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Identity map and injectivity

  1. Feb 25, 2008 #1
    The question is to prove for finite dimensional T: V to W,
    T is injective iff there exists an S: W to V such that ST is the identity map on V.
    I can't quite make the connection between injectivity and the identity map.
    any suggestions?
    thanks in advance.
     
  2. jcsd
  3. Feb 25, 2008 #2
    What's the definition of injectivity? How could you use that to define a function from W to V?
     
  4. Feb 25, 2008 #3

    Hurkyl

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    Have you tried working with any specific examples to get ideas? Can you think of any examples of injective linear maps?

    (I'm assuming V and W are supposed to be vector spaces, and T and S are supposed to be linear transformations)

    (What does it mean for T to be "finite dimensional"?)
     
  5. Feb 25, 2008 #4
    Suppose ST = I
    To show T is injective:
    Let T(x)=T(y) , f.s. x,y in V
    => S(T(x)) = S(T(y)) [permitted since S is well defined from W to V, and T is of course from V to W]
    => (ST)(x) = (ST)(y)
    => I(x) = I(y)
    => x = y
    proving, T is injective.


    I think the converse is true only when Dim V = Dim W


    Now, suppose T is injective.
    Let x be arbitrary in N(T)
    T(x) = 0 (zero of W) = T(0)
    => x = 0
    So, your N(T) = {0}
    Now Dim N(T) + Dim R(T) = Dim V (Sylvester's law)
    But N(T) = 0
    => Dim R(T) = Dim V = Dim W
    Also R(T) is contained in W
    Therefore, R(T) = W
    => T is surjective.

    Now T is bijective => T is invertible, so there exists an S from W to V s.t ST = TS = I

    Q.E.D.
     
  6. Feb 25, 2008 #5

    HallsofIvy

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    You are assuming that U and V are finite dimensional vector spaces and T is a linear transformation from U to V?

    Saying that there exist S such that ST= I means that T has an inverse. That is not possible unless T is, in fact, bijective: both injective and surjective. It is fairly easy to prove that if a linear transformation is injective, then it is also surjective, and vice-versa.
     
  7. Feb 25, 2008 #6
    erm, where did my reply go ?
    this is strange.
     
  8. Feb 25, 2008 #7
    I don't think this is true unless you assume U and V have the same dimension. For T to be injective, it must be the case that dim V [itex]\geq[/itex] dim U, but equality doesn't have to hold. However, there does have to be an S such that ST is the identity on U (namely you just take Sx to be the preimage under T if x is in the image, and take Sx = 0 otherwise).

    Of course, knowing me I'm probably overlooking something painfully obvious, so please be gentle if I'm wrong here.
     
  9. Feb 25, 2008 #8

    HallsofIvy

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    No, you are completely right. I was thinking of U and V having the same dimension without even realizing it.
     
  10. Feb 27, 2008 #9
    thanks for all the hints. To clear up: S and T are Linear Transformations. and W is finite dimensional (V and W are vector spaces though nothing mentioned about the dimension of V) . I remember from a logic/proof class that the way to prove if and only if statements is to go both ways meaning:
    1) suppose T is injective. Show that there exists a linear transformation S: W -> V such that ST is the identity map on V.
    2) Suppose there exists a linear transformation S: W -> V such that ST is the identity map on V. Prove T is injective.
    I know that the question is hinting at an inverse of T (namely S), and I know that invertible functions are bijective. so is this proposition false? then what would be the counter example? of if true, how do I go from T is injective to an inverse S???
    thanks again
     
  11. Feb 27, 2008 #10
    the entire proof is there !
    scroll up to my reply.
     
  12. Feb 27, 2008 #11
    By the way, my example map is completely wrong. That's what happens when you answer algebra questions while doing complex analysis I suppose.
     
  13. Feb 26, 2009 #12
    what if V is not finite dimensional? how do you prove that T is injective implies that there exists S such that ST is an identity map?
     
  14. Mar 4, 2009 #13
    Hi,

    The first half of sihag proof is right but the assumption that:
    Dim V = Dim W,
    is wrong.

    The proof of the other way is:
    If T:V->W is injective define
    S(x) = y if T(y)=x or,
    S(x) = 0 if for all y in V T(y) <> x
    Hence
    S(T(y)) = S(x) = x => ST is the identity for all element in V.

    Just remember that ST = I does not mean TS = I. This is explained as
    T: V->W and S:W->V,
    ST: V -> V and TS: W -> W
    they are different maps.
     
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