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Jordan Normal Form of Matrices

  1. Feb 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the Jordan Normal Form of

    3,0,8
    3,-1,6
    -2,0,-5


    2. Relevant equations



    3. The attempt at a solution

    I got the eigen values as lambda=-1

    found 2 eigen vectors:

    -2
    0
    1

    and

    0
    1
    0

    now when i try and do (A+I)^2 I get a zero matrix so cannot seem to find the last column of the P matrix to then calculate P-1AP=J
     
  2. jcsd
  3. Feb 7, 2010 #2
    anyone?
     
  4. Feb 7, 2010 #3

    Dick

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    Homework Helper

    If (A+I)^2=0 then ANY vector satisfies (A+I)^2. What you really want to get to the Jordan form is three vectors satisfying (A+I)v1=0, (A+I)v2=0, (A+I)v3=v2. Start by picking v3 outside of the eigenspace spanned by {v1,v2}. That determines the eigenvector v2. Now pick v1 to be any other linearly independent eigenvector.
     
  5. Feb 7, 2010 #4
    ok im not quite sure what you mean. when I set (A+I)v2=v3

    using my v2= (2 0 -1)

    i get a matrix that doesnt multiply to get a jordan form....
     
  6. Feb 7, 2010 #5

    Dick

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    Homework Helper

    (A+I)v2=0 if v2=(2,0,-1). You have a two dimensional eigenspace spanned by {(-2,0,1),(0,1,0)}. You want to pick three vectors for the basis that satisfy the relations I gave in the last post. Since you want (A+I)v3=v2, you want to pick a v3 OUTSIDE of that eigenspace. Now define v2=(A+I)v3. Since (A+I)^2=0 that will automatically be in the eigenspace. Now pick the final v1 to be another linearly independent vector in the eigenspace. That's a Jordan basis. Just start doing it. You'll see.
     
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