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  1. #1
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    null

    given that f: V->W and g: W->Z. to show that null (gf) less than or equal to null (g) + null(f),

    can i say that null(gf) = null(g) intersect rank (f) < null(g) < null(g) +null(f)?
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  2. #2
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    by null(g) do you mean the nullspace, or the nullity of g (because one is a vector space, the other is an integer)?

    the dimension of ker(gf) might not be dim(ker(g)∩f(V)). for example suppose V = W = Z = R^2,

    and g(x,y) = (x,y), and f(x,y) = (x,0). then ker(gf) = ker(f) = {(0,y) : y in R}, which has dimension 1, but ker(g)∩f(V)

    = {(0,0)} ∩ {(0,y): y in R} = {(0,0)}, which has dimension 0.

    use the rank-nullity therorem instead. it may be helpful to consider the linear transformation g*:f(V)-->Z g*(w) = g(w),

    (g restricted to the range of f) and to prove first that null(g*) ≤ null(g).
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  3. #3
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    I mean nullity
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  4. #4
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    Is this working for nullity or nullspace? By proving that null(g*)<null(g), we are showing that null(gf) <null(g) right?

    Sorry but im quite confused about how to go about using linear transformation here
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  5. #5
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    yes but it really is ≤. for all we know, f might be onto, in which case g* = g.

    sure, null(gf) ≤ null(g), but that is taking "too big a bite" all at once. we want to get some idea of how much smaller null(gf) is than null(g).

    null(h), for any linear transformation h is dim(ker(h)), where ker(h) is the nullspace.

    now f may "shrink down the size" of the space of vectors v available for gf to annihilate, whereas g has the whole of W to look at.

    gf will take to 0 anything in ker(g)∩f(V), but g may take even more vectors of W to 0.

    the only thing we have to guage "how much smaller" null(gf) is than null(g) is the behavior of f.

    unfortunately, g is defined on all of W, not just on f(V)∩W = f(V).

    which amongst f,g and gf have common domains? gf, and f. the rank-nullity theorem says:

    rank(f) + null(f) = rank(gf) + null(gf) = dim(V).

    now, we want to keep the "null" parts of this equation, but we need to somehow get rid of the "rank" parts.

    we know that rank(g) + null(g) = dim(W), but that's not very helpful.

    but if we consider g*, we have: rank(g*) + null(g*) = dim(f(V)) = rank(f).

    note that rank(g*) = rank(gf) (because g* only looks at w's with w = f(v)).

    so rank(gf) + null(g*) = rank(f). now, if we combine our two equations we have:

    rank(gf) + null(g*) + null(f) = rank(gf) + null(gf)...can you see what comes next?
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  6. #6
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    Since null(g*) is less than or equal to null(g), we get null(gf) less than or equal to null(g) + null (f).

    But what about when it means nullity. Isit the same method?
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  7. #7
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    i am using null(f) to mean the nullity of f, i am referring to the nullspace as ker(f) to emphasize the distinction.
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  8. #8
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    Re: null

    how to show null(g*)<= null g?
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