)

  1. Oct 11, 2011 #1
    quick plz..diagonalize matrix (row reduce)

    1. The problem statement, all variables and given/known data

    http://www.math4all.in/public_html/linear algebra/chapter10.1.html
    10.1.5 Examples: (ii)
    this part:

    -----------------------------------------------------------------------------------------
    april95.gif
    are both eigenvector for the eigenvalue λ = 3. Similarly, for λ = 6, since
    -----------------------------------------------------------------------------------------

    Does it matter if I reduce it to rref?




    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 11, 2011 #2

    Mark44

    Staff: Mentor

    Re: quick plz..diagonalize matrix (row reduce)

    Your question doesn't make much sense. The matrices you show are NOT eigenvectors. In the link you gave, it shows two vectors that are the eigenvectors for the eigenvalue 3.

    What do you mean by "reduce it to rref"?
     
  4. Oct 11, 2011 #3
    Re: quick plz..diagonalize matrix (row reduce)

    oh i I figure it out...
    I was just asking
    Do I get the same eigenvector (which is 0,1,1) if I
    transform
    -3,0,0
    0,1,-1
    0,0,0
    to rref

    In reduced row-echelon form (RREF), the matrix above is
    1 0 0
    0 1 -1
    0 0 0

    The only thing that changed is row 1.
    You'll get exactly the same eigenvector from this matrix as from the unreduced one.
     
    Last edited by a moderator: Oct 11, 2011
  5. Oct 11, 2011 #4
    Re: quick plz..diagonalize matrix (row reduce)

    And using this calculator
    http://wims.unice.fr/wims/en_tool~linear~matrix.html
    I found that
    Value Multiplicity Vector
    6 1 (0,1,1)
    3 2 (1,0,1), (0,1,-1/2)

    (1,0,1) and (0,1,1) is right, but (0,1,-1/2) is different to the website , which is (1/2,1,0)
    I m confused..
     
  6. Oct 11, 2011 #5

    Mark44

    Staff: Mentor

    Re: quick plz..diagonalize matrix (row reduce)

    It's easy to check. Substitute each of these two vectors (i.e., <0, 1, -1/2> and <1/2, 1, 0> in the equation Ax = 3x. If they both make a true statement, they're both eigenvectors.
     
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