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Homework Help: Linear Algebra: Linear transformation and Isomorphism

  1. Mar 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Let T: V [tex]\rightarrow[/tex] Z be a linear transformation of a vector space V onto a vector space Z. Define the mapping

    [tex]\bar{T}[/tex]: V/N(T) [tex]\rightarrow[/tex] Z by [tex]\bar{T}[/tex](v + N(T)) = T(v)

    for any coset v+N(T) in V/N(T).

    a) Prove that [tex]\bar{T}[/tex] is well-defined; that is, prove that if v+N(T)=v'+N(T), then T(v)=T(v').

    b) Prove that [tex]\bar{T}[/tex] is linear.

    c) Prove that [tex]\bar{T}[/tex] is an isomorphism.

    2. Relevant equations
    Proving part (a) will prove that the transformation is a one-to-one transformation. This is important for part (c).

    In order to prove part (b), use the fact that for T: V[tex]\rightarrow[/tex]W, T is linear if and only if T(cx + y) = cT(x)+T(y) for all x,y [tex]\in[/tex] V and c [tex]\in[/tex] F.

    To prove that a transformation is an isomorphism, part (c), there must exist a linear transformation [tex]\bar{T}[/tex] that is invertible ([tex]\bar{T}[/tex][tex]^{-1}[/tex]). [tex]\bar{T}[/tex][tex]^{-1}[/tex] exists iff [tex]\bar{T}[/tex] is one-to-one and onto.

    3. The attempt at a solution
    (a) If v+N(T) = v'+N(T), then v+N(T)-N(T) = v'+N(T)-N(T). Thus v = v'. The vectors v and v' are the same without the nullspace of T.
    If v=v', then T(v) = T(v'). This also means it is one-to-one.

    I'm not sure if what I have is a valid proof for part (a). The part that I'm not certain about is if I am actually able to subtract the nullspace of T from both sides.

    (b) If [tex]\bar{T}[/tex] = T(v), let cx+y = v+N(T). Then [tex]\bar{T}[/tex](cx+y) = T(cx+y-N(T)) = cT(x)+T(y)-T(N(T)). This is the first part of the proof.

    I am having trouble with the next part of the proof. I have to prove that c[tex]\bar{T}[/tex](x) + [tex]\bar{T}[/tex](y) = cT(x)+T(y)-T(N(T)) (the right hand side is from the first part of the proof), but I'm not sure how to plug x and y into [tex]\bar{T}[/tex].

    (c) A transformation is an isomorphism if it is one-to-one and onto.
    To prove one-to-one: Since v+N(T)=v'+N(T), then v=v'. If v=v', then T(v)=T(v'). T(v)=T(v') implies that v=v', this is the definition of one-to-one.

    I'm having trouble proving that [tex]\bar{T}[/tex] is onto. From the question, we know that T is a linear transformation that is onto, but how do I relate it to [tex]\bar{T}[/tex]?

    Any and all help is much appreciated!
  2. jcsd
  3. Mar 12, 2009 #2
    Your part a is incorrect. You can't take away N(T) from both sides, its a set. You could have that v+n=v'+m for m not equal to n.

    For b) you really only need to show that [itex]\overline{T}(cx+ky+N(T))=c\overline{T}(x+N(T))+k\overline{T}(y+N(T))[/itex] which uses the linearity of T.

    c) you are making the same mistake again. Plus to show its injective you need to take [itex]\overline{T}(x+N(T))=\overline{T}(y+N(T))[/itex] and show that [itex]x+N(T)=y+N(T)[/itex].

    I hope some of that made sense. Been a long day.
  4. Mar 12, 2009 #3
    N(T)-N(T) is not 0, it is N(T), check the definition of the quotient vector space. It is in general not true that v=v', but you can still conclude that T(v)=T(v') by using the definition of N(T).

    No, one-to-one means that if T(v)=T(v') then v=v', it is the other way around. Besides, if T were 1-to-1, then N(T) would contain only the 0 vector.

    You have to show that [tex]\bar{T}[/tex](cx+y+N(T))=c[tex]\bar{T}[/tex](x+N(T))+[tex]\bar{T}[/tex](y+N(T)). Remember that you should only apply [tex]\bar{T}[/tex] to cosets, not the original vectors. To prove this part, use that [tex]\bar{T}[/tex](v+N(T))=T(v) for any v and linearity of T.

    You cannot conclude that v=v', same reason as above.

    That T is onto implies that for any v in Z you can find a w in V with T(w)=v. Now you want to find a coset u+N(T) such that [tex]\bar{T}[/tex](u+N(T))=v. What would be a good choice for u?
  5. Mar 15, 2009 #4
    Hello again. Thanks for all the help and advice. This is the new proof I have now.

    a) Prove that [tex]\bar{T}[/tex] is well defined; that is prove that if v+N(T)=v'+N(T), then T(v)=T(v').

    By definition, N(T) is the set of all vectors x in V such that T(x)=0.
    If v+N(T)=v'+N(T), both cosets have vectors v and v' added to the same Nullspace of T.
    v+N(T) = v'+N(T)
    [tex]\bar{T}[/tex](v+N(T)) = [tex]\bar{T}[/tex](v'+N(T))
    T(v)= T(v')

    Is this enough for the first proof? I took the [tex]\bar{T}[/tex] transformation of both v+N(T) and v'+N(T). Since they were equal, the [tex]\bar{T}[/tex] transformation should also be equal.

    b) Prove that [tex]\bar{T}[/tex] is linear.

    [tex]\bar{T}[/tex](cx+y+N(T)) = [tex]\bar{T}[/tex](cx+N(T)) + [tex]\bar{T}[/tex](y+N(T))
    = T(cx+y)
    = T(cx) + T(y)
    = cT(x) + T(y)

    [tex]\bar{T}[/tex](cx+N(T)) + [tex]\bar{T}[/tex](y+N(T))
    = T(cx) + T(y)
    = cT(x) + T(y)
    cT(x) + T(y) = cT(x) + T(y)

    I think my work for part (b) is enough to prove that [tex]\bar{T}[/tex] is linear. Is there anything else I need?

    c) Prove that [tex]\bar{T}[/tex] is an isomorphism.

    [tex]\bar{T}[/tex] is an isomorphism iff [tex]\bar{T}[/tex] is one-to-one and onto.
    One-to-One: Let T(x)=T(y) for any x,y [tex]\in[/tex] V.
    [tex]\bar{T}[/tex](x+N(T)) = [tex]\bar{T}[/tex](y+N(T))
    If the transformation of the coset of x+N(T) is equal to the transformation of the coset of y+N(T), the coset x+N(T) must be equal to the coset of y+N(T). As stated in part (a), N(T) is the same in both cosets (because the Nullspace does not change). So, x=y. If T(x)=T(y) implies x=y, then [tex]\bar{T}[/tex] must be one-to-one.

    Onto: Let [tex]\bar{T}[/tex](x+N(T)) = T(x) for any x [tex]\in[/tex] V/N(T).
    [tex]\bar{T}[/tex] is onto if there exists a vector y in Z such that T(x) = y for any y [tex]\in[/tex] Z.
    So, [tex]\bar{T}[/tex](u+N(T)) = T(u) = y
    y = T(x)
    Then, T(u) = T(x)
    [tex]\bar{T}[/tex](u + N(T)) = [tex]\bar{T}[/tex](x + N(T))
    u = x
    Therefore, for any vector y [tex]\in[/tex] Z, there exists a vector x [tex]\in[/tex] V; this means it is onto.

    Since it is both one-to-one and onto, [tex]\bar{T}[/tex] is an isomorphism.

    I'm not sure if the proof I've given is correct or if it is adequate enough. That is my biggest worry that I do not provide enough work.
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