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?
 A: 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 $\ldots$
A: 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.
A: In a row reduced echelon form, each row has a leading one and rest of the elements are zeros.
Hence, you can't create any row by a linear combination of other rows, because other rows contain zeros where this row contain a one. Hence this row is linearly independent to other rows. Follow this reasoning for all non zero rows.
Similar reasoning can be given for pivot columns as being independent.
