TITLE: Is the rank of a matrix the same of its transpose? If yes, how can I prove it?
QUESTION [64 upvotes]: I am auditing a Linear Algebra class, and today we were taught about the rank of a matrix. The definition was given from the row point of view: 

"The rank of a matrix A is the number
  of non-zero rows in the reduced
  row-echelon form of A".

The lecturer then explained that if the matrix $A$ has size $m 
\times n$, then $rank(A) \leq m$ and $rank(A) \leq n$. 
The way I had been taught about rank was that it was the smallest of 

the number of rows bringing new information
the number of columns bringing new information. 

I don't see how that would change if we transposed the matrix, so I said in the lecture:
"then the rank of a matrix is the same of its transpose, right?" 
And the lecturer said: 
"oh, not so fast! Hang on, I have to think about it". 
As the class has about 100 students and the lecturer was just substituting for the "normal" lecturer, he was probably a bit nervous, so he just went on with the lecture.
I have tested "my theory" with one matrix and it works, but even if I tried with 100 matrices and it worked, I wouldn't have proven that it always works because there might be a case where it doesn't.
So my question is first whether I am right, that is, whether the rank of a matrix is the same as the rank of its transpose, and second, if that is true, how can I prove it?
Thanks  :)

REPLY [23 votes]: There are several simple proofs of this result. Unfortunately, most textbooks use a rather complicated approach using row reduced echelon forms. Please see some elegant proofs in the Wikipedia page (contributed by myself):
http://en.wikipedia.org/wiki/Rank_%28linear_algebra%29
or the page on rank factorization:
http://en.wikipedia.org/wiki/Rank_factorization
Another of my favorites is the following:
Define $\operatorname{rank}(A)$ to mean the column rank of A: $\operatorname{col rank}(A) = \dim \{Ax: x \in
\mathbb{R}^n\}$. Let $A^{t}$ denote the transpose of A. First show that $A^{t}Ax = 0$
if and only if $Ax = 0$. This is standard linear algebra: one direction is
trivial, the other follows from:
$$A^{t}Ax=0 \implies x^{t}A^{t}Ax=0 \implies (Ax)^{t}(Ax) = 0 \implies Ax = 0$$
Therefore, the columns of $A^{t}A$ satisfy the same linear relationships
as the columns of $A$. It doesn't matter that they have different number
of rows. They have the same number of columns and they have the same
column rank. (This also follows from the rank+nullity theorem, if you
have proved that independently (i.e. without assuming row rank = column
rank)
Therefore, $\operatorname{col rank}(A) = \operatorname{col rank}(A^{t}A) \leq \operatorname{col rank}(A^{t})$. (This
last inequality follows because each column of $A^{t}A$ is a linear
combination of the columns of $A^{t}$. So, $\operatorname{col sp}(A^{t}A)$ is a subset
of $\operatorname{col sp}(A^{t})$.) Now simply apply the argument to $A^{t}$ to get the
reverse inequality, proving $\operatorname{col rank}(A) = \operatorname{col rank}(A^{t})$. Since $\operatorname{col rank}(A^{t})$ is the row rank of A, we are done.