Linear Algebra - Direct Sums [SOLVED]

[pezola]

[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 \oplusK2\oplus ... \oplusKp
and
W2 = M1 \oplusM2 \oplus...\oplusMq

Prove that if W1 \capW2 = {0}, then W1 + W2 = W1 \oplusW2 = K1 \oplusK2\oplus...\oplus Kp \oplus M1 \oplusM2 \oplus...\oplusMq

The Attempt at a Solution

Can we not just say W1 + W2 = W1 \oplusW2 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 \oplusW2 = K1 \oplusK2\oplus...\oplus Kp \oplus M1 \oplusM2 \oplus...\oplusMq ?

 

[quasar987]

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 W_1+W_2 =W_1 \oplus W_2 is justified.

And the last part is just a statement about the associativity of \oplus.

 

[Hurkyl]

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

W_1 + W_2 \neq W_1 \oplus W_2

W_1 + W_2 \cong W_1 \oplus W_2


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