Basis of column space for row echelon form I want to prove following statement:

The columns of R containing leading ones are a basis of col R(R is a row echelon form)

Let $cj_1, cj_2, ..., cj_r$ denote the columns of R containing leading $1s$. We must show $\alpha_1cj_1 + \alpha_2cj_2 + \dots + \alpha_rcj_r = 0$ leads to $\alpha_1 = \alpha_2 = \dots = \alpha_r = 0$ and also $col R =span \{ cj_1, cj_2, ..., cj_r\}$. I failed to get result in both of these statements.
 A: These columns of $R$ are all of the form $[\star~…~\star~1~0~…~0]^T$, where $\star$ is some possibly nonzero entry and the position of the separating “$1$” ranges from $1$ to $\operatorname{rnk} R$, that is they are of the form:
$$\begin{bmatrix}1 \\ 0 \\ 0\\ \vdots \\ 0 \\ \vdots\\ 0\end{bmatrix}, \begin{bmatrix}\star \\ 1 \\ 0\\ \vdots \\ 0 \\ \vdots\\ 0\end{bmatrix}…, \begin{bmatrix}\star \\ \vdots\\ \star\\ 1\\ 0\\ \vdots\\ 0 \end{bmatrix}.$$
Viewing it this way, it’s quite easy to see that the claim is true. To show that these columns are linearly independent and generate the span of all columns, you can perform a base change on these replace them by linear combinations. Start by substracting some multiple of the first column from the second, then continue to subtract a linear combination of the resulting first two columns from the third and so on, to get a system of columns
$$\begin{bmatrix}1 \\ 0 \\ 0\\ \vdots \\ 0 \\ \vdots\\ 0\end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 0\\ \vdots \\ 0 \\ \vdots\\ 0\end{bmatrix}…, \begin{bmatrix}0 \\ \vdots\\ 0\\ 1\\ 0\\ \vdots\\ 0 \end{bmatrix},$$
which still generates the same span and which is linearly independent if and only if the original system of columns is.
