Creating a matrix of rank r from r number of rank 1 matrices? I am told that all matrices of Rank $r$ can be formed out of the combinations of $r$ number of Rank 1 matrices. So that's the original matrix can be broken down into $r$ number of rank 1 matrices. But I don't understand and see how this is possible.
Say for a matrix of this form:
$$
A=\begin{bmatrix}
1 & 3 & 2 & 6\\ 
3 & 0 & 1 & 4\\ 
2 & 1 & 1 & 4
\end{bmatrix}
$$
The $rank(A)= 3$.
So if the claim was right, then I can form back the same matrix $A$ with the combination of 3 of Rank 1 matrices. I tried to "emulate" that idea but I just don't totally get how I could do it.
Thanks for any help on this!
 A: 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.
A: You can easily write SVD http://en.wikipedia.org/wiki/Singular_value_decomposition in the form of what you need.
