Suppose that $A$ is a $m \times n$ matrix then show that row rank $(A) =$ column rank $(A)$.
My attempt $:$
Suppose $R$ is the row reduced echelon form of $A$. Then $A$ is row equivalent to $R$ i.e. $R$ can be obtained by applying finite number of row operations on $A$ and the effect of each row operation can be viewed as pre-multiplying $A$ by a suitable elementary matrix. Suppose $R$ is obtained from $A$ by applying $n$ simultaneous row operations and suppose that the corresponding elementary matrices are respectively $E_1,E_2, \cdots , E_n$.Clearly each $E_i$ is a $m \times m$ matrix. Then $R$ can be wriiten as $R =PA$ where $P=E_1E_2 \cdots E_n$ is a $m \times m$ matrix. Since each $E_i$ is invertible so is $P$. Now $R^t=A^tP^t= A^tQ$ where $Q = P^t$. Suppose $Q =(q^1,q^2, \cdots, q^m)$ and $R^t=(r^1,r^2,\cdots,r^m)$. Then by the above equation we get $r^i=A^tq^i$ for $1 \le i \le m$. i.e. each $r_i$ is in the column space of $A^t$ i.e. in the row space of $A$. So column space of $R^t$ is a subspace of row space of $A$ i.e. row space of $R$ is in the row space of $A$. Again since $P$ is invertible $A=RP^{-1}=RS$ where $S=P^{-1}$. From this equation we similarly have row space of $A$ is a subspace of row space of $R$. Therefore we have row space of $A$ is same as the row space of $R$. Now if $R$ has $r$ non zero rows then row rank of $R$ is $r$ and cosequently row rank of $A$ is also $r$.
Now if I can show that column rank of $A$ is also $r$ then we are done. But I find difficulty to prove this part. Please help me.
Thank you in advance.
