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October 20

Associativity axiom in groups

Obviously the associativity axiom for groups could be stated as 'for all n > 0, each product of group elements of length n can be rebracketed as you like', rather than the usual case when n = 3 (i.e. (ab)c = a(bc)), from which the former statement can be derived. My question is, is there a reason other than efficiency/elegance of definition that it is usually stated how it is, rather than 'for all n'? I can't see why there should be, as the statements are equivalent, but I have a vivid memory of a lecture on introductory group theory, in which it was stated that there was some 'tangible' (bad word) reason that it is stated like it is. Not that I'm downplaying the importance/beauty of an elegant definition, it's just nagging at me that there might be another reason too. Thanks, Icthyos (talk) 16:04, 20 October 2011 (UTC)[reply]

One reason (probably not what you're thinking of) is that it's harder to state for general n. Your "rebracketed as you like" is not a good technical statement of the property for general n. It's very easy to state for n=3 as you wrote. Also, it's customary to choose your axioms to be as simple as possible to build the structure you need. For the same reason a commutativity property is usually just stated as ab=ba, rather than discussing all possible reorderings of n elements, distributive properties are written c(a+b) = ca+cb, etc. Staecker (talk) 16:51, 20 October 2011 (UTC)[reply]

Hahaha, yes, that's why I ended up just saying 'rebracket as you like'. I suppose that might be what it was, since that at least is a practical reason to do it, rather than just appealing to elegance. Wish I could remember... Icthyos (talk) 18:28, 20 October 2011 (UTC)[reply]

In general it's better to make axioms as simple as possible. Suppose you have a set with operation and wish to show it's a group. It's a lot easier to prove the n=3 case of associativity than for all n, and since one implies the other it's pointless to do the harder one. Btw, as I recall Huppert's classic book on finite groups gives a rigorous definition of the general associativity law and proves that it's implied by the n=3 case. It might be in Bourbaki as well, or at least it's that kind of thing you'd find in Bourbaki.--RDBury (talk) 21:46, 20 October 2011 (UTC)[reply]

I think that might've been it, actually - proving something satisfies the associative law. Clearly this is far easier when n = 3. Thanks! Icthyos (talk) 13:04, 21 October 2011 (UTC)[reply]

Multiple singular eigenvalues

What is the fastest way to determine if a matrix has 2 or more eigenvalues of 0? I'm looking at the number of connected components in a Laplacian matrix. 184.98.145.18 (talk) 17:30, 20 October 2011 (UTC)[reply]

The fact that the Laplacian matrix is symmetric is relevant; in particular, this implies that the geometric and algebraic multiplicities are equal, making your question unambiguous, and it may be the basis for a better method than I suggest.

A good way (I don't know if the best) to find the (geometric) multiplicity of 0 is with the LU decomposition. If

A=LU

and L has has 1's on the diagonal, then it is invertible and so A and U have the same rank, and hence the same nullity. By doing Gaussian elimination on U you can find its rank.

I don't think you can do significantly better for the specific question of whether the multiplicity is at least 2. -- Meni Rosenfeld (talk) 20:16, 20 October 2011 (UTC)[reply]

Wouldn't the fastest way be to calculate the characteristic polynomial, say κ(λ), and see how many times λ divides into it? Wouldn't that be simpler and computationally easier to do than to calculate a decomposition? — Fly by Night (talk) 16:05, 21 October 2011 (UTC)[reply]

Behaviour of a function (modified)

f(a)g(a) < 0

f(a)g(a) > 0

Hi all,

I have a problem which can be regarded as below:

Given f(x) is an increasing function, and g(x) is a decreasing function with domain x∈[a;b]; and another function h(x) to which its range is d≤h(x)≤D. Given both of f(x) and g(x) are positive within their domain

Find d and D i. h(x)=f(x)g(x)

ii. h(x)=(f(x))/(g(x))

iii. h(x)=|f(x)g(x)-(f(x))/(g(x))|

I understand that in order to solve this problem we have no choice but consider a certain number of cases under various circumstances.

For instance, in part (i), see top image.

If f(a)g(a)<0 and f(b)g(b)>0 then f(a)g(a)=d as minimum, and f(b)g(b)=D as maximum. This may sound logical but in another case like the bottom image.

If f(a)g(a)>0 and f(b)g(b)>0 then d=f(a)g(a) and D=f(b)g(b); but even in this case we cannot tell whether this is true or not, let’s say f(a)=1, g(a)=5 then f(a)g(a)=5, and in some neighborhood we have f(a1)=2, g(a1)=4 then f(a1)g(a1)=8 which will contradict with the result of f(a)g(a)=5 as minimum above!

And the same trends are still applied for the next parts. I have tried to find some information relating to the subject in the internet, but unfortunately I got no clue until now.

Any suggestion would be appreciated. Thanks.Torment273 (talk) 20:42, 20 October 2011 (UTC)[reply]

The number of cases is huge, because you can easily get extremes in the middle of the domain. I don't think this is a reasonable problem. Where did it come from? Looie496 (talk) 16:29, 21 October 2011 (UTC)[reply]

Thanks for your comment. This is one of the problems of last year paper of the entrance exam into Nanyang Technological University in Singapore. Anyways, do you think that those Profs over there just want to make it as challenging as possible so that it will look like an IQ test? In this case i have no choice but consider all of the possible circumstances regarding to the question, right?Torment273 (talk)

N.B. The fact that f is increasing (resp. strictly increasing) means that df/dx ≥ 0 (resp. > 0) for all a ≤ x ≤ b, and the fact that g is decreasing (resp. strictly decreasing) means that dg/dx ≤ 0 (resp. < 0) for all a ≤ x ≤ b. — Fly by Night (talk) 18:47, 21 October 2011 (UTC)[reply]

Thanks you all. My current approach is by inequality properties:

As f(x) is increasing in [a,b], we have f(a)<=f(b); and g(x) is decreasing in [a,b], we also have g(a)>=g(b). Consequently, we obtain f(a)g(b)<=f(b)g(a). Given d≤h(x)≤D, d will be f(a)g(b), and D is f(b)g(a). But this is still puzzling me as i don't know how to obtain h(x)=f(x)g(x) in this case (i mean under which conditions the equal sign "=" will take place so that d=h(x)=D).Torment273 (talk) 16:48, 23 October 2011 (UTC)[reply]