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Math Help - bijective

  1. #1
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    bijective

    let T:V->W be a linear transformation. to prove that T is bijective,i need to show onto and injective. but how do i show that.

    does showing that dim (v)= dim(W) help?
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  2. #2
    Member HappyJoe's Avatar
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    (For completeness, I assume we are dealing with vector spaces of finite dimension.)

    If T is a bijective linear transformation, then it must necessarily be so that dim(V)=dim(W). The converse implication is not generally true, that is, even though we know that dim(V)=dim(W), the linear transformation may not be bijective.

    To see why this is so, you can take V=W=R (where R is the real line), so V and W are both vector spaces of dimension 1. Then the map T : V -> W defined by T(v) = 0 for all v in V is a linear transformation, and it is certainly not a bijection.

    If you have a more specific context, maybe we can help further.
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  3. #3
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    to prove that T is injective, it is sufficient to show that the null space of T is the 0-vector of V.

    why? well suppose T(u) = T(v). then T(u) - T(v) = 0 (the 0-vector of W).

    but, since T is linear, T(u) - T(v) = T(u - v). now if the null space of T consists only of {0}, then we must have that

    u - v = 0, so u = v, and therefore T is injective.

    this is the same as saying dim(V) = dim(T(V)) (that is the image of V under T has the same dimension as V,

    assuming, of course, that dim(V) is finite).

    but T(V) may not be all of W. if it is, however, than of course we have dim(V) = dim(T(V)) when T is injective,

    and if T is surjective as well, and thus bijective, T(V) = W, so dim(V) = dim(W).

    the number dim(T(V)) is often called the rank of T (and is the rank of any matrix for T, relative to bases for V and W).
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