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Inverse linear transformation (think I got it, but don't have answer to compare)

  1. Jul 30, 2009 #1
    1. The problem statement, all variables and given/known data

    Define [tex]T:P_1 \rightarrow P_1[/tex] by T(a+bx)=(a-b)+ax. Check as to whether T has an inverse or not and if it has, find [tex]T^{-1}[/tex]

    2. Relevant equations

    [tex]T^{-1}(T(v))=v[/tex]

    3. The attempt at a solution

    The range of T is P1, so T is one-to-one. The inverse of T is therefore

    [tex]T^{-1}((a-b)+ax)=(a+bx)[/tex]

    Is my reasoning sound?

    Thanks for the help!
    phyz
     
    Last edited: Jul 30, 2009
  2. jcsd
  3. Jul 30, 2009 #2
    Yes, T is a linear transformation from P1 to P1, so if it is injective or surjective it is invertible. Should you show T is one-to-one using the definition or by showing null(T)={0}?

    Your T-1 needs to be actually defined. As defined, T's range is P2. So let c+dx+ex2 be in P2. Define T-1:P2→P1 by T-1(c+dx+ex2)=d+(d-c)x. Now check that T-1 is actually the inverse of T by showing T-1(T(a+bx))=a+bx.
     
  4. Jul 30, 2009 #3
    Hi there...sorry, but I made a typo :redface:

    Should be [tex]T:P_1 \rightarrow P_1[/tex] not [tex]T:P_1 \rightarrow P_2[/tex]. Fixed it now though. They don't say how you should determine whether T has an inverse or not.
     
  5. Jul 30, 2009 #4
    Oh okay, no problem. That makes more sense, as that confused me a little bit. Then just remove the ex2 from everywhere it appears above.

    Read http://books.google.com/books?id=BNsOE3Gp_hEC&lpg=PP1&dq=sheldon axler linear algebra&pg=PA53" for invertibility of linear maps. Look at Theorem 3.21. To show T is invertible, all you need to do is show that it is injective or onto since it is a linear map from P1 to P1. I would recommend showing that if T(a+bx)=T(c+dx), then a+bx=c+dx to show it is one-to-one. Then just define the inverse of T as above, and you're done.
     
    Last edited by a moderator: Apr 24, 2017
  6. Jul 31, 2009 #5
    Thanks for that! Have a great weekend! :smile:
     
  7. Jul 31, 2009 #6
    Haha. No problem, and the same to you!
     
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