Show that the non zero rows of an echelon form matrix form a linearly independent set.

That’s what it is?

(1) (0) (0) = (0)

c1(0) + c2 (1) +c3 (0) = (0) so c1=0, c2=0, c3=0

(0) (0) (1) = (0)

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- Nov 21st 2009, 01:49 PMewanderLinear Algebra question
Show that the non zero rows of an echelon form matrix form a linearly independent set.

That’s what it is?

(1) (0) (0) = (0)

c1(0) + c2 (1) +c3 (0) = (0) so c1=0, c2=0, c3=0

(0) (0) (1) = (0)

- Nov 21st 2009, 01:53 PMewander
again

1 0 0 0

c1 0 + c2 1 + c3 0 = 0 so c1 = 0, c2=0 and c3=o----->independent

0 0 1 0 - Nov 22nd 2009, 04:08 AMHallsofIvy
I think you are trying to write

$\displaystyle c_1\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}+ c_2\begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}+ c_3\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}= 0$

and derive from that that those vectors are independent. But that is the identity matrix, not a general "row echelon matrix". A three by three row echelon matrix would be of the form:

$\displaystyle \begin{bmatrix}a & b & c \\ 0 & d & e \\ 0 & 0 & f\end{bmatrix}$

so you should be looking at

$\displaystyle c_1\begin{bmatrix} a \\ 0 \\ 0\end{bmatrix}+ c_2\begin{bmatrix}b \\ d \\ 0\end{bmatrix}+ c_3\begin{bmatrix}c \\ e \\ f\end{bmatrix}$.

Also, I see nothing in your question that restricts this to 3 by 3 matrices. You should be able to do a general proof.