# Linear Algebra - Direct Sums

1. Apr 22, 2008

### pezola

[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 $$\oplus$$K2$$\oplus$$ ... $$\oplus$$Kp
and
W2 = M1 $$\oplus$$M2 $$\oplus$$...$$\oplus$$Mq

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

3. The attempt at a solution

Can we not just say W1 + W2 = W1 $$\oplus$$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 $$\oplus$$W2 = K1 $$\oplus$$K2$$\oplus$$...$$\oplus$$ Kp $$\oplus$$ M1 $$\oplus$$M2 $$\oplus$$...$$\oplus$$Mq ?

2. Apr 22, 2008

### 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 $$\oplus$$.

3. Apr 22, 2008

### Hurkyl

Staff Emeritus
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)