Question

It is said that all matrices of any rank=\[r\] can be created from a \[r\] number of rank 1 matrices. But I still don't understand how I can do that? Say for a matrix like this: \[\begin{bmatrix}
1 & 3 & 2 & 6\\
3 & 0 & 1 & 4\\
2 & 1 & 1 & 4
\end{bmatrix}\]. I don't see how it is possible.

Answer 1

oh just to add a little more to my question, by "create", I meant the original matrix of rank r can be formed out of the combinations of 4 number of rank 1 matrices.

Answer 2

It has been a while since I've studied linear algebra, but it seems that you are being asked to write a basis for the null space (here R_n denotes the nth row). The following is an example of how to find this basis and hence determine the rank of the matrix, but it is not a general theorem (which I will leave up to you to find & confirm). You gave the following matrix:

\begin{bmatrix}1&3&2&6\\3&0&1&4\\2&1&1&4\end{bmatrix} \[-2R_1 + R_3 \rightarrow\]
\begin{bmatrix}1&3&2&6\\3&0&1&4\\0&-5&-3&-8\end{bmatrix} \[-3R_1 + R_2 \rightarrow\]
\begin{bmatrix}1&3&2&6\\0&-40&-25&-120\\0&0&2&-48\end{bmatrix}
\begin{bmatrix}1&3&2&6\\0&1&5/8&3\\0&0&1&-24\end{bmatrix}

Now, let \[x_4=t\] and according to the last row \[x_3 = 24t\] hence \[x_2=-18t\] and \[x_1=0\]

So:

\[x=\begin{bmatrix}0\\-18t\\24t\\t\end{bmatrix}=\begin{bmatrix}0\\-18\\24\\1\end{bmatrix}t\]

This matrix is rank 1. Also, try reading through the following if this does not help:
http://tutorial.math.lamar.edu/Classes/LinAlg/FundamentalSubspaces.aspx