1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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))
    [tex]\bar{T}[/tex](cx+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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Linear Algebra: Linear transformation and Isomorphism
Loading...