My other thread might be useful:

http://www.mathhelpforum.com/math-he...subspaces.html
Sifting $\displaystyle (1,2,0,1)$ and $\displaystyle (1,1,1,0)$ gives those two vectors. Hence $\displaystyle dim(W_1)=2$.

Similarly $\displaystyle dim(W_2)=1$.

Since $\displaystyle (1,2,0,1)+(1,1,1,0)=(2,3,1,1)$ we know that $\displaystyle W_2$ is a subspace of $\displaystyle W_1$.

Therefore $\displaystyle dim(W_1 \cap W_2)=1$ and $\displaystyle dim(W_1+W_2)=2$.

Is this it? This question was worth 10 marks but my answer is remarkably small!