# 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!

-
What do you mean "created" - "added", "mulitiplied", some other operation? – Mark Bennet Jul 4 '11 at 18:19
oh.. I meant like say for a matrix of rank 3, it can be formed from a combination of 3 of rank 1 matrices. – xenon Jul 4 '11 at 18:28
BUt are you combining the matrices by adding them, multiplying them or doing something different? – Mark Bennet Jul 4 '11 at 18:51
If you are adding: Take the matrix you get by taking the first row of A and making all the other entries zero, add it to the matrix you get by taking the second row of A and making the other entries zero, and add to the matrix you get from the third row of A with other entries zero. Each of the three matrices trivially has rank 1, and the sum is the matrix you first thought of. – Mark Bennet Jul 4 '11 at 18:52
But this would not be the combination of rank 1 matrices, would it? Actually, I am not sure if the operation between the matrices would be adding or multiplying. I only know that the original matrix can be broken down into $r$ number of rank $r$ matrices and I thought this could be some kind of definition or something of that sort, which I am not aware of. – xenon Jul 4 '11 at 19:20

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.

-
I don't understand why the first row of the text is that. I wrote it between "and" and "." – Thomas Santoli Jul 4 '11 at 18:41
Done, now it's ok. – Thomas Santoli Jul 4 '11 at 18:43
Give me a few mins. I am trying out your answer and will get back to you in a bit. Thanks! – xenon Jul 4 '11 at 18:55
I don't understand why I followed your steps but I still don't get it. Could you do an example based on the matrix I posted in my question? thanks! – xenon Jul 4 '11 at 19:13
thank you so much for your example! :) – xenon Jul 4 '11 at 20:50

You can easily write SVD http://en.wikipedia.org/wiki/Singular_value_decomposition in the form of what you need.

-