# Spans

Now that we have a better idea of what a space is and what linear independence is, we can expand our definition to a span.

A span just describes the space reachable by linear combinations of some given vectors. In fact, it is the set of all vectors reachable by linear combinations of vectors in the span.

What makes this slightly annoying to think about is that that a span describes an infinite set and infinite sets are a little hard to reason about.

But if we think about it geometrically, it is actually a case that we have seen before. Say for instance we have the following vectors in 2D space:

$\left[\begin{array}{c}1\\ 0\end{array}\right]$ $\left[\begin{array}{c}0\\ 1\end{array}\right]$

The span of these two vectors is all of 2D space. The reason for that is that you can give me any 2D point and I can tell you two scalar multiples of these two vectors that will give you that point.

For example, say you wanted the point $\left(4,1\right)$; I could say, well, $\left(4,1\right)$ is really just $4×\left(1,0\right)+1×\left(0,1\right)$

Now, what if you had the vectors $\left[\begin{array}{c}1\\ -1\end{array}\right]$ and $\left[\begin{array}{c}-1\\ -1\end{array}\right]$. These two vectors are not linearly dependent. Can we reach any point in 2D space using just combinations of those two? What about $\left(4,1\right)$?

$\left[\begin{array}{c}1\\ -1\end{array}\right]$
$\left[\begin{array}{c}-1\\ -1\end{array}\right]$

The answer, if you look at the diagram is that, yes, you can. I can take$1.5$$\left[\begin{array}{c}1\\ -1\end{array}\right]$ and $-2.5$$\left[\begin{array}{c}-1\\ -1\end{array}\right]$ ; adding both gives me $\left[\begin{array}{c}4\\ 4\end{array}\right]$

What if I have the vectors $\left[\begin{array}{c}1\\ -1\end{array}\right]$ and $\left[\begin{array}{c}-1\\ 1\end{array}\right]$?

$\left[\begin{array}{c}1\\ -1\end{array}\right]$
$\left[\begin{array}{c}-1\\ -1\end{array}\right]$

The answer by the diagram is that you cannot. The two vectors are parallel and so it is impossible to combine them in such a way to make $\left(4,1\right)$

In fact, because of the fact that we cannot reach that vector (or any other vector outside of that line), the we can only say that the span of those two vectors is the line defined by $\left(x,y\right):-x=y,.$

Of course, you can generalize to higher dimensions, for instance, the span of three vectors may be all of 3D space, or it might only be a plane if there was a linear dependence amongst the vectors.
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