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Similarity proof

  1. Dec 12, 2007 #1
    I need help with this proof. Can anyone lead me in the right direction?

    Let A be an nxn matrix such that A^2=C.
    Prove that if B~A, then B^2=C.
     
  2. jcsd
  3. Dec 12, 2007 #2

    Dick

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    Let A=diag(1,2), B=diag(2,1). A^2=diag(1,4), B^2=diag(4,1). A~B. A^2 is not equal to B^2. Is there something you aren't telling us about C or do you want to prove A^2~B^2?
     
  4. Dec 12, 2007 #3
    I gave you exactly what the book says
     
  5. Dec 12, 2007 #4

    Dick

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    If that's exactly what the book says, then you can't prove it. Because it's false.
     
  6. Dec 12, 2007 #5
    what if C were to be 0. Would that proof make sense?
     
  7. Dec 12, 2007 #6

    Dick

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    It makes all of the difference in the world. If A^2=0 and A~B then B^2=(PAP^(-1))(PAP^(-1)). What's that?
     
  8. Dec 12, 2007 #7
    that would be:
    B^2=P^(-1)A^(2)P
    B^2=P^(1)0P=0
     
  9. Dec 12, 2007 #8

    Dick

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    It would also be very different if C were any other multiple of the identity matrix. Wouldn't it?
     
  10. Dec 12, 2007 #9
    THe book really has instead of A^2=C its A^2=O.But I cant tell if its zero or the letter O. THe O is at a slant if that means anything
     
  11. Dec 12, 2007 #10

    Dick

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    Yes.
     
  12. Dec 12, 2007 #11

    Dick

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    I don't know. But as I said any multiple of the identity would work as well as 0.
     
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