Can someone explain span and basis in matrices? I did not really understand my textbook. So would someone mind simply explaining what exactly is span and basis and how do you find it?
Thanks guys :)
 A: The every day use of the word span can be applied to a bridge say, spanning a river from one side to the other, it fills the gap. Applied to a set of vectors in linear algebra, those vectors span a vector space, that is they fill it out, but they fill it out by first calculating all possible linear combinations of those vectors.
So, looking at one vector first, for example $(1, 1, 0)$ in $\mathbb R^3$. When this is multiplied by any real number it fills out a line through the origin and through $(1, 1, 0)$, so $W = \{s(1, 1, 0)\,|\,s \in \mathbb R\}$ is the set which is the span of $(1, 1, 0)$ and is in fact the line in the $xy$-plane for which $x=y$ and $z=0$.
This set, $W$, is a one dimensional subspace of $\mathbb R^3$. The one element set $\{(1, 1, 0)\}$ is a basis for, and spans, this subspace. It's a basis in the sense that only one vector is required to span, or completely fill out, a one dimensional subspace.
Now consider the two element set $\{(1, 1, 0), (1, 0, 1)\}$. The vector $(1, 0, 1)$ spans the one dimensional subspace of $\mathbb R^3$ which is the line through the origin and $(1, 0, 1)$ in the $xz$-plane with $x=z$ and $y=0$. If we now look at all possible linear combinations of $(1, 1, 0)$ and $(1, 0, 1)$, we get $U = \{s(1, 1, 0) + t(1, 0, 1)\,|\, s, t \in \mathbb R\}$ which is the plane through the origin that contains both the aforementioned lines.
This set, $U$, is a two dimensional subspace of $\mathbb R^3$. The two element set $\{(1, 1, 0), (1, 0, 1)\}$ is a basis for, and spans, this subspace. It's a basis in the sense that at least two linearly independent vectors are required to span a two dimensional space and it is in fact exactly two linearly independent vectors. This is how linear independence is used to define dimension.
If we looked instead at the two element set $\{(1, 1, 0), (2, 2, 0)\}$, these two vectors only span $W$ because $(2, 2, 0) = 2(1, 1, 0)$ and they are not linearly independent. They are also not a basis because, as we showed above, $W$ is one dimensional and can therefore only have a one element basis.
A: So, very generally. Let's suppose you work in $\mathbb{R}^2$. Here, $V = \mathbb{R}^2$ and $F = \mathbb{R}$. You want to find a basis for V, that is to say a subset of V = $(v_{1},...)$ that :


*

*is Linearily independant : it should not be possible to express one of the $v_{i}$ with a linear combination of the others $v_{j}$. In other words, for all $(\lambda{}_{1},...)$ where every $\lambda{}_{i}$ is in $F$, $\sum{\lambda{}_{i}v_{i}} \neq 0$.

*Spans V : every element of V can be expressed with a linear combination of the $v_{i}$.


You can for example try the very basic basis $((1,0),(0,1))$. It's quite obvious that you can't manage to get $(0,1)$ from a multiplication of $(1,0)$ (that gives the linear independance condition), and every $(x,y)$ in $\mathbb{R}^2$ can be expressed as $x*(1,0)+y*(0,1)$ (it gives the span condition).
What is very important to understand is that there are several basis which work. For example, $((0,1),(1,0))$,$((1,0),(1,1))$,$((0,10),(45,0))$...
An example that doesn't work is $((1,1),(2,1),(0,1))$, because $2*(1,1)-(2,1) = (0,1)$, so this subset is not a basis, because it is not lineary independant (with my $\lambda{}$ notation, here we have $\lambda{}_{1} = 2, \lambda{}_{2} = -1, \lambda{}_{3} = -1$, you can verify that the sum is $(0,0)$).
Now, can you try to give a basis of $\mathbb{R}^3$ ? :)
