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 🙂
Best Answer
The answer is yes. This statement often goes under the name "row rank equals column rank". Knowing that, it is easy to search the internet for proofs.
Also any reputable linear algebra text should prove this: it is indeed a rather important result.
Finally, since you said that you had only a substitute lecturer, I won't castigate him, but this would be a distressing lacuna of knowledge for someone who is a regular linear algebra lecturer.