Linear Algebra - Direct Sums

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[SOLVED] Linear Algebra - Direct Sums

Homework Statement




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

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 ?
 

Answers and Replies

  • #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].
 
  • #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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