row rank= column rank, alternative proof I am studying the theorem that states that the row rank of a matrix is the same as the column rank.
I understood the proof and managed to use it in specific examples using a matrix. 
I am now, trying to find an informal proof/explanation about the fact that row rank=columns rank but I am not sure what is suppose to do by given an informal proof or explanation.
Can anyone help me on this or show me where can I find more about this topic? 
Thanks  
 A: Here's my preferred proof:
First, note that $A$ and $A^TA$ have the same rank, since these matrices have the same nullspace.  In particular: clearly, $Ax = 0 \implies A^TAx = 0$.  On the other hand,
$$
A^TAx = 0 \implies x^TA^TAx = 0 \implies \|Ax\|^2 = 0 \implies Ax = 0
$$
Then, note that for any product of matrices, $rk(PQ)  \leq rk(P)$.  Thus, we have $$
rk(A) = rk(A^TA) \leq rk(A^T)
$$
on the other hand, applying the same argument to $A^T$ yields
$$
rk(A^T) = rk(AA^T) \leq rk(A)
$$
So, we can conclude that $rk(A) = rk(A^T)$, as desired.

Another approach:
Begin by showing that the (column-)rank of and $m \times n$ $A$ is the smallest $r$ such that there exist matrices $P \in \Bbb R^{m \times r}$ and $Q \in \Bbb R^{r \times n}$ such that $A = PQ$; we can call $A = PQ$ a rank-$r$ factorization of $A$.  
Now, the row-rank of $A$ is the column rank of $A^T$, which can be defined similarly.  However, whenever $A = PQ$, we have
$$
A^T = (PQ)^T = Q^TP^T
$$
which is to say that if $A$ has a rank $r$ factorization, then $A^T$ also has a rank $r$ factorization.  Conclude that $rank(A) \leq rank(A^T)$.  Since we also have $rk(A^T) \leq rk(A^{TT}) = rk(A)$, we can conclude that the ranks are equal.
A: Here is a simple conceptual proof.
1) Row operations do not change the row rank. 
I think you already believe this so I offer no proof.
Simpler in fact is:
2) Column operations do not change the row rank.
This is because if you think of the rows as vectors the column operations are just changing the representation of these vectors to another basis, but the linear relations stay the same. 
Similarly 
3) Column operations do not change the column rank.
4) Row  operations do not change the column rank.
So both row and column rank are unchanged by the row and column operations.
Using both row and column operations any matrix can be reduced to 
$$\begin{pmatrix}
1&0&0&0&0\\
0&1&0&0&0\\
0&0&1&0&0\\
0&0&0&0&0\\
0&0&0&0&0\\
\end{pmatrix}$$
Clearly the row and column ranks  of such a matrix are equal.
A: The isomorphism of the rowspace $\operatorname{im}{A^T}$ and the columnspace $\operatorname{im}{A}$ of a matrix $A$ is a consequence of the isomorphisms among the four fundamental subspaces associated to a matrix (rowspace, columnspace, (right) nullspace, and left nullspace). There are two paths through the isomorphisms among these spaces which can get you the result.
Taking the first route through the codomain, we can argue that the rowspace of an $m\times n$ matrix $A\colon k^n\to k^m$ is span of the row vectors, i.e. $\operatorname{im}{A^T}$. By the first isomorphism theorem that is isomorphic to $(\ker A^T)^\perp$. So if the rowspace has dimension $r$, then so does $(\ker A^T)^\perp$ and so $\dim\ker A^T=m-r.$ But $\ker A^T \cong (\operatorname{im}{A})^\perp,$ and so $\dim\text{colspace} = \dim(\operatorname{im}(A)) = m-(m-r) = r$.
Alternatively, focusing on the dimension of the domain, if the rowspace $\operatorname{im}{A^T}$ has dimension $r$, then its orthogonal complement $(\operatorname{im}{A^T})^\perp$ has dimension $n-r$. But this is $\ker A$, and by the first isomorphism theorem $n-\dim\ker A=r.$ Hence $\dim(\text{rowspace})=n-(n-r).$
Both routes use the first isomorphism theorem for linear maps $\operatorname{dom}(f)/\ker(f)\cong\operatorname{im}(f)$ and the formal fact that $\ker(f^*)\cong(\operatorname{cod}(f)/\operatorname{im}(f))^*\cong\operatorname{im}(f)^o.$ With an inner product, you have isomorphisms between spaces and their duals, dual maps and transposes, and quotients and orthogonal complements, so these two isomorphisms become $\ker(f)^\perp\cong\operatorname{im}(f)$ and $\ker(f^T)\cong\operatorname{im}(f)^\perp.$
