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How do I prove that the rank of a matrix in reduced row echelon form is equal to the number of non-zero rows it has?

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  • $\begingroup$ StackExchange is not intended to do your homework for you. At the very minimum, post what you've already tried. $\endgroup$
    – Ephraim
    Nov 29 '15 at 18:14
  • $\begingroup$ I've edited my question to be significantly more general. $\endgroup$
    – George
    Nov 29 '15 at 18:41
  • $\begingroup$ The row rank of a matrix is defined as the dimension of the row space. A row reduced matrix is one in which the set of nonzero rows are linearly independent. So the only thing you need to prove is that each of the three elementary row operations (the operations that you use to get from a given matrix to its RREF) preserve the row space of a matrix. $\endgroup$ Nov 29 '15 at 18:44
  • $\begingroup$ Are you looking for a formal proof, or an intuitive explanation? $\endgroup$
    – Ephraim
    Nov 29 '15 at 18:44
  • $\begingroup$ Either would help. $\endgroup$
    – George
    Nov 29 '15 at 18:47
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By definition of row echelon form, all the non-zero rows are linearly independent. (Think on all the non-zero leading coefficients which are "aligned" ).

So the non-zero rows form a basis for the row space (i.e the subspace spanned by all the rows). Therefore their number equals to the dimension of this space, as required.

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