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To find the rank of the matrix why it is enough to find the row echelon form only? why the REF gives us the maximum number of linearly independent row vectors in the matrix? could anyone explain this for me please?

I understood from this question that we can use both RREF and REF: Calculating the rank of a matrix , reduced row echelon or row echelon?

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The key point is that two vectors like

  • $v_1=(a_1,b_1,c_1,\cdots)$
  • $v_2=(0,b_2,c_2,\cdots)$

can't be linearly dependent for $a_1\neq 0$ because we can't never obtain the zero vector by linear combinations.

Therefore in the RREF we can show that row vectors are lineraly independent.

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  • $\begingroup$ My question is why it is enough to calculate the REF and not the RREF? I see that your are saying I should use RREF. $\endgroup$ – Mathstupid Oct 2 '19 at 19:15
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    $\begingroup$ @Smart REF suffices for the motivation I've given here. The scaling to obtain 1for the pivot doesn't change the conditions for independence. $\endgroup$ – user Oct 2 '19 at 19:22
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We know that elementary row operations doesn't change the rank of a matrix.

Also, we know that row echelon form / reduced row echelon form can be obtained by finite number of elementary row operations, hence REF and RREF preserve the rank of the original matrix.

Hence, by obtaining the REF or RREF, we can tell the rank of the original matrix, which is equal to the number of non-zero rows.

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  • $\begingroup$ My question is why it is enough to calculate the REF and not the RREF? I see that your are saying I should use RREF. $\endgroup$ – Mathstupid Oct 2 '19 at 19:15
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    $\begingroup$ note that number of non-zero rows for both REF and RREF is the same. And for the purpose of computing rank, it suffices to stop at REF, since the rank is jsut equal to the number of non-zero rows. $\endgroup$ – Siong Thye Goh Oct 3 '19 at 0:56

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