For a nonzero matrix in row echelon form, is it always true that the non zero rows are always linearly independent?

For a nonzero matrix in row echelon form, is it always true that the non zero rows are always linearly independent?

I thought for a long time but still couldnt come out with an answer. Im just too weak in my concepts. Can anyone help clarfy?

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Yes, it is true. For a proof consider a linear combination of the rows and consider what happens to the combination for each coordinate. – Shahab Oct 18 '12 at 12:53
Ok. Is it when i change any generic matrix in ref form to column form, then let the right argument be 0 vector, i will always have a trivial solution because there will always be pivot columns? – Yellow Skies Oct 18 '12 at 12:57
By the way, these rows form a basis for the row space for the original matrix. If you are interested in a basis for the column space, it consists of the columns of the original matrix corresponding to the pivot containing columns of the row echelon form matrix. – Shahab Oct 18 '12 at 15:23

Yes. $$\begin{pmatrix} 1 & M_{12} & M_{13} & \cdots & M_{1m} & \cdots & M_{1n}\\ 0 & 1 & M_{23} & \cdots & M_{2m}& \cdots & M_{2n}\\ \vdots & & \ddots & & & & \vdots\\ 0 & 0 & 0 & \cdots & 1 & \cdots & M_{mn}\\ \end{pmatrix}$$
Write $v_1$ for the first row, $v_2$ for the second row, etc. Suppose $$a_1v_1 + \cdots + a_nv_n = 0.$$ We have to prove that each $a_i = 0$. Well, the expression is just $$\begin{pmatrix} a_1 & M_{12}a_1 + a_2 & M_{13}a_1 + M_{23}a_2 + a_3 & \ldots \end{pmatrix}$$ the $i$th entry in this vector looks like $$a_i + \text{stuff that is 0 if we know that } a_j \text{ is } 0 \text{ for } j < i$$ Working left to right, we see that $a_1$ must be 0; therefore $a_2$ must be zero, therefore $\ldots$