Proof of rank of an Unimodular Matrix

[Nemanjavuk69]

Hello everyone

Just as the title says, is there a proof, theorem or doctrine which talks about the rank of an "Unimodular Matrix"?

I would very much appreciate it if someone could provide me with one. Bonus point for the person if you can tell me where it is cited from so I can read up on it.


Have a wonderful day/ night forward!

 

[blamocur]

Do you know the definition of unimodular matrix? Have you looked it up on internet?

 

[Nemanjavuk69]

Do you know the definition of unimodular matrix? Have you looked it up on internet?

Dear @blamocur

Yes. As far as I know a "Unimodular Matrix" is a square matrix which has a determinant of [imath]\pm1[/imath]. I also know, that for a given matrix if the determinant [imath]\neq 0[/imath] and there exists an invertible for the square matrix than the rank has to be full. However, I am looking for a theory, proof or doctrine for this. Are you perhaps in possesion of such one?

 

[blamocur]

I also know, that for a given matrix if the determinant ≠0\neq 0=0 and there exists an invertible for the square matrix than the rank has to be full.

Do you mean that you know this but don't how to prove this? I'd expect most textbook to have such proofs.
Here is a simple proof by contradiction that comes to mind: if the rank is less than full then the matrix's columns (as well as rows) are linearly dependent, which means that the determinant is 0.

 

[Nemanjavuk69]

Do you mean that you know this but don't how to prove this? I'd expect most textbook to have such proofs.
Here is a simple proof by contradiction that comes to mind: if the rank is less than full then the matrix's columns (as well as rows) are linearly dependent, which means that the determinant is 0.

Exactly, I don't have a proof in my textbook, so I want to know the proof. I am not expected to know the proof in my class, however, out of curiosity and because I want to know why the rank is full or not, I want to read a proof, theorem or doctrine if such one exists. Hope it is not too much of a burden to ask you for such.

 

[stapel]

Here is a simple proof by contradiction that comes to mind: if the rank is less than full then the matrix's columns (as well as rows) are linearly dependent, which means that the determinant is 0.

...I want to know the proof.... Hope it is not too much of a burden to ask you for such.

Um... you've already been *given* a proof...?

 

[Steven G]

I'll 2nd that you have been given a proof.

 

[topsquark]

If jonah2.0 were here he'd be about 120 proof!

-Dan