If $A=(A_{1}, \dots A_{n}), A_{i} \in K^{m}$, then the rank of $A$ is $r=dim(Span(A_{1}, \dots, A_{n}))$. Let $\{v_{1}, \dots v_r\}$ be a basis for it. So, $\forall i=1\dots n, \exists a_{i1}, \dots , a_{ir} \in K$ such that $A_{i}=a_{i1}v_1+\dots + a_{ir}v_r $ and
$A=(a_{11}v_1+\dots + a_{1r}v_r, \dots , a_{n1}v_1+\dots + a_{nr}v_r)$ .
Now you can take $B_j=(a_{1j}v_j, \dots, a_{nj}v_j)$ $\forall j=1, \dots, r$ and observe that $rk(B_j)=1$ and $A=\sum_{j=1}^{r}B_j$.
Here's the example:
$A=(A_1 , A_2, A_3, A_4)$ with $A_1=\begin{bmatrix}1 \\ 3 \\ 2 \end{bmatrix}$ , $A_2=\begin{bmatrix}3 \\ 0 \\ 1 \end{bmatrix}$ , $A_3=\begin{bmatrix}2 \\ 1 \\ 1 \end{bmatrix}$ , $A_4=\begin{bmatrix}6 \\ 4 \\ 4 \end{bmatrix}$ .
$Span(A_1,A_2,A_3,A_4) = R^3$ , and so we can take the standard basis $\{e_1=\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}, e_2=\begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix}, e_3=\begin{bmatrix}0 \\ 0 \\ 1 \end{bmatrix} \}$
Now, $A_1=1e_1+3e_2+2e_3$ , $A_2=3e_1+0e_2+e_3$ , $A_3=2e_1+1e_2+1e_3$ , $A_4=6e_1+4e_2+4e_3$ ,
and so I make:
$B_1=\begin{bmatrix} 1 & 3 & 2 & 6\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}$
$B_2=\begin{bmatrix} 0 & 0 & 0 & 0\\ 3 & 0 & 1 & 4\\ 0 & 0 & 0 & 0 \end{bmatrix}$
$B_3=\begin{bmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 2 & 1 & 1 & 4 \end{bmatrix}$
and $A=B_1+B_2+B_3$.
N.B. The matrixes $B_i$ are so "well done" in this case because we could take the standard basis as basis of the columns.