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

In summary, the space $\mathbb{R}^2$ can be spanned by the sets of vectors $\left\{i, j\right\}$, $\left\{i, j, i+j\right\}$, and $\left\{0, i, -i, -j, i+j\right\}$, which can be written as linear combinations of the vectors $i$ and $j$. However, in order to span $\mathbb{R}^2$, the vectors must be linearly independent, meaning they are not scalar multiples of each other.
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I'm given the example that the space $\mathbb{R}^2$ is spanned by each of the following set of vectors: \(\displaystyle \left\{i, j\right\}\), \(\displaystyle \left\{i, j, i+j\right\}\), and \(\displaystyle \left\{0, i, -i, -j, i+j\right\}\).

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?
 
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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.
 

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

How do you prove the span of $\mathbb{R}^2$ using sets of vectors?

To prove the span of $\mathbb{R}^2$ using sets of vectors, you need to show that any vector in $\mathbb{R}^2$ can be written as a linear combination of the given set of vectors. This can be done by setting up a system of equations and solving for the coefficients of the linear combination.

What is the significance of proving the span of $\mathbb{R}^2$?

Proving the span of $\mathbb{R}^2$ is significant because it shows that the given set of vectors can generate all possible vectors in $\mathbb{R}^2$. This is important in linear algebra and other areas of mathematics where vector spaces and linear transformations are studied.

Can the span of $\mathbb{R}^2$ be proven using only two vectors?

Yes, the span of $\mathbb{R}^2$ can be proven using only two vectors. This is because any vector in $\mathbb{R}^2$ can be written as a linear combination of two vectors, as long as the two vectors are linearly independent.

Are there multiple ways to prove the span of $\mathbb{R}^2$?

Yes, there are multiple ways to prove the span of $\mathbb{R}^2$. Some common methods include using Gaussian elimination, finding a basis for the set of vectors, or using the definition of span in terms of linear combinations. The specific method used may depend on the given set of vectors and the desired outcome.

Can the span of $\mathbb{R}^2$ be proven for higher dimensions?

Yes, the span of $\mathbb{R}^2$ can be proven for higher dimensions, such as $\mathbb{R}^3$ or $\mathbb{R}^n$. The same principles apply, but the calculations may become more complex as the dimension increases.

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