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Linear Transformations

  1. Nov 20, 2005 #1
    A question reads:

    Let T: V-->W be a linear transformation.

    a) If T is one-to-one and TR=TR1 for transformations R and R1: U -->V, show that R = R1
    b) If T is onto and ST=S1T for transformations S and S1: W -->U, show that S=S1

    I am sooo very lost here, and no idea where to start:(

    for part a) what does it mean by TR and TR1?
    just T(R): U --->V and T(R1): U ---V?

    If its onto, dosn't that just mean T(R) = o and then show that R = 0?

    I wont even bother with part be given that im confused with part A

    Can somebody help me please

    Thanks
     
  2. jcsd
  3. Nov 20, 2005 #2

    AKG

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    Definitions:

    If A is a map from X to Y, and B is a map from Y to Z, then BA is a map from X to Z defined by:

    BA(x) = B(A(x))

    for all x in X. So, if f(x) = 3x, and g(x) = x+5, then gf(x) = g(f(x)) = g(3x) = 3x+5. Note that fg(x) = f(g(x)) = f(x+5) = 3(x+5) = 3x+15, so gf is not the same as fg.

    If A is a map from X to Y and B is also a map from X to Y, then A = B if and only if, for all x in X, A(x) = B(x).

    A map A from X to Y is one-to-one if and only if, for all x and x' in X, A(x) = A(x') if and only if x = x'. In simpler terms, a one-to-one map never sends two different things to the same element; each element of the co-domain gets its own image. The map f on the reals defined by f(x) = x² is not one-to-one, because it sends both 1 and -1 to the same thing, namely 1.

    A map A from X to Y is onto if and only if, for all y in Y, there exists some x in X such that A(x) = y. In simpler terms, a map is onto if it reaches everything in the codomain. The map f in the previous example is not onto, since it will only map into the non-negative half of the reals.
     
  4. Nov 22, 2005 #3
    no progress so far :(
     
  5. Nov 23, 2005 #4

    HallsofIvy

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    No, R takes a vector in U to one in V, then T takes it to a vector in W.
    TR: U-> W That is, if x is in U, apply R to it to get R(x) in V. Now apply T to that to get T(R(x)) in W. It is essentially composition of functions.
    No, "one-to-one" means that if T(x)= T(y) then x= y: two different vectors in V cannot[\b] be mapped to the same vector in W. For T a linear transformation, T(0)= 0 so "one-to-one" for a linear transformation is the same as saying T(x)= 0 if and only if x= 0.
    TR= TR1 means that for every vector x in U, TR(x)= TR1(x). That is,
    T(R(x))= T(R1(x)). Since T is one-to-one, what does that tell you about
    R(x) and R1(x)? Since x could be any vector in U, what does that tell you about R and R1?

    "onto" means that the range of T is all of W: if y is any vector in W, then there is some x in V so that T(x)= y. (for example, x2 is not "onto" the set of all real numbers since no x2 is equal to -1. It is "onto" the set of all non-negative numbers.)

    Again ST= S1T1 means that S(T(x))= S1(T(x)) for all x in V.
    Let y be any vector in W. Then there exist x such that y= T(x). For that x, S(T(x))= S(y)= S1(T(x))= S1(y). That is, for every y in W, S(y)= S1(y).
     
  6. Nov 23, 2005 #5
    Since T is one-to-one, that tells me that R(x) = R1(x)
    And since X is any vector in U, then that tells me that R = R1
     
    Last edited by a moderator: Nov 23, 2005
  7. Nov 26, 2005 #6
    Okay im back again with a similar question.

    Let V---(T)--->U---(S)---W be linear transformations.
    a) If ST is one-to-one, show that T is one-to-one and that dimV </= dim U

    Here is how i attempted it:

    Assume T is NOT one-to-one

    if x is a vector in V, then for ever x not equal to 0 such that T(x) = 0

    then
    ST(x) = 0
    =S(T(x))
    =S(0)
    therefore, T is one-to-one?

    How does that sound?

    Now how do i show that dimV is </= dim U?
     
  8. Nov 26, 2005 #7

    HallsofIvy

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    I'm not sure what you mean by this: "if x is a vector in V, then for ever x not equal to 0 such that T(x) = 0"
    It doesn't have to be for "ever(y) x not equal to 0". It is true that if T is not one-to-one, then there must exist some x, not equal to 0, such that T(x)= 0. Then, of course, ST(x)= S(0)= 0 which contradicts the fact that S is one-to-one.

    Now, apply T to every basis vector of V. How many vectors does give you? What does that tell you about the dimension of T(V) (which is a subspace of U)?
     
  9. Nov 27, 2005 #8
    Okay im so lost. Ill start again.

    Let x be a basis vector in V, not equal to 0, such that T(x) = 0
    Then ST(x) = S(0) = 0 contradicts that S is one-to-one?
    How does this show that T is one-to-one

    If T is one-to-one, ker(T) = 0. therefore, dimV = dimImage(T) and Image(T) = U... so dimV = dimU

    How does that sound?
     
  10. Nov 28, 2005 #9
    oh i don't know how i missed that, i see why it contradicts one-to-one! im so silly.

    Now for the dimension,

    If you apply T to ever basis in V, since T is one-to-one, you will get that many vectors. And since V is a subspace of U, its dimension cannot be greater than that of U.

    How does that sound? (If anyone actually seems to care :S)
     
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