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Prove that if $R$ is a matrix in echelon form , then a basis for $\mathcal{R}(R)$ consists of the non zero rows of $R$

My attempt : For basis we must show the span and linearly independent properties

let $r_1,r_2 ,..., r_k$ be the non zero rows of $R$ starting from the $1$st row to the $k-$th row

now we will prove that $\mathcal{R}(A)= \operatorname{span} \{ r_1,r_2,....., r_k\}$

for any vector $v \in \mathcal{R}(A)$, we can write

$ v= c_1r_1 +....+c_kr_k +c_{k+1}0 +......+c_m0= c_1r_1 +....+c_kr_k$

so $v \in \operatorname{span}\{r_1,r_2,........,r_k\}$

Therefore $\mathcal{R}(R) \le\operatorname{span} \{r_1,....,r_k\}$

now trivially we also have $\operatorname{span}\{r_1,....,r_k\} \le \operatorname{span}\{r_1,r_2,...,r_k,0,0,...,0\} =\mathcal{R}(R)$

This implies that $\operatorname{span}\{r_1,....,r_k\}=\mathcal{R}(R)$

My doubts : How to show that $r_1, r_2,.., r_k$ are linearly independent

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linear independence By echelon form, you have ordered the rows by the number of zeros on the left. This number of zeros is strictly increasing. Take they order from the maximum number of zeros to the minimum (so in your row-echelon matrix you go from the lowest one up to the first row).

Let say you have $r_1,\dots, r_m$ non-zero rows. Now $r_1\neq 0$ so it is lineary independ. Now you add $r_2$, this can't by scalar multiple of $r_2$ because it have non-zero on the position where the $r_1$ has zero.

Now you have proved that $r_1,\dots r_k$ are independ. $r_{k+1}$ has non-zero on the position where all $r_1,\dots r_k$ have zero so it cant be spanned by them.

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