Talk:Rank (linear algebra)

From Wikipedia, the free encyclopedia

You have new messages (last change).

I was rather surprised by the statement in the article that the rank "is usually denoted rk(A)". So I checked some books, with the following result:

  • Most books actually do not introduce a notation for the rank of a matrix.
  • Five books use rank A, namely Linear Algebra and Geometry by Bloom, Topics in Matrix Analysis by Horn and Johnson, Linear Algebra by Friedberg et al., Linear Algebra by Satiste, and Berkeley Problems in Mathematics by De Souza and Silv.
  • Three books use rk(A), namely Elements of Linear Algebra by Cohn, Linear Algebra by Jänich, and Linear Algebra by Kaye and Wilson.
  • Two books use r(A), namely Linear Programming by Hartley, and Linear Algebra with Applications by Scheich.

I changed the article accordingly. -- Jitse Niesen 23:44, 21 Aug 2003 (UTC)

Contents

[edit] Rank of the product of two matrices

The Rank (linear algebra) page states:

  • If B is an n-by-k matrix with rank n, then AB has the same rank as A.
  • If C is an l-by-m matrix with rank m, then CA has the same rank as A.

Does anyone have a proof (or reference to a proof) for this? Maybe it's obvious and I'm just not seeing it. Connelly 15:49, 7 September 2005 (UTC)

It's not that obvious. Sketch of the proof: Think of the matrices as linear transformations. If B is an n-by-k matrix with rank n, then the function x |-> Bx is surjective, hence the range of the function x |-> ABx is the same as the range of the function x |-> Ax, hence the ranks are equal. I'll see whether I can find a reference (rectangular matrices always confuse me). Let me know if you want me to elaborate. PS: Thanks for your edit to Hermitian matrix. -- Jitse Niesen (talk) 16:25, 7 September 2005 (UTC)
It follows from (0.4.5c) in Horn & Johnson, Matrix Analysis, which states (without proof): If A is m-by-n and B is n-by-k then
\operatorname{rank} \, A + \operatorname{rank} \, B - n \le \operatorname{rank} \, AB \le \min \{ \operatorname{rank} \, A, \operatorname{rank} \, B \}.
If rank B = n, then this becomes rank A ≤ rank AB ≤ rank A. -- Jitse Niesen (talk) 19:49, 7 September 2005 (UTC)
Wow, thanks! I didn't expect a response so soon. Your proof works for me, but I'll check out the Matrix Analysis book too. I'm actually trying to show a more complex result, but I needed to check the validity of the Wikipedia statement first. I can post up your linear transformation proof on Wikipedia if you think that's a good idea (not really sure where to put it...maybe the Rank page or as a separate page linked to from Rank?). - Connelly 23:35, 7 September 2005 (UTC)
I think the proof would make a nice addition if it's kept short, because it explains the concept of rank and how to handle it. It's probably more important to mention the double inequality for rank AB (by the way, how hard would it be to prove that?). Generally, proofs on Wikipedia are a contentious issue and need to be considered on a case-by-case basis (how important is the proof and how much does it disrupt the flow of the article?). You can read a discussion about it on Wikipedia:WikiProject Mathematics/Proofs, which also has a proposal for putting proofs on a separate page. -- Jitse Niesen (talk) 12:16, 8 September 2005 (UTC)

[edit] Matrix rank definition with minor

Another definition of matrix rank:

The matrix A has rank r if it has a minor of size r which is different from zero and every minor of size r + 1 is equal to zero.

[edit] Ring question

The article says There are different generalisations of the concept of rank to matrices over arbitrary rings. In those generalisations, column rank, row rank, dimension of column space and dimension of row space of a matrix may be different from the others or may not exist. It doesn't distinguish between rings and commutative rings. Is it true that the generalisation to just commutative rings also has all of these issues? (it seems likely, and if it is true I think it would be useful to mention it since it would make the statement much stronger) A5 18:28, 19 March 2006 (UTC)

[edit] How about "rank deficient"

I think the page should mention the term "rank deficient" .

MusicScience 23:32, 12 January 2007 (UTC)

Term sounds familiar, and has the benefit of being self-explanatory. However, a quick Google search for 'intitle:"matrix algebra" "rank deficient"' finds only 7 distinct websites. Sources? References? Textbooks? -- JEBrown87544 18:19, 17 January 2007 (UTC)
Retrieved from "http://en.wikipedia.org../../../r/a/n/Talk%7ERank_%28linear_algebra%29_6254.html"