As a reference to this and this and similarly many sources, it has been sait that the rank of a matrix, say $A$, is the number of non-zero rows of the row-reduced form of $A$.
However, why is this the case ? How can we prove it ?
I mean in the second question that I have linked, the answerer says the non-zero row form a basis etc. which I think does not connect to the rank of matrix.
If you agree that row operations don't affect the rank of the matrix, then this follows from one of the definitions of the rank of a matrix, which is the dimension of the row space of the matrix.
The rows of that matrix span a vector space, and the dimension of that vector space is equal to the number of non-zero vectors that span it.
The dimensionality is clearly not greater than the number of rows. But its also not less: there are no redundant vectors, because every row contains an entry where it is the only non-zero element in its column, so it is impossible to build that row from a linear combination of the other rows.