MHB Proving Span of $\mathbb{R}^2$ Using Sets of Vectors

[Guest2]

I'm given the example that the space $\mathbb{R}^2$ is spanned by each of the following set of vectors: $$\left\{i, j\right\}$$, $$\left\{i, j, i+j\right\}$$, and $$\left\{0, i, -i, -j, i+j\right\}$$.

However, it's not obvious to me how. Let $i = (s, t)$ and $j= (u, v)$ then $\left\{i, j\right\}$ means $a(s,t)+b(u, v) $ for $a,b \in \mathbb{R}$, which belongs to the span of $\mathbb{R}^2$. Is this correct? If so, then I could also say since $i = (s, t)$ and $j= (u, v)$ then $i+j = (s+u, t+v)$ so $\left\{i, j, i+j\right\}$ means $a(s,t)+b(u, v)+c(s+u, t+v) $ for $a,b,c \in \mathbb{R}$, which belongs to the span of $\mathbb{R}^2$. Similarly, for the last one $\left\{0, i, -i, -j, i+j\right\}$ $= a(0, 0)+b(s, t)+c(-s, -t)+d(-u, -v)+e(s+u, t+v)$ for $a,b,c,d,e \in \mathbb{R}$ which also belongs to the span of $\mathbb{R}^2$.

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I'm starting to think it means that any $(x, y) \in \mathbb{R}^2$ can written in the form $a(s, t)+b(u,v)$ for $a,b \in \mathbb{R}$

But how do I prove this?

 

[Euge]

Usually, $i = (1,0)$ and $j = (0,1)$, in which case $(x,y) = (x,0) + (0,y) = x(1,0) + y(0,1) = x i + yj$ for all $x,y\in \Bbb R$. So then $\Bbb R^2$ is spanned by $i$ and $j$. In particular, $\Bbb R^2$ is spanned by $i$, $j$, and $i + j$ ($(x,y) = xi + yj + 0(i + j))$. Finally, given $(x,y)\in \Bbb R^2$, $(x,y) = 1(0) + x i + 0(-i) - y(-j) + 0(i + j)$; this implies $\Bbb R^2$ is spanned by the vectors $0$, $i$, $-i$, $-j$, $i + j$.

You can't have in general that $\Bbb R^2$ is spanned by any two vectors. For example, $\Bbb R^2$ is not spanned by $(1,0)$ and $(2,0)$, since the point $(1,1)$ is not in the span of $\{(1,0), (2,0)\}$ (which is the $x$-axis). It turns out that you need are two linearly independent vectors to span $\Bbb R^2$, i.e., two vectors that are not scalar multiples of each other.