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Nilpotent / Diagonalizable matrices

  1. Jun 9, 2012 #1
    Hey guys
    I hope i'm in the right place...
    I have this question i've been trying to solve for too long:

    Let A be an nxn matrix, rankA=1 , and n>1 .
    Prove that A is either nilpotent or diagonalizable.


    My best attempt was:
    if A is not diagonalizable then det(A)=0 then there is a k>0 such that A^k = 0 then A is nilpotent.

    But I'm quite sure that's not good...

    Anyone can help?

    Thanks a lot
     
  2. jcsd
  3. Jun 9, 2012 #2

    micromass

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    This is incorrect. This should be clear since you did not use the rank=1 property.

    Also, why should a non-diagonalizable matrix have det(A)=0??
    And more crucial, why should a matrix with det(A)=0 have A^k=0 for a k???

    What does diagonalizable mean?? What does it say about the eigenvectors??

    Do you know anything about Jordan canonical forms??
     
  4. Jun 9, 2012 #3
    alright so everything i said is wrong i guess.
    i just tried to get to SOMEthing... because i came to a dead end...
    and yes i know about Jordan forms...
    How can I use that for the proof?
     
  5. Jun 9, 2012 #4

    micromass

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    Well, try to write down the Jordan canonical forms of a matrix with rank 1. Can you find some eigenvalues of such a matrix?
     
  6. Jun 9, 2012 #5
    if rank = 1 then all eigenvalues are zero, right ?
     
  7. Jun 9, 2012 #6

    Ray Vickson

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    What is the rank of [tex] A = \pmatrix{1&0\\0&0}?[/tex] What are its eigenvalues?

    RGV
     
  8. Jun 9, 2012 #7
    t1=1 , t2=0
     
  9. Jun 9, 2012 #8

    micromass

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    So this is a counterexample to your post 5, right?
     
  10. Jun 9, 2012 #9
    yeah, and its jordan form is just as the original matrix.
    I tried another rank 1 matrix to see if i come across a rule for jordan forms for rank 1 matrices... but i didn't find anything special...
     
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