Answer
See below
Work Step by Step
Let $S$ be the subspace of $R^2$ spanned by the vectors $v_1=(1,1)\\v_2=(-1,1) \in R^2$.
a) Let $v=(x,y)\in R^2$
We can see that $\frac{x+y}{2}(1,1)+\frac{y-x}{2}(-1,1)\\=(\frac{x+y}{2},\frac{x+y}{2})+(\frac{x-y}{2},\frac{y-x}{2})\\=(\frac{x+y}{2}+\frac{x-y}{2},\frac{x+y}{2}+\frac{y-x}{2})\\=(\frac{2x}{2},\frac{2y}{2})\\=(x,y)\\=v$
Hence, any vector in $R^2$ can be written as a linear combination of $v_!$ and $v_2$
Consequently, $\{v_1,v_2\}$ spnas $R^2$
b) Let $a,b$ be scalars such that
$a(1,1)+b(-1,1)=(0,0)$
Obtain $(a-b,a+b)=(0,0)\\\rightarrow a-b=0,a+b=0$
then $a-b+a+b=0 \\
\rightarrow 2a=0\\
\rightarrow a=0\\
\rightarrow b=0$
The linear combination of $v_1$ and $v_2$ is trivial and therefore set $\{v_1,v_2\}$ is linearly independent in $R^2$
c) From exercise b), $\{v_1,v_2\}$ is a basic for $R^2 \rightarrow \dim[R^2]=2 $ (by Corollary 4.6.13)