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I Prove that V is the internal direct sum of two subspaces

  1. Feb 6, 2017 #1
    Let V be a vector space. If U 1 and U2 are subspaces of V s.t. U1+U2 = V and U1 and U1∩U2 = {0V}, then we say that V is the internal direct sum of U1 and U2. In this case we write V = U1⊕U2. Show that V is internal direct sum of U1 and U2if and only if every vector in V may be written uniquely in the form v1+v2 with v1∈U1 and v2∈ U2.

    What does it mean to be unique? Does it matter if it is unique?
     
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  3. Feb 6, 2017 #2

    andrewkirk

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    It means that for any vector ##u\in V##, if ##w_1+w_2=u=v_1+v_2##, for ##w_1,v_1\in U_1## and ##w_2,v_2\in U_2##, then ##v_1=w_1## and ##v_2=w_2##.

    Yes, the uniqueness matters. If the two subspaces overlap, there is more than one way of writing a vector in ##V## as a sum of two vectors one from each subspace. And vice versa, if the representation as a sum is not unique, the two subspaces must overlap.
     
  4. Feb 6, 2017 #3
    So for your example if it was not unique U2 = U1 and you can write w1 in U2 and w2 in U1? therefore w2+w1 is not unique anymore because the same things came from different subspaces?
     
  5. Feb 6, 2017 #4

    Stephen Tashi

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    Perhaps an example of non-uniqueness will make this clear. You can represent 12 as the product of two factors ##12 = x_1 x_2## but the two factors are not unique since they and be chosen to have various different values - e.g. (12)(1), (4)(3), (2)(6) etc.

    As whether uniqueness matters - that's a subjective question. Uniqueness is often a convenient property to have. For example, the rigorous approach to defining "a" zero z_1 in arithmetic would be to define it by the property that for all numbers x, z1 + x = x. It is convenient that there is a unique number, denoted by "0", that has this property. If there were several unequal zeroes, arithmetic would be more complicated.
     
  6. Feb 6, 2017 #5

    andrewkirk

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    No, that would be going too far. U2 and U1 will overlap nontrivially, but that doesn't mean they are identical. All it means that their intersection is a vector space of dimension at least 1.
     
  7. Feb 6, 2017 #6
    Thanks!! That was a good example.
     
  8. Feb 6, 2017 #7
    What do you mean by at least a vector space of dimension 1? I don't think I've gotten that far to understand it completely. Could you elaborate?
     
  9. Feb 6, 2017 #8

    andrewkirk

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    A vector space that is not just ##\{0_V\}##.
     
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