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Homework Help: Find a Linear Transformation such that T squared = T

  1. Feb 13, 2010 #1
    Hi, I had to prove that if T2=T then the direct sum of im(T) and ker(T) is a vector space, and I did that, but now I am suppose to find such a T that isn't the zero operator (I'm not even sure what that is, just a transformation that makes any vector zero?) or the identity operator. Problem is I can't think of any :confused:. I am just trying to think of a transformation T2=T, but none of what I have thought of works, any help please?
     
  2. jcsd
  3. Feb 13, 2010 #2
    T is a projection operator on arbitrary subspace.
     
  4. Feb 13, 2010 #3

    Hurkyl

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    That's right.

    Then let's solve for one. If we can't figure things out with matrix algebra, then break things into components, and turn the problem into one of number algebra!

    I'll get you started: let's suppose T is a 0x0 matrix. Oh, this can't work -- the only 0x0 matrix is the zero matrix. (Which is also an identity matrix), so nevermind.

    Well, let's try 1x1, so that T = [a]. So in terms of a, what do we want?
     
  5. Feb 13, 2010 #4
    What if I write Tv= a 2x2 matrix where the top row entries are v and the bottom row entries are zero.
    Sorry that is suppose to be a matrix. Anyway does that make sense as an answer?
     
  6. Feb 13, 2010 #5
    I don't think what I wrote even makes sense :S
     
  7. Feb 13, 2010 #6

    Hurkyl

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    It makes sense if the top row is v. (It doesn't make sense for the entries to be v -- at least not in the arithmetic framework you're using)

    But then what does T2 work out to be?
     
  8. Feb 13, 2010 #7
    Sorry I didn't see the end of your post haha, well we want an a that preserves addition and scalar multiplication for one, and when you transform it again gives you a still.
     
  9. Feb 13, 2010 #8
    T2 ends up being a matrix where the top row is a matrix of V? I'm not sure...
     
  10. Feb 13, 2010 #9

    Dick

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    Why don't you pick a basis, any basis. {x1...xn}. Now define T(xi)=ai*xi where ai is equal to zero or one as you wish. That defines a T. Is T^2=T?
     
  11. Feb 13, 2010 #10
    Yes but isn't that T the zero or identity operator in that case?
     
  12. Feb 13, 2010 #11

    Dick

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    It's zero if you pick ALL of the ai's=0. It's the identity if you pick ALL of the ai's=1. If you don't want either of those then pick some of them to be 0 and some to be 1.
     
  13. Feb 13, 2010 #12
    Ah so can I write Tx1,...xn=a1x1,...,aixi where there is at least one a=0 and one a=1 (but a can only be 0 or 1)?
     
  14. Feb 13, 2010 #13

    Dick

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    That's the idea.
     
  15. Feb 13, 2010 #14
    Thanks!
     
  16. Feb 13, 2010 #15
    How were you able to prove that the direct sum of im(T) and ker(T) is a vector space?
     
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