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?

by rank-nullity theorem dim $kerf$+dim $im f$=dim $\mathbb R^3$  
since dim $im f=2 $ dim $kerf$ ought to be $1$ i.e it should contain only one linearly 
independent vector of $\mathbb R^3$ .hence we can assign only one element out of the 3 basis 
elements of $\mathbb R^3$ to zero
