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Math Help - Prove a direct sum...

  1. #1
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    Prove a direct sum...

    Here's the question as an image because it's easier to format:


    Please note I'll just write + to mean the plus with the circle around it (direct sum). + is just a normal addition.

    My work so far

    V = im(T) + ker(S) means that im(T) ∩ ker(S) = {0} and that im(T) + ker(S) = V.

    If ST = 1v, then TS = 1w (Identity transformations). Thus w = T(v) and v = S(w).
    S[w - TS(w)] = S(w) - STS(w) = v - ST(v) = v - S(w) = v -v = 0, therefore it's in ker(S).

    Now I'm stuck. I don't know how to use this to do the proof... I think showing the intersection might go:
    im(T) ∩ ker(S) = T(v) ∩ w - TS(w) = w ∩ 0 = 0. But I'm not sure.

    I have no idea about the im(T) + ker(S) part though.
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  2. #2
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    edit: disregard me, I'm an idiot

    Spoiler:

    ****preserved for reference****

    Suppose we can write a vector v\in V as the sum of s+T(t) where S(s)=0 and t\in V ie as the sum of an element of ker S and of im T. Apply S to the equation v=s+T(t) to get S(v)=S(s) + ST(t) = 0 + t, so t=S(v). Therefore s = v - TS(v). So we have the form of the required decompositon,  v = (v - TS(v)) + T(S(v)) . Does this help you?
    Last edited by maddas; April 12th 2010 at 09:42 PM. Reason: modding out stupidity
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  3. #3
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    Hmm, I'm not sure I got all of that. I'll repeat to make sure.

    So we get to the point v = (v - TS(v)) + T(S(v)). The first element is in ker(S), and the second element is in im(T).

    Then ker(S) + im(T) = v.

    I'm confused though. Does this hold for all v in V? Part of what was tripping me up before was that w-TS(w) is a specific element, so aren't you just proving one case?
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  4. #4
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    edit: disregard me, I'm a moron

    Spoiler:

    *****preserved for referene*****

    Well the relation is only formal since s is in W. You need to find a bijection between formal sums of elements of ker S and im T, and vectors in V. You should see v = (v - TS(v)) + TS(v) as "suggestive" of this bijection.

    edit: This was really vague. Will fix in a minute.
    edit2: Note that this is formal. It doesn't really hold per se since v is not in the domain of S so the compositions don't really make sense. It was meant to suggest why the term w-TSw will be useful. To find the bijection I was talking about, break V into the ker T and its complement and look at how the complement relates to im S.
    Last edited by maddas; April 12th 2010 at 09:42 PM. Reason: removing stupidity
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  5. #5
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    I don't really understand what you mean... am I supposed to show that it's one to one?
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  6. #6
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    Thanks! It definitely shows how w-TS(w) will be useful.

    I looked up "complement" in the index of my book and it's not there... but based on what I remember from stats, it would be everything that is NOT in ker(S). Which I'm guessing should turn out to be equal to im(T).

    So I understand how this works in theory, just not sure how to apply it.
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  7. #7
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    Wait, disregard all of that. If ST=1, don't V and W have to be isomorphic? I have no idea anymore. Disregard me! Sorry!
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