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 :)

^{T}, note that one decomposition is expressible in terms of the other, and then show that the two diagonal matrices resulting from the two decompositions have the same rank (and nullity too). – J. M. ain't a mathematician Aug 13 '10 at 01:11