Determine a base and the dimension of:
a) U and W
b) U + W
c) U $\cap$ W $$ U = \left\{ (x,y,z) \in R^{3} : x+y+z = 0 \right\} \\ W = \left\{ (x,y,0) \in R^{3} : x,y \in R \right\} $$
I see that both sets can be described as points in a plane...
Basis:
For U: $$ \alpha = (1,1,-2)\\ \beta = (0,1,-1)\\ \gamma = (2,-2,0)\\ \alpha,\beta,\gamma \in x+y+z = 0\\ \text{ }\\ \alpha - \beta = (1,0,-1)\\ \gamma - \alpha = (1,-3,2)\\ \text{ }\\ B_{u} = \left\{ (1,0,-1), (1,-3,2) \right\} $$
For W: $$ B_{w} = \left\{ (1,0,0), (0,1,0) \right\} $$
a) $$ dim(u) = dim(w) = 2 $$
b) $$ dim(u+w) = dim(u) + dim(w) - dim(u \cap w) =\\ dim(u+w) = 2 + 2 - dim(u \cap w)\\ $$Since U and W are planes with different inclinations, based on their normal vectors: $(1,1,1),(0,0,1)$ respectively, we can affirm that their intersection is going to be a line, with $dim = 1$ $$ dim(u \cap w) = 1 $$ Hence: $$ dim(u+w) = 2 + 2 - 1 = 3 $$ c) $$ dim(u \cap w) = 1 $$
I don't know the best way to get the basis and dimension in this case... But there was what I've done. Please tell me if there's an easier way to do it. What if they were not planes, but the all $R^{3}$ space, I would have had a little bit of trouble on solving it!
Thanks
Your work is fine for parts $(a)$ and $(b)$. Notice that $U\cap W$ is the set of all vectors whose third component is zero and are orthogonal to the vector $(1, 1, 1)$. That is, $(x, y, 0)\cdot(1,1,1) = x + y = 0$ or the set of all vectors such that $x = -y$.
In other words, $U\cap W = \{(-y, y, 0)\,\colon y\in \Bbb R\}$. Can you find a basis for this space? Notice that the dimension is in general the number of vectors in a basis, so once you have determined a basis, picking up the dimension is straightforward.