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Showing that V is a direct sum of two subspaces

  1. Apr 8, 2014 #1
    Hi guys, I have this general question.

    If we are asked to show that the direct sum of ##U+W=V##where ##U## and ##W## are subspaces of ##V=\mathbb{R}^{n}##, would it be possible for us to do so by showing that the generators of the ##U## and ##W## span ##V##? Afterwards we show that their intersection is a zero-vector. For example:

    ##U## is a subspace generated by ##(0,1)## and ##W## is a subspace generated by ##(2,2)##. Clearly those generators span two dimensional ##V##, and their intersection is ##(0,0)##. Therefore the conclusion can be made that their direct sum is ##V##. Is this kind of reasoning okay?
     
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  3. Apr 8, 2014 #2

    jbunniii

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    Yes, this reasoning is valid. In general, ##V## is the direct sum of ##U## and ##W## if and only if ##V = U+W## and ##U \cap W = \{0\}##. However, this is not the definition of direct sum, it's a (simple) theorem which you should try to prove.

    However, be aware that this only works for two subspaces. If you have three or more subspaces, say ##\{U_i\}_{i=1}^{N}##, then it's possible to have ##V = U_1 + \ldots U_N## and ##U_i \cap U_j = \{0\}## for all ##i \neq j##, but the sum is not direct. It's a good exercise to construct an example where this occurs.
     
  4. Apr 9, 2014 #3
    Ah ok thanks for your explanation jbuniii. If it's not too much, I would also like to ask why this is true? ##V_1 - V_2 = V_1 + V_2## if we define subtraction to be ##\{ x-y | x \in V_1, y \in V_2 \}.## This doesn't feel intuitive at all, I thought the subtraction of two subspaces will somehow reduce the elements of the new subspace, why it could be equal to the sum of subspaces?
     
  5. Apr 9, 2014 #4

    jbunniii

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    Since ##y \in V_2## if and only if ##y \in -V_2##, it doesn't matter whether you add or subtract elements of ##V_2##, you get the same result in both cases: ##V_1 - V_2 = V_1 + V_2##.

    Admittedly, the notation ##V_1 - V_2## can seem a bit misleading. Fortunately, since it is the same as ##V_1 + V_2##, there's no reason to use ##V_1 - V_2##. :biggrin:

    Note that you could even define ##aV_1 + bV_2 = \{a x + b y | x \in V_1, y \in V_2\}##, where ##a## and ##b## are nonzero scalars. Then ##V_1 - V_2## is just a special case with ##a=1## and ##b=-1##.

    Once again we have ##aV_1 + bV_2 = V_1 + V_2## for any nonzero ##a,b##, because ##x \in aV_1## if and only if ##x \in V_1## and similarly, ##y \in V_2## if and only if ##y \in bV_2##. So we can just stick with the notation ##V_1 + V_2## since all the other "linear combinations" of ##V_1## and ##V_2## give the same result.
     
    Last edited: Apr 9, 2014
  6. Apr 9, 2014 #5

    jbunniii

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    By the way, here's a bit of extra info in case it is helpful. Suppose we start with a collection of subspaces ##\{U_i\}_{i=1}^{N}## of ##V##. We may be interested in combining the ##U_i##'s to form a larger subspace. The naive thing to do would be to form the set theoretic union ##\cup_{i=1}^{N}U_i##. But this won't generally be a subspace.

    So what we really want is the smallest subspace containing all of the ##U_i## (or equivalently, the smallest subspace containing ##\cup_{i=1}^{N}U_i##). This turns out to be exactly ##U_1 + \ldots + U_N##. Proof: certainly ##U_1 + \ldots + U_N## is a subspace containing each ##U_i##. If ##S## is another such subspace then it must contain all elements of the form ##u_1+\ldots+u_N## with ##u_i \in U_i##. Thus ##U_1 + \ldots + U_N \subset S## so ##U_1 + \ldots + U_N## is the smallest such subspace.

    Now in general, the ##U_i##'s may not be "linearly independent" of each other: maybe there is some nonzero element of ##U_1## which can be expressed as a linear combination of elements of the other ##U_i##'s. But if they ARE linearly independent, then we say that the sum is a direct sum, and we write it as ##U_1 \oplus \ldots \oplus U_N## instead of ##U_1 + \ldots + U_N##. We can obtain a basis for a direct sum ##U_1 \oplus \ldots \oplus U_N## by simply selecting a basis ##B_i## for each ##U_i## and taking the union: ##B = \cup_{i=1}^{N} B_i##. Indeed, this union is disjoint because there are no common elements among the ##B_i##'s. In the finite-dimensional case, this immediately tells us that
    $$\text{dim}(U_1 \oplus \ldots \oplus U_N) = \sum_{i=1}^{N}\text{dim}(U_i)$$
    This is not true if the sum is not direct. In that case, all we can say is that
    $$\text{dim}(U_1 + \ldots + U_N) \leq \sum_{i=1}^{N}\text{dim}(U_i)$$
     
    Last edited: Apr 9, 2014
  7. Apr 10, 2014 #6
    Wow thanks for your detailed explanation. I just want to make sure for this last time that I understand this correctly.

    Can I also intuitively think that ##V_2## = ##-V_2## if ##V## is a subspace, because if you multiply every elements in the subspace with -1, then you will basically get the same subspace because every element has an additive inverse? (In other words the negatives will become the positives and vice-versa)
     
  8. Apr 10, 2014 #7

    jbunniii

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    Sure, the fact that ##V_2## is a subspace and therefore contains its additive inverses is exactly the reason why ##y \in V_2## if and only if ##y \in -V_2##, and this latter statement is the definition of set equality: we conclude that ##V_2 = -V_2##.
     
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