# Reducing a matrix of rank $r$ to a product of two matrices

In problem set 2.2, question 28 of the book Linear Algebra and its Applications by Gilbert Strang, we are asked to express a matrix $$A$$ of rank $$r$$ as a product of two matrices:

1. A matrix containing the pivot columns of $$A$$ and
2. A matrix containing the first $$r$$ rows of $$R$$.

For instance,

$$\begin{bmatrix}1 & 3 & 3 & 2\\2 & 6 & 9 & 7\\ -1 & -3 & 3 & 4\end{bmatrix} = \begin{bmatrix}1 & 3\\2 & 9\\ -1 & 3 \end{bmatrix} \begin{bmatrix}1 & 3 & 0 & -1\\0 & 0 & 1 & 1\end{bmatrix}$$

Why is this possible and what is the significance of such a representation?

• $R$ being the reduced row echelon form of $A$? – AnyAD Dec 7 '18 at 12:12
• @AnyAD Yes, it is the row reduced echelon form. – Mohideen Imran Khan Dec 7 '18 at 12:24

The columns that have pivots are the linearly independent columns of $$A$$. So all other columns of $$A$$ are linear combinations of these. The matrix with non-zero rows from the reduced form contain the relevant coefficients of linear dependence (for each column that can be expressed as a linear combination of the linearly independent/pivot set). The pivot columns in the matrix when premultiplied by the matrix of pivot columns simply extract the pivot columns.