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Math Help - dimension

  1. #1
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    dimension

    let f: V--> W. if the dim V = dim W, prove that f might not be an isomorphism.

    my working:
    in this case, if the dimV= rank f + nullity f= dim W,
    it means that nullity f = 0.

    for f to be an isomorphism, it means that it is bijective and linear.
    so in this case, we need to prove that it is not linear.

    let f(x) = (x)^2 + 1. then its proven.

    can it be done like this?
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  2. #2
    Member HappyJoe's Avatar
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    To me it's not clear exactly what you are asking.

    Are you asking: If f\colon V\rightarrow W is a linear map between vector spaces V and W with \text{dim}(V) = \text{dim}(W), show that f need not be an isomorphism?

    If so, just let f be the linear map sending all elements of V to the zero vector of W (which won't be an isomorphism, whenever \text{dim}(V)>0).

    And why should the equations \text{dim}(V) = \text{rank}(f) + \text{nullity}(f) = \text{dim}(W) imply that \text{nullity}(f) = 0? The map f could easily have rank 1, and thus non-zero nullity (at least as far as \text{dim}(W)>1).

    About non-linear maps not being isomorphisms of vector spaces: Well, yeah, it is very much a true statement, but I really think that the map f is implicitly assumed to be linear, so the author of the problem would probably like a linear example of a non-isomorphism.
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  3. #3
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    i just took it that dim W = rank W = dim V and forgot the fact that the that is only true if f is an isomorphism.

    in this case, from the example that you have given,
    assuming that the dimV = n and all elements in V gets sent to the zero vector in W, does it mean that nullity is n and rank is 0 thus dimW= n?
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  4. #4
    Member HappyJoe's Avatar
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    In the example with the map sending everything from V (with dimV = n) to the zero vector, you are right that the nullity of the map is n, and that the rank is 0. You then write "thus dimW = n", which I don't quite get. Remember that you assumed to begin with that dimV = dimW, so if dimV=n, then dimW=n is immediately true as well.
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