# Short proof that rows-rank=column-rank?

• Jun 25th 2009, 05:09 PM
arbolis
Short proof that rows-rank=column-rank?
Assuming that if $\displaystyle A$ is an $\displaystyle m\times n$ matrix whose rank is $\displaystyle k$, then $\displaystyle A^{T}$ (the transpose of $\displaystyle A$) is an $\displaystyle n\times m$ matrix whose rank is also $\displaystyle k$.
From this I conclude that if the rank of the rows of $\displaystyle A$ is not equal to the rank of the column of $\displaystyle A$ then what I first assumed is not true, thus it's a contradiction.

If what I wrote is right then this proof is quite simpler than the one my professor gave us! And also way simpler than the one given in the Hoffman's book.
Hope to hear your comments soon. (I'm not confident in myself enough not to post this here...)
• Jun 25th 2009, 07:32 PM
Bruno J.
The two statements are logically equivalent. If $\displaystyle M$ is the statement "the row-rank of $\displaystyle A$ is equal to the column-rank of $\displaystyle A$", and $\displaystyle N$ is the statement "the row-rank of $\displaystyle A$ is equal to the row-rank of $\displaystyle A^T$ then $\displaystyle M \Leftrightarrow N$ easily. You haven't proved much; only that, assuming $\displaystyle N$, $\displaystyle M$ is easy to prove.

Try proving $\displaystyle N$ now! (obviously without using $\displaystyle M$)
• Jun 25th 2009, 07:44 PM
arbolis
Quote:

Originally Posted by Bruno J.
The two statements are logically equivalent. If $\displaystyle M$ is the statement "the row-rank of $\displaystyle A$ is equal to the column-rank of $\displaystyle A$", and $\displaystyle N$ is the statement "the row-rank of $\displaystyle A$ is equal to the row-rank of $\displaystyle A^T$ then $\displaystyle M \Leftrightarrow N$ easily. You haven't proved much; only that, assuming $\displaystyle N$, $\displaystyle M$ is easy to prove.

Try proving $\displaystyle N$ now! (obviously without using $\displaystyle M$)

Yeah I just realized this. I wasn't accurate when I said
Quote:
I should have said , etc.
Ok, I'll try to do the proof.

Edit : I have it proved for invertible matrices : If A is invertible, so is $\displaystyle A^{T}$ and so both have full rank and N is proved without using M.
• Jun 25th 2009, 10:37 PM
Bruno J.
Notice that the row-rank is not affected by multiplication on the left or on the right by an invertible matrix (prove this, it's not very hard).

But using both row and column operations we can reduce $\displaystyle A$ to a matrix

$\displaystyle D=\begin{pmatrix} {I_r}&{O}\\ {O}&{O} \end{pmatrix}$

where the O's are all of appropriate size. This is equivalent to writing $\displaystyle A=PDQ$. But $\displaystyle A^T = Q^TD^TP^T = Q^TDP^T$; but then $\displaystyle \mbox{row-rank }A^T = \mbox{row-rank }Q^TDP^T = \mbox{row-rank }D = \mbox{row-rank }A$.
• Jun 26th 2009, 07:37 AM
arbolis
Thanks a lot/bunch!!!! I like to see/learn different proofs of the same theorem.
I'll print your post and show it to my friends so we might fully understand it together. By the way I think I already prefer your proof than the one in the book.
• Jun 26th 2009, 09:44 AM
Bruno J.
Quote:

Originally Posted by arbolis
Thanks a lot/bunch!!!! I like to see/learn different proofs of the same theorem.
I'll print your post and show it to my friends so we might fully understand it together. By the way I think I already prefer your proof than the one in the book.

Careful! There are flaws in my proof, I realized that as soon as I got into bed last night (finished writing at 3am... girlfriend wasn't happy when I got in the bed (Giggle)). See if you can find the flaws!
I'll think about it a bit more.
• Jun 26th 2009, 10:02 AM
Bruno J.
Edited my proof above!