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Consider reduced row echelon matrix $$ R= \begin{pmatrix} r_1 \\ \ldots \\ \ldots \\ r_n \\ 0 \\ \ldots \end{pmatrix} $$ where $\{r_1, r_2, r_3, \ldots, r_n\}$ are the nonzero dependent row vectors of $R$. Then $r_k = c_1r_1 + c_2r_2 + \ldots + c_{k-1}r_{k-1}$ where $1 \le k \le n$ and so $r_k - (c_1r_1 + c_2r_2 + \ldots + c_{k-1}r_{k-1}) = 0$ which means there occurs a zero vector before $r_n$.

The quote above is a part of a proof. I am stuck at the bolded part. Can someone elaborate how the conclusion in bold follows?

Edit:

Statement: if $R$ is a reduced row echelon matrix then the nonzero rows of $R$ are linearly independent.

Proof: The quote above is a contradiction ($k > n$) to the fact that $0$ vectors occurs after $r_n$.

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    Context missing. Normally, in a reduced row echelon matrix, the nonzero rows should be independent, and an $r_k = \cdots$ equation like in your case cannot hold. But hell knows what this particular author means by "reduced row echelon matrix". – 2017-01-03
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    @ darij grinberg, I edited my OP. – 2017-01-03
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    Then I don't see why the conclusion holds. The author seems to have goofed here; read any other text (or fill in the argument yourself: what can you say about the column of the pivot of row $r_i$, where $i$ is the smallest nonzero index satisfying $c_i \neq 0$ ?). – 2017-01-03
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    Just to be sure I am not the one who goofed here I link the proof as is : https://s29.postimg.org/7lgaun8s7/Capture.png If it still makes no sense I'll look elsewhere – 2017-01-03
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    Perhaps what we need to do is take a look at this particular author's defininition of reduced row echelon form. – 2017-01-03
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    Ah! The inclusion of row operations is actually quite interesting, and somewhat of a game-changer. I'm saying "somewhat" because I don't know how he defines a row-echelon form, and depending on the definition it's not clear whether it is required to be unsimplifiable by any row operations. – 2017-01-03

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