How to prove that $m \times n$ matrix $A$ can be represented by multiplication of pivot columns of $A$ and first $r$ rows of reduced row echelon form $A$ is any $m \times n$ matrix.
$R$ is the reduced-echelon form matrix of $A$.
$B$ is a matrix consisting of pivot columns of $A$.
$C$ is a matrix consisting of first $r$ rows of $R$ where $r$ is rank of $A$.
How do I prove that $A = BC$?
 A: First, we find whether there exist a matrix $X$ such that $BX=A$.
Let $r$ be the no. of pivot column in $R$(which is also the number of non-zero row in $R$ and no. of column of $B$)
Let $X_k,A_k$ be the kth column of matrix $X $and $A$ respectively
$BX=A$ is equivalent to $BX_k=A_k$ for $k=1,2,....n$
Consider the linear system of $ LS(B,A_k)$
Let $P$ be the sequence of row operations that convert $A$ into $R$.
As $ B$ is consists of the pivot columns of $A$,performing the sequence $P$ on $ B$ gives a matrix with all $r$ columns being pivot column ,ie. the first $r$ row be an identity matrix and remaining row be zero rows, and performing the sequence on $A_k$ gives $R_k$,which the last $m-r$ rows is zero rows(there are only r non-zero rows in $R$). Therefore the system is consistent and the solution of the linear system is unique and equal to a column of consists of first row of $R_k$.
Therefore such matrix $X$ exists equals to the matrix consists of the first $r$ rows of $R$,which is $C$.
This shows that $BC=A$.
