Suppose I know the rank $r$ of a matrix $A$ by reducing it to echelon form $U$ ( Rank is defined as the number of pivots).
Since the row space of $U$ and $A$ is the same, and that the dimension of row space of $U$ is $r$, I therefore know that row space of $A$ has dimension $r$ as well . Therefore Any basis for the row space of A will have $r$ vectors in it because the number of vectors in a basis is independent of the basis itself.
Now can I use this knowledge to find the number of linearly independent rows of $A$?
Attempt:
The independent rows of A will form a basis for it's row space. But every basis should have $r$ vectors, therefore there are $r$ independent rows in $A$.
Hence the rank = number of independent rows.
Is the attempt correct.