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I wanted to check if I understand this correctly, or maybe it can be explained in a simpler way: why is matrix rank equal to the number of linearly independent rows?

The simplest proof I can come up with is: matrix rank is the number of vectors of the basis of vector space spanned by matrix rows (row space). All bases of a given vector space have the same size.

Elementary operations on the matrix don't change its row space, and therefore its rank.

Then we can reduce it to row echelon form (reduced row echelon form is not necessary, because I think the non-zero rows in row echelon form are linearly independent already). So we might pick only the rows that are non-zero and still get the same row space (adding or removing arbitrary number of zero rows don't change a thing), and because these rows are linearly independent, they are basis for the row space. As mentioned above, all bases have the same size, so number of linearly independent vectors is equal to matrix rank (the dimension - size of basis - of row space).

Is it correct? Didn't I make it too complicated?

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Two facts about elementary row operations are useful to resolve this question:

  • Elementary row operations alter the column space but do not alter the linear dependences among the columns. For example, if column 10 is $4$ times column 3 minus $7$ times column 5, then after doing any elementary row operation, column 10 is still be $4$ times column 3 minus $7$ times column 5. It's not too hard to figure out why that's true. Therefore elementary row operations do not alter the number of linearly independent columns.
  • Elementary row operations do not alter the row space. It's also not hard to see why that is true. Therefore elementary row operations do not alter the number of linearly independent rows.

After a matrix is fully reduced, it's not hard to see that the number of linearly independent columns is the number of pivot elements, and the number of linearly independent rows is the number of pivot elements. Therefore the number of linearly independent rows. Since the row operations don't change the number of linearly independent columns or the number of linearly independent rows, those two quantities must be the same in every matrix.

Consequently one can define “rank” of a matrix either

  • as the dimension of the column space; or as
  • as the dimension of the row space,

and it's the same thing either way.

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  • $\begingroup$ I understand why row operations don't change row space, but why does it imply that the number of linearly independent rows doesn't change? Is it because: all the rows span the row space, so we can just pick the linearly inependent rows and they will still span the same space. Therefore these vectors are basis. So we can do the same thing for any other matrix obtained by row operations, and the row space will stay the same. Using the fact that all bases have the same number of elements we see that the number of linearly independent rows doesn't change under row operations. Correct? $\endgroup$ – user5539357 Feb 13 '16 at 18:02
  • $\begingroup$ Yes. The fact that all bases have the same number of elements is the essential point in that part of the argument. $\qquad$ $\endgroup$ – Michael Hardy Feb 13 '16 at 18:55
  • $\begingroup$ @MichaelHardy Michael, Can you check my proof here based on what you said.(math.stackexchange.com/questions/2295243/…) $\endgroup$ – Bijesh K.S May 25 '17 at 8:58

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