Talk:Overdetermined system

Some of the expansion here belongs at system of linear equations, really. The use of the linear case was to be able to illustrate how to count constants, in a less abstract setting. Charles Matthews 20:57, 13 June 2007 (UTC)[reply]

I think that the example is necessary for the reader who doesn't know how to count constants already. I think I wiki-linked to system of linear equations in the article, but it didn't really give the example that I was going for, so I wrote my own.--Cronholm¹⁴⁴ 07:43, 25 June 2007 (UTC)[reply]

RE (Inhomogeneous case) : "M equations* and N unknowns*, such that M>N and all M are linearly independent. This case yields no solution." The number of linearly independent rows of an MxN matrix (M equations, N unknowns) equals the rank of the matrix. However, from (Strang, G. Linear Algebra and Its Applications Brooks Cole, 1988) p.105: the row rank = column rank. Thus how it is possible to have M linearly independent rows, thus rank = M but M > N ? —Preceding unsigned comment added by Cpanagio (talk • contribs) 01:12, 28 October 2008 (UTC)[reply]

Shouldn't ART solutions be included or at least linked to for solutions to overdetermined equations? —Preceding unsigned comment added by 199.88.91.106 (talk) 17:04, 23 March 2011 (UTC)[reply]

Independent equations or independent sets of coefficients of unknowns?

The section "Non-homogenous case" currently categorizes systems of more linear equations than unknowns as follows:

M equations and N unknowns, such that M > N and all M are linearly independent. This case yields no solution.
M equations and N unknowns, such that M > N but all M are not linearly independent. There exist three possible sub-cases of this:
- M equations and N unknowns, such that M > N but all M are not linearly independent, but when the D linearly dependent equations are removed M - D > N. This case yields no solutions.
- M equations and N unknowns, such that M > N but all M are not linearly independent, but when the D linearly dependent equations are removed M - D = N. This case yields a single solution.
- M equations and N unknowns, such that M > N but all M are not linearly independent, but when the D linearly dependent equations are removed M - D < N. This case yields infinitely many solutions, which can be described as F which is the entirety of the field in which the equations are operating.

It appears to me that part of the above passage would be right if it referred to the number of equations with independent rows of coefficients of unknowns, rather than the number of independent equations, but that as is it is wrong. The difference is exemplified by the system x=1, x=2, in which the coefficients are not independent but the equations are independent because there is no constant that you can multiply one of them by to get the other one.

The first case listed in the above-quoted passage, and the first sub-case of the second case, I believe, are misleading as stated because there can be more independent equations (equations, not rows of the unaugmented matrix) than unknowns only if M − D = 1 N+1, as in the system x=1, x=2 (since these equations can't be derived from each other) But I think it's impossible if M − D > 1 N+1. And if independent equations is replaced by independent rows of the coefficient matrix, then these cases are impossible.

Similarly, I think the second sub-case of the second case is wrong, since there is not necessarily a single solution when the number of independent equations equals the number of unknowns -- for example, consider the system x+y=1 and x+y=2, which has no solution, but the equations are mutually independent since you can't multiply the first one by any constant to get the second one. This case would be right if independent equations is replaced by independent rows of the coefficient matrix.

And I believe the last sub-case is wrong, since if there are fewer independent equations than unknowns, there are not necessarily any solutions at all -- for example, the system x+y+z=1 and x+y+z=2, with three unknowns and fewer than three independent equations. This case too would be right if independent equations is replaced by independent rows of the coefficient matrix.

Comments and suggestions for revising? Duoduoduo (talk) 22:22, 20 November 2012 (UTC)[reply]

Firstly, there is another flaw in this formulation, that is "the D linearly dependent equations", which suggest that these D equations are well defined, although, in most cases, any set of D equations may be removed.

Secondly, I personally do not like this way of presenting the cases, because, in practice, linear dependences are not immediate to detect. If I have had written this article from scratch, I would have presented the things in terms of the ranks of the non augmented and of the augmented matrices of coefficients.

The last sub-case contains another mistake: it seems to say that the dimension of the set of solutions is one, Which is wrong if M - D < N - 1. A correct formulation for the last case would be: M equations and N unknowns, such that M > N but all M are not linearly independent, but when D linearly dependent equations are removed M - D < N, that is the system is an underdetermined system. This case yields either infinitely many solutions or no solutions.

It seems that, in your post, M − D = 1 should be replaced by M − D = N + 1, and M − D > 1 by M − D > N + 1. With this modification, you are right, because, the rank of the augmented matrix may not exceed the number of its columns. To correct the classification, it suffices thus to replace, in the first case, M > N by M = N + 1, in the first sub-case M - D > N by M - D = N + 1, and to say somewhere that the number of independent equations may not exceed N + 1.

--D.Lazard (talk) 03:27, 21 November 2012 (UTC)[reply]

Thanks for your reply. Could you please double-check the revisions I've put in to this section? Also, could you edit if necessary the last paragraph of the section? I'm not sure what k is supposed to be, and I don't know enough to edit that paragraph myself. Thanks. Duoduoduo (talk) 00:53, 22 November 2012 (UTC)[reply]

Your revision seems correct. About the last paragraph: I do not think that its content is easier to understand. IMO it is too technical. In fact, one may not hope that geometry in a space of dimension higher than 3 makes sense for most readers of this article. Therefore, I suggest to remove this paragraph, and to replace it by something like:

"These results may be easier to understand by putting the augmented matrix of the coefficients of the system in row echelon form by using Gaussian elimination. This row echelon form is the augmented matrix of a system of equations that is equivalent to the given system (it has exactly the same solutions). The number of independent equations in the original system is the number of the non zero rows in the echelon form. The system is inconsistent (none solution) if and only if the last non zero row in the echelon form has only one non zero entry that is in the last column (giving an equation 0 = c where c is a non zero constant). Otherwise, there is exactly one solution when the number of non zero rows in the echelon form is equal to the number of unknowns and there is infinitely many solutions when the number of non zero rows is lower than the number of variables."

One may argue that linking to Gaussian elimination is also too technical. IMO, this is not the case, as Gaussian elimination is the easiest way to get the number of independent equations and talking of this number of independent equations to someone that has no mean to knowing it is a non sense.

Reading again what I suggest, it appears that it has nothing specific to overdetermined systems. In fact, overdetermined systems have nothing else that is specific than a high probability of not having any solution. In other words, "overdetermined system" is not a topic, but a dictionary definition. This suggest of merging this article (and underdetermined system) into simultaneous equations

--D.Lazard (talk) 16:39, 22 November 2012 (UTC)[reply]