## The row rank of a reduced row echelon form matrix is the number of nonzero rows

Let $A$ be a matrix over a field $F$ in reduced row echelon form. Prove that the row rank of $A$ is the number of nonzero rows of $A$.

Recall that the row rank of a matrix is the maximal number of linearly independent rows in it. We proceed by induction on the number of rows of $A$. (Recall our recursive description of reduced row echelon form.) In fact, we can prove this more generally for matrices in row echelon form, and we will prove that the nonzero rows are a maximal linearly independent subset.

For the base case, Let $A$ be a $1 \times m$ matrix. If $A = 0$, then the set of rows of $A$ is simply $\{ 0 \}$. There are no linearly independent subsets among the rows of $A$, so that the row rank is 0. Likewise, the number of nonzero rows is 0. Suppose now that $A \neq 0$. Then certainly there is one nonzero row of $A$, which forms a maximal linearly independent subset.

For the inductive step, suppose that the result holds for every row echelon form matrix having $n$ rows. Let $A = \left[ \begin{array}{c|c|c} 0 & 1 & V \\ \hline 0 & 0 & A_2 \end{array} \right]$ be a row echelon form matrix having $n+1$ rows, where $A_2$ is also in row echelon form. By the induction hypothesis, the row rank of $A_2$ is the number of nonzero rows. Moreover, the rows of $A_2$ are a maximal linearly independent subset of the rows of $A_2$. Clearly these remain linearly independent as rows of $A$. Moreover, we can see that if $\sum \alpha_i R_i = 0$, where $R_i$ are the nonzero rows of $A$, then we must have $\alpha_1 = 0$. Since the nonzero rows of $A_2$ are linearly independent (as rows of $A$) we have $\alpha_i = 0$ for all $i$. Thus the set of nonzero rows of $A$ is a linearly independent set. Certainly this subset is maximal, since every other row is 0. Thus the row rank of $A$ is the number of nonzero rows.