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Linear Algebra - Direct Sums

  1. Apr 22, 2008 #1
    [SOLVED] Linear Algebra - Direct Sums

    1. The problem statement, all variables and given/known data


    Let W1, W2, K1, K2,..., Kp, M1, M2,..., Mq be subspaces of a vector space V such that
    W1 = K1 [tex]\oplus[/tex]K2[tex]\oplus[/tex] ... [tex]\oplus[/tex]Kp
    and
    W2 = M1 [tex]\oplus[/tex]M2 [tex]\oplus[/tex]...[tex]\oplus[/tex]Mq

    Prove that if W1 [tex]\cap[/tex]W2 = {0}, then W1 + W2 = W1 [tex]\oplus[/tex]W2 = K1 [tex]\oplus[/tex]K2[tex]\oplus[/tex]...[tex]\oplus[/tex] Kp [tex]\oplus[/tex] M1 [tex]\oplus[/tex]M2 [tex]\oplus[/tex]...[tex]\oplus[/tex]Mq

    3. The attempt at a solution

    Can we not just say W1 + W2 = W1 [tex]\oplus[/tex]W2 since their intersection is empty?

    Then, by the definition of direct sum, the subspaces inside W1 and W2 cannot intersect each other.

    Then can we say
    W1 [tex]\oplus[/tex]W2 = K1 [tex]\oplus[/tex]K2[tex]\oplus[/tex]...[tex]\oplus[/tex] Kp [tex]\oplus[/tex] M1 [tex]\oplus[/tex]M2 [tex]\oplus[/tex]...[tex]\oplus[/tex]Mq ?
     
  2. jcsd
  3. Apr 22, 2008 #2

    quasar987

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    I guess the relevance of the question would depend on the definition of direct sum you've seen, but I agree that it is quite trivial. If the intersection of W1 and W2 is trivial, then every element in W1+W2 can be written uniquely as a sum of the form w1+w2 (wi in Wi), and thus the notation [tex]W_1+W_2 =W_1 \oplus W_2[/tex] is justified.

    And the last part is just a statement about the associativity of [tex]\oplus[/tex].
     
  4. Apr 22, 2008 #3

    Hurkyl

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    For the record... in the case where [itex]W_1 \cap W_2 \cong \{ 0 \}[/itex],

    [tex]W_1 + W_2 \neq W_1 \oplus W_2[/tex]

    [tex]W_1 + W_2 \cong W_1 \oplus W_2[/tex]


    Also for the record, if [itex]W_1 \cap W_2 = \{ 0 \} [/itex], then their intersection is nonempty. (It contains the element 0)
     
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