Let $\displaystyle W_{1},W_{2},K_{1},K_{2},...,K_{p},M_{1},M_{2},..., M_{q} $ be subspaces of a vector space V such that $\displaystyle W_{1}=K_{1} \oplus K_{2} \oplus \cdot \cdot \cdot \oplus K_{p}$ and $\displaystyle W_{2} = M_{1} \oplus M_{2} \oplus \cdot \cdot \cdot \oplus M_{q} $

Prove that if $\displaystyle W_{1} \cap W_{2} = \{ 0 \} $, then $\displaystyle W_{1}+W_{2} = W_{1} \oplus W_{2} = K_{1} \oplus K_{2} \oplus \cdot \cdot \cdot \oplus K_{p} \oplus M_{1} \oplus M_{2} \oplus \cdot \cdot \cdot M_{q}$

Proof:

Now, $\displaystyle W_{1} + W_{2} = W_{1} \oplus W_{2} $ is quite simple since their intersection is empty.

Now, since all the subspaces inside W1 and W2 doesn't intersect with one another by the defintion of direct sum, that means $\displaystyle W_{1} \oplus W_{2} = K_{1} \oplus K_{2} \oplus \cdot \cdot \cdot \oplus K_{p} \oplus M_{1} \oplus M_{2} \oplus \cdot \cdot \cdot M_{q}$

This looks too simple, is that right?