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Math Help - Direct Sum of Subspaces - Question

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    Direct Sum of Subspaces - Question

    Let U and W be subspaces of V . Show that the following are equivalent:
    (i) for each vector v \in V there exist unique vectors u \in U and w \in W such that v=u+w;
    (ii) dim V \leq dim U + dim W and U\cap W = {0}.

    I am currently stuck showing i) implies ii). So far all I have is that Sp(U+W)=V, but can't see how this will help. Any tips on where to start?
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    Re: Direct Sum of Subspaces - Question

    suppose B = {u1,u2,...,um} is a basis for U, and C = {w1,w2,...,wn} is a basis for W.

    is BUC a spanning set for U+W? is the size of a basis for V:

    a) equal to, b) less than, c) greater than, d) less than or equal to, or e) greater than or equal to a spanning set for V?

    once you have answered THAT question, what can we say about U∩W if B∩C = ?

    why does the uniqueness of v = u+w imply something about the intersection of the two bases?

    (hint: write the 0-vector in this form).
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    Re: Direct Sum of Subspaces - Question

    Ok, by writing u and v as linear combinations of the elements of B and C, I can show that Sp(BUC)=U+W, and I know that Sp(U+W)=V. So from this (I think) I can then say Sp(BUC)=V. Now intuitively I can see that the inequality is correct, but how do I write it formally?
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    Re: Direct Sum of Subspaces - Question

    again, when you have a basis A for a vector space V, dim(V) = |A|. that is how you know the dimension, it's the size of any basis.

    now if a set is a spanning set, it contains a basis (but it may be too big, because there may be some linearly dependency).

    if a set is linearly independent you can extend it to a basis (because it may be too small, and does not span).

    the concepts are stacked like this: linearly independent set ≤ basis ≤ spanning set.

    so, since BUC is a spanning set for V, we have dim(V) ≤ |BUC| = |B| + |C| - |B∩C| ≤ |B| + |C| = dim(U) + dim(W)

    (the middle equality for |BUC| is also known as the inclusion-exclusion principle: we count elements common to both sets twice, so we have to subtract them out).

    saying sp(U+W) = V is overkill. we already know U+W = V (since we can write every v as u+w). certainly every vector space spans itself: sp(V) = V.

    but generally, you want to do MUCH better than that, you want to find as small a spanning set as possible. such a set is called a basis. BUC is a LOT smaller than U+W.

    and smaller spanning sets are better (just like bigger linearly independent sets are better). if BUC spans U+W, then it certainly spans V

    since V EQUALS U+W. and since B is a basis for U, and C is a basis for W, |B| = dim(U), |C| = dim(W). isn't that what we're after? trying to express

    some sense of relative size between the dimensions? when ever you see a question involving "dimension" you should think: "i need a basis".

    now, we know that the 0-vector is in V. and since 0 = 0+0, and 0 is in U, and 0 is in W, this must be the ONLY way we can write 0 as a sum of something

    in U, and something in W. so using our bases B and C, suppose we have 0 = a1u1 + a2u2 +...+ amum + b1w1 + b2w2 +....+ bnwn.

    then a1u1 + a2u2 +...+ amum and b1w1 + b2w2 + ...+ bnwn have to be both 0 (why? the answer is above).

    but that means that BUC is a _____ ______ set, which means that dim(U+W) = _______ + ________ ,

    which means in particular that |B∩C| = _______ , which in turn means that U∩W = ______ (because it has an _____ basis).
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    Re: Direct Sum of Subspaces - Question

    Ok. first (i) \RightarrowGiven (i) then you know that V \subseteq U \oplus W. We also know that dim(U\oplus W) = dim(U) + dim(W). It is known that the dimesion of a subspace is lesser than or equal to the dimesion of the larger space. Hence we have shown that
    $$dim(V) \le dim(U\oplus W) = dim(U ) + dim(W)$$
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    Re: Direct Sum of Subspaces - Question

    Please excuse my previous post as I accidentally pressed the submit button instead of review button. English is not my first language but I'm doing my best. I hope that the following helps, I like to avoid using any vectors whatsoever in proofs like this.


    Ok. first (i) \Rightarrow(ii). Given (i) then you know that V \subseteq U \oplus W. We also know that dim(U\oplus W) = dim(U) + dim(W). It is known that the dimesion of a subspace is lesser than or equal to the dimesion of the larger space. Hence we have shown that
    $$dim(V) \le dim(U\oplus W) = dim(U ) + dim(W)$$.
    Now (ii) \Rightarrow (i). Given (ii) we automatically know that
    $$ dim(V) \le dim(U) + dim(W)$$, and we know that U\cap W = \{0\}. Next, we know that since U and W are both subspaces of V then their sum (that is U+W) is also a subspace of V, hence the following holds:
    $$ dim(U+W) \le dim(V).
    Finally, since we have the equation dim(U+W) = dim(U) + dim(W) - dim(U\cap W), which gives us that dim(U+W) = dim(U) + dim(W) since dim(U\cap W) = dim(\{0\}) = 0. Effecetively we have shown that
    $$dim(V) \ge dim(U) + dim(W)$$
    also holds. Then we must have dim(V) = dim(U) + dim(W) = dim(U + W) . This gives us that V = U+W and, since U\cap W = \{0\}, therefore V = U \oplus W which is equivalent to (i).
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    Re: Direct Sum of Subspaces - Question

    Quote Originally Posted by obd2 View Post
    Please excuse my previous post as I accidentally pressed the submit button instead of review button. English is not my first language but I'm doing my best. I hope that the following helps, I like to avoid using any vectors whatsoever in proofs like this.


    Ok. first (i) \Rightarrow(ii). Given (i) then you know that V \subseteq U \oplus W. We also know that dim(U\oplus W) = dim(U) + dim(W). It is known that the dimesion of a subspace is lesser than or equal to the dimesion of the larger space. Hence we have shown that
    $$dim(V) \le dim(U\oplus W) = dim(U ) + dim(W)$$.
    Now (ii) \Rightarrow (i). Given (ii) we automatically know that
    $$ dim(V) \le dim(U) + dim(W)$$, and we know that U\cap W = \{0\}. Next, we know that since U and W are both subspaces of V then their sum (that is U+W) is also a subspace of V, hence the following holds:
    $$ dim(U+W) \le dim(V).
    Finally, since we have the equation dim(U+W) = dim(U) + dim(W) - dim(U\cap W), which gives us that dim(U+W) = dim(U) + dim(W) since dim(U\cap W) = dim(\{0\}) = 0. Effecetively we have shown that
    $$dim(V) \ge dim(U) + dim(W)$$
    also holds. Then we must have dim(V) = dim(U) + dim(W) = dim(U + W) . This gives us that V = U+W and, since U\cap W = \{0\}, therefore V = U \oplus W which is equivalent to (i).

    Next time you can just press on 'Edit post'...
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