How do you sum these given subspaces? $$S_1=\{(x,y) \in R^2 | x=y\}$$$$S_2=\{(x,y) \in R^2 | x=-y\}$$ The book that I am currently learning from gives the answer to be $R^2$, but how do you get there?
Why does $S_1+S_2=R^2$? It also says that the sum is a direct one. What does that mean?
Similarly, for $S_3=\{(x,y) \in R^2 | 3x-2y=0\}$ and $S_4=\{(x,y) \in R^2 | 2x-y=0\}$, how do you show that $S_3 \oplus S_4=R^2$? The $\oplus$ from what I understand means that this is a direct sum (although I don't know what that is). The problems seem the same. Is it just a matter of writing $x$ in terms of $y$ for both $S_3$ and $S_4$ and nothing more? I guess if I will understand the first one, I will be able to understand the second one, too.
Also, I should point out that I know a little about linear combinations and linear independence, and I found somewhere that you can solve it with those, but I am at a chapter before the one with linear combinations in the book, so I think it can be solved without them. If not, I will be satisfied with any explanation. And please excuse my possible English mistakes.
Since $S_1$ is generated by $b_1=(1,1)$ and $S_2$ by $b_2=(1,-1)$ and both are linearly independent in $\Bbb R^2$ then $S_1+S_2=\Bbb R^2$.
For $S_3$ the restriction $3x-2y=0$ implies that $$(x,y)=(x,\frac{3}{2}x)$$ after solving for $y$. The geometrical meaning is a parameterization (in terms of $x$) of a line which passes through the origin. This is you subspace $S_3$ and is generated by any nonzero vector therein. I chose $(\frac{1}{3},\frac{1}{2})$.