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Can I define the rank of a matrix(A) as the number of non zero rows in RREF(A)? Here's my reason: Let number of zero rows be $x$
Then these rows are the linearly dependent rows of A and $x=dim(left null space)=m-r$.
So number of non zero rows is equal to $rows-x=m-(m-r)=r$.

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    $\begingroup$ Yes, what you state is a well known result. $\endgroup$ Sep 23 '17 at 17:03
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Yes, this also follows immediately from the fact that the Gauss-algorithm leaves the rank of a matrix unchanged. Since the rank $\operatorname{RREF}(A)$ is the number of its non-zero rows the claim follows.

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