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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

2 Answers 2

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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 $\ldots$

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First, the restriction to non-zero matrices isn't necessary, since a zero matrix has no non-zero rows and the empty set is linearly independent, since the only linear combination forming the zero vector is the one in which all coefficients are zero (since there aren't any coefficients).

The non-zero rows are linearly independent since you can go through them from the top and successively conclude that the entry where they are non-zero whereas all rows below them are zero forces their coefficient in a linear combination yielding a zero row to be zero.

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