There has been some confusion how to show that the maximum number of linearly independent rows is equal to the maximum number of linearly independent columns. Here are some hints.

Let $A$ have size $m \times n$. Let the maximal number of linearly independent rows be $r$ and the maximal number of linearly independent columns be $c$. We want to show that $r=c$.

• Argument 1
  Let’s collect a set of $r$ linearly independent rows into a $r \times n$ matrix $R$. Then the rows of $A$ are spanned by the rows of $R$. Therefore $A = BR$, where $B$ is an $m \times r$ matrix. Then we have $c \le r$ (why?) By a similar argument (do it!), we have $r \le c$. Hence $r=c$.

• Argument 2
  This is the hint I gave in class. We permute the rows and columns of $A$ to form the partitioned matrix
  \(\begin{pmatrix} C & D \\\\ E & F \end{pmatrix},\) where $C$ is $r \times c$, $D$ is $r \times (n-c)$, $E$ is $(m-r) \times c$, and $F$ is $(m-r) \times (n-c)$. Rows of the matrix $(C, D)$ are linearly independent; columns of the matrix $\begin{pmatrix} C \\ E \end{pmatrix}$ are linearly independent. Without loss of generality, we assume $r < c$. Now columns of the matrix $C$ have length $r$, which is strictly less than $c$, therefore they must be linearly dependent. That is there exists a non-zero vector $c$ such that $C c = 0$. Since the rows of $(E, F)$ are linearly dependent on the rows in $(C, D)$, there exists a matrix $T$ such that $E = TC$. From this we can deduce that $\begin{pmatrix} C \\ E \end{pmatrix} c = 0$, contradicting with the fact that columns of $\begin{pmatrix} C \\ E \end{pmatrix}$ are linearly independent. Therefore $r=c$.