How to find the basis of the Col($A$) from the basis of Null($A$)? In a linear algebra course, the textbook proves a version of the Dimension Theorem. More specifically, If $A$ is a $m \times n$ matrix, then rank $A$ + dim Null($A$) = $n$.
The proof essentially says that if $B$ is the basis for $ \mathbb{R}^n$ and $B = \{{\vec v_1},...,{\vec v_n},{\vec v_{n+1}},...,{\vec v_k}\}$ and $\{{\vec v_1},...,{\vec v_n}\}$ is the basis for Null($A$), then $\{{\vec v_{n+1}},...,{\vec v_k}\}$ is the basis for Col($A$). Then it shows linear dependency for $\{{A\vec v_{n+1}},...,{A\vec v_k}\}$, and Span$\{{A\vec v_{n+1}},...,{A\vec v_k}\}$ = Col($A$) to complete the proof. 
What I don't understand is that after the proof it says you can find the basis for Col($A$) from the basis for Null($A$) without example or explanation. What I think is the proof implies that since we have the basis for Null($A$), $B - \{{\vec v_{n+1}},...,{\vec v_k}\}$ gives us Col($A$), but it's unclear to me how to find $B$ and how that set subtraction works.
 A: Pick $\mathfrak B=\{v_1,\cdots,v_s\}$ a basis of $\operatorname{Null}A$. Complete $\mathfrak B$ to a basis $\mathfrak D=\{v_1,\cdots,v_s,w_1,\cdots,w_{n-s}\}$ of the whole $\Bbb R^n$. You should have learned a standard procedure to do it, for instance, pick any basis $\mathfrak C=\{c_1,\cdots,c_n\}$ of $\Bbb R^n$ and eliminate all the vectors $c_i$ such that $\operatorname{rk}(v_1,\cdots,v_s,c_i)\le s$. Perhaps not the smartest one, but without a doubt it is explicit, and it works in every field.
Then, $\{Aw_1,\cdots,Aw_{n-s}\}$ is a basis of $\operatorname{Col}A$. In fact, you know by rank-nullity theorem that $\dim\operatorname{Col}A=n-s$; those vectors are linearly independet because $$\alpha_1Aw_1+\cdots +\alpha_{n-s}Aw_{n-s}=A(\alpha_1w_1+\cdots+\alpha_{n-s}w_{n-s})=0\iff \\ \iff \alpha_1w_1+\cdots+\alpha_{n-s}w_{n-s}\in \operatorname{Null}A$$
And, by construction, the latter is true if ans only if $\alpha_1=\cdots=\alpha_{n-s}=0$.
A: $B$ is any basis for $\mathbb{R}^n$, meaning that the span of this basis is equal to all of $\mathbb{R}^n$. For example, in $\mathbb{R}^3$ the standard basis is $\{(1,0,0),(0,1,0),(0,0,1)\}$. Set difference is defined as $A-B=\{a | a\in A,a \notin B\}$. Given this information, does it clear up what you need to do?
