Find a linear transformation $F:\mathbb{R}^3\to\mathbb{R}^4$ that $\mbox{span}([1,1,2,1],[2,1,0,1])$ My book solves an exercise that asks to find a linear transformation such that its image is:
$$\mbox{span}([1,1,2,1],[2,1,0,1])$$
The solution:

Since $\mbox{dim Im}(F) = 2$, then $\mbox{Ker}(F) = 1$. Then we can
  take $F:\mathbb{R}^3\to\mathbb{R}^4$ such that  $$F(0,0,1) =
 (0,0,0,0), F(0,1,0) = (1,1,2,1), F(0,0,1) = (2,1,0,1)$$ so
$$(x,y,z) = x(1,0,0) + y(0,1,0) + z(0,0,1)\implies\\F(x,y,z) =
 xF(1,0,0) + yF(0,1,0) + zF(0,0,1)\implies\\F(x,y,z) = y(1,1,2,1) +
 z(2,1,0,1) = (y+2z, y+z, 2y, y+z)$$

I understood what the book did, but I don't understand why $\mbox{Ker}(F) = 1$ is important. It has something to do with $F(1,0,0) = (0,0,0,0)$? What if $\mbox{Ker}(F) > 1$? Could somebody explain to me what's happening?

I would say that that statement is simply confusing; though the rank-nullity theorem does imply that the kernel of $F$ must have dimension $1$ (since $F$ is a surjective linear map from a 3 dimensional space to a 2 dimensional one, and thus projects some line down to the zero vector), this is really not a helpful statement towards finding an answer.
What you really need is that every element of a basis in $\mathbb R^3$ maps to something in $\text{span}([1,1,2,1],[2,1,0,1])$ and that the image of the basis in $\mathbb R^3$ spans that set. The first solution that comes to my mind is sending the canonical basis of $\mathbb R^3$ to $(1,1,2,1)$, $(2,1,0,1)$, and $(2,1,0,1)$ yielding
$$F(x,y,z)=(x+2y+2z,\,x+y+z,\,2x,\,x+y+z).$$
However, their solution is essentially to note that if we map one basis element to $(1,1,2,1)$ and $(2,1,0,1)$, then, since this already spans the desired image, we can choose to map the last one to any element we choose in the desired span - and they choose the zero vector to satisfy this, even though any other element of the span suffices. The advantage of choosing $0$ is merely that it makes the kernel of the map the set of vectors of the form $(x,0,0)$, which is easy to visualize, I guess (the map I suggested has kernel of the form $(0,x,-x)$ and any such map will project some line to $0$, and have a kernel of dimension $1$)
