# How to show that $r_1, r_2,.., r_k$ are linearly independent

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

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.