# Basis for row space of A

Assume that A is not reduced form, and R is the REF of A. I have understand that the set of nonzero rows in R is the basis for the row space of A. My question is why can't the corresponding rows of A be the basis for the row space of A? Any explanations?

e.g.:

A = $\begin{pmatrix} 1 & 3 & 2 \\ 2 & 7 & 4 \\ 1 & 5 & 2 \end{pmatrix}\to \begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$ = R

basis for the row space of $A = \{ (1,0,2), (0,1,0) \}$

why not $\{ (1,3,2), (2,7,4) \}$?

Actually, $\{ \underbrace{(1,3,2)}_{a_1}, \, \underbrace{(2,7,4)}_{a_2}\}$ can be a basis for the row space. Assume that your basis of the row space is the set $\{ \underbrace{(1,0,2)}_{e_1},\, \underbrace{(0,1,0)}_{e_2}\}$.
Take any $\vec x \in$ rowspace$(A)$. Then $\vec x = k\cdot \vec e_1 + \lambda\cdot \vec e_2.$ But $\vec e_1 = 7 \vec a_1 -3\vec a_2$ and $\vec e_2 = -2\vec a_1 + \vec a_2$. Thus, $$\vec x = k\cdot (7\vec a_1 -3\vec a_2) + \lambda \cdot (-2\vec a_1 + \vec a_2)=(7k+\lambda)\vec a_1 + (-3k+\lambda) \vec a_2,$$ which means that the set $\{\vec a_1, \vec a_2\}$ spans the rowspace of $A$. Also, $\vec a_1, \vec a_2$ are linearly independent, hence $\{\vec a_1, \vec a_2\}$ can form a basis of the rowspace of $A$.
Actually, any $2$ linearly independent vectors in the rowspace of $A$ can form a basis.
It’s not that the corresponding rows of $A$ can’t be used as a basis. Since $R=SA$, where $S$ is a nonsingular square matrix, they are linear combinations of the non-zero rows of $R$ and are linearly independent. It’s more a matter of convenience. The basis vectors from $R$ will look almost like the standard basis vectors—lots of zeros and a single $1$ until you get to the last few coordinates, so it’s usually much easier to work with them than with the rows of $A$.