Talk:Dimension theorem for vector spaces

message: Could someone write about how to prove the uniqueness of the dimension for infinite cardinals? - Gauge 06:02, 18 December 2005 (UTC)
replies: text: I know its been a year and a half since anyone replied, but done.  jbolden1517Talk 15:19, 12 May 2007 (UTC)

message: In the proof it is assumed that there is a relation between the cardinalities of the two bases (that one is greater than the other). It seems to me that this relies on the Well-Ordering Principle. However I'm not familiar with the Ultra-filter Lemma and its applications... is this where it is used? There is a proof here: http://www.dpmms.cam.ac.uk/~tf/cam_only/cleverblass.pdf that relies only on propositional compactness and Hall's Matching Theorem.
:You (who?) are right to the extent that trichotomy (any two cardinals are comparable) is equivalent to the axiom of choice, as stated in that article. So indeed the very start of the proof (assume one basis bigger than the other) is in blatant contradiction with the claim in the lead that the theorem only needs the ultrafilter lemma. Marc van Leeuwen (talk) 05:43, 20 March 2011 (UTC)
replies: text: Although this particular proof does use full AC, it is nevertheless true that the dimension theorem (also known as Löwig's theorem) follows from the ultrafilter lemma. I have checked the primary source on this kind of issues: "Consequences of the axiom of choice" by Howard & Rubin. You can also check it online at http://consequences.emich.edu/file-source/htdocs/conseq.htm .Godelian (talk) 05:09, 11 October 2011 (UTC)
replies: text: I've put up a somewhat rough proof at ProofWiki. That proof uses Hall's marriage theorem for infinite sets of finite sets, which I prove here from the finite form of the marriage theorem along with the Cowen-Engeler Lemma, an equivalent of the Ultrafilter Lemma. --Dfeuer (talk) 04:04, 24 May 2013 (UTC)

text: Please put a (more detailed) reference in the article as well. Thanks a lot. Marc van Leeuwen (talk) 08:45, 11 October 2011 (UTC)
replies: text: Done. Godelian (talk) 23:57, 11 October 2011 (UTC)
thread_title:  AC in Proof? 

message: The argument says that "... the matrix has n columns so it's column rank cannot exceed n." Isn't this using that the cardinality of a basis cannot exceed the cardinality of a spanning set, which is basically what the case is trying to prove? The argument also says that "This [the last assertion] means that its m rows cannot be linearly independent." Isn't this using that the cardinality of a linearly independent set cannot exceed the cardinality of a basis, which is basically what the case is trying to prove? This seems circular to me.  
I see that the Alternative Proof says that "The [first] proof above uses several non-trivial results. If these results are not carefully established in advance, the [first] proof may give rise to circular reasoning." But how is one supposed to establish the results need for Case 2 in advance without first giving the alternative proof? It seems like Case 2 should just be completely replaced by the Alternative Proof.
— Preceding unsigned comment added by LoonetteTheClown (talk • contribs) 16:56, 10 July 2018 (UTC)
replies: text: Good point. This proof may probably made correct if one use a definition of the rank of a matrix that is independent of the concept of vector space dimension (for example the number of non-zero rows in the row echelon form of the matrix). But here, rank, although not explicitly linked refers to rank (linear algebra), where it is defined as the dimension of the row space (a vector space) of the matrix. Thus the first proof is clearly circular, and must be removed. D.Lazard (talk) 17:57, 10 July 2018 (UTC)
replies: text:  I'm glad you agree about replacing Case 2. You mentioned that the proof may be made correct "if one uses the definition of the rank of a matrix that is independent of the concept of vector space dimension (for example the number of non-zero rows in the row echelon form of the matrix)." This change will certainly work to deal with my complaint about the assertion "... the matrix has n columns so it's column rank cannot exceed n." But it's not clear to me how changing the definition will deal with my complaint about the assertion "... its m rows cannot be linearly independent." Indeed, matrix rank must be connected to bases in some way if one wants to use matrix rank to prove something about bases.  — Preceding unsigned comment added by LoonetteTheClown (talk • contribs) 19:20, 10 July 2018 (UTC)
replies: text:  The fix D.Lazard made looks good. Further to my last comment, it seems there is a way to handle the finite case using linear equations (matrix) theory. See the proof of Theorem 4 of Chapter 2 of Linear Algebra, 2 ed. by Hoffman and Kunze. It's page 44.  — Preceding unsigned comment added by LoonetteTheClown (talk • contribs) 02:25, 12 July 2018 (UTC)
replies: text: It was my guess in my above post. However, it appears that something equivalent with Steinitz exchange Lemma is always needed: the rank of a matrix or of a system of linear equations may be defined as the number of nonzero rows in a row echelon form, but one has to prove that this does not depend on the way of computing this row echelon form. Similarly, if the rank is defined as the maximal size of a square submatrix with nonzero determinant, one has to prove that the determinant is (up to the multiplication by a constant) the unique alternating n-linear form in n variables. It may also be argued that the proof of Steinitz exchange lemma is essentially an abstract description of the Gaussian elimination algorithm for computing a reduced row echelon form. However, explicating this requires some developments that make Steinitz exchange lemma the simplest way for proving the dimension theorem. D.Lazard (talk) 07:40, 12 July 2018 (UTC)
thread_title:  Case 2 is Circular Logic?