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Let $Ax=b$ be a system of $r$ nonzero equations in $n$ unknowns.
Suppose that $rank(A)=rank(A|b)$ and that $(A|b)$ is in RRE.
Then $rank(A)=r$.

We want to prove that.
The book says since $(A|b)$ is in reduced row echelon form, $(A|b)$ must have $r$ nonzero rows.

I can't follow the logic.
RRE form can have zero rows in the last position, then why it must have $r$ nonzero rows in here?

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The rows of $(A\mid b)$ just reflect the equations, and it was given that $r$ of them were nonzero. –  Marc van Leeuwen Apr 9 '13 at 7:59
But if we transform that matrix into reduced row echelon form then it can have zero rows, I think. –  noname Apr 9 '13 at 8:01
Your question is unclear. Just after stating the equations you say "Since $(A\mid b)$ is in RRE", so one must assume that somehow they were already so. If you have changed the equations, then you should not call them by an unchanged name. –  Marc van Leeuwen Apr 9 '13 at 8:03
I stated the whole sentence. Please check that. –  noname Apr 9 '13 at 8:11

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