How do I show that {$x_1,...,x_k, b_1,..., b_l$} is a basis of $\mathbb{R}^n$? 
Let A be an $n×n$ matrix such that A = A·A. Let {$x_1,...,x_k$} be a basis of Nul(A),
  and let {$b_1,..., b_l$} be a basis of Col(A). Show that {$x_1,...,x_k, b_1,..., b_l$} is a basis of $\mathbb{R}^n$.

What I basically have thought is that by the rank theorem we have that $$rank(A) = n = dim(Nul(A)) + dim(Col(A)) = k + l.$$ I really don't know where to continue from there... How do I prove that {$x_1,...,x_k, b_1,..., b_l$} is a basis of $\mathbb{R}^n$?
 A: Define the matrices $X = [x_1\ldots x_k]$ and $B = [b_1\ldots b_l]$. Our goal is to show that $Xu + Bv = 0 \implies u,v = 0$. Suppose that $Xu + Bv = 0$. Since $Xu \in \operatorname{null}(A)$ it follows that $0 = AXu + ABv = ABv$. Since $Bv\in\operatorname{col}(A)$, there exists a vector $y$ such that $Ay = Bv$. Recalling that $A^2 = A$, we have that $Ay = ABv\implies Bv = ABv$. Now because $B$ has linearly independent columns $ABv = Bv = 0 \iff v = 0$. Similarly, because $X$ has linearly independent columns $Xu = 0 \iff u = 0$. Thus, $\{x_1,\ldots,x_k,b_1,\ldots,b_l\}$ is linearly independent.
A: To follow @Bye-World's comment, consider for the scalar $u_i$ and $w_j$,
$$
u_1x_1+\cdots+u_kx_k+w_1b_1+\cdots+w_lb_l=0\tag{1}
$$
You want to show that (1) imples $u_i,w_j=0$ for all $i=1,\cdots,k$ and $j=1,\cdots,l$. But once you show that $w_j=0$ then you automatically have $u_i=0$ since $\{x_1,\cdots,x_k\}$ is a basis for the null space of $A$. 
Applying $A$ on both sides of (1), you have
$$
A(w_1 b_1+\cdots+w_lb_l)=0\tag{2}
$$
But $w_1 b_1+\cdots+w_lb_l$ is a vector in the column space of $A$, hence equal to $Aw$ for some $w\in\mathbb{R}^n$. (Note that $Aw$ gives you a linear combination of the columns of $A$.) Therefore,
$$
w_1 b_1+\cdots+w_lb_l=Aw=A^2w=A(Aw)=A(w_1 b_1+\cdots+w_lb_l)=0
$$
which implies that $w_j=0$ for all $j$, since $\{b_1,\cdots,b_l\}$ is a basis.
