Abstract Algebra

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Another way you could prove that is to write and then use the binomial theorem to expand that.

err could you explain how that works? I can't see how that explains it

[zzdfa]

Well after typing it up I see that its not as quick as I thought it was, I thought the last step (*) would be more straightforward. I'll post it anyway since the technique is occaisonally useful.

We want to show that A^n is never equal to the identity.
which is the same as showing that (I+B)^n is never the identity , where

By the binomial theorem, so that if (I+B)^n = I then





(*) and so the problem reduces to showing that the expression is never 0. I can't think of any quick way to do this for this particular B (apart from induction, which would make it too similar to your proof). Oh well, this would work with a more convenient b, (say the matrix with a  single 1 in the top left).

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Ah, thanks zzdfa, that certainly looks useful

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In the video lectures, the guy takes the determinant of the matrix:



By multiplying together the determinants of

and

Is there a general rule for this sort of stuff? I don't remember seeing it before

(I know how to take the determinant by interchanging the top two rows and multiplying along the diagonal, but is he using a different principle?)



Also, how would you go about proving that for is the ONLY centre of ?

i.e. for all square matrices of dimension .

Thanks

[zzdfa]

For your first question, you probably saw the example and probably realized that 'the det of a block diagonal matrix is the product of the det of the blocks'. This is true and easy to prove (using Leibniz's formula).

The natural generalization is 'the det of a block matrix is the det of the matrix of the det of the blocks'. (see how the diagonal case is a special case of this because the det ofa  diagonal matrix is the product of its entries). Unfortunately this is not true. See
http://en.wikipedia.org/wiki/Determinant#Block_matrices

And you were asking about 'general rules', another 'general rule' you could use is that the matrix you gave above is a permutation matrix (this particular matrix swaps entries 1 and 2 when you let it act on a vector) and the det of a permutation matrix is just the sign of the permutation, which is -1 in this case since it swaps 2 elements.

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thanks zzdfa

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A homomorphism is proposed in the lecture:

, where under composition.



But isn't the homomorphism meant to be a function of alone? How come we have elements of here?

The lecture then proceeds to show it's a homomorphism by doing this:



?

Thanks

[zzdfa]

here's a sketch of how you'd do the 2nd question. The idea is to, given an nonidentity matrix A, construct a matrix B that doesn't commute with A.
Given an A, since it is not the scalar multiple of an identity,
1. there is a vector v such that Av is not the scalar multiple of v. In particular Av is linearly independent to v.
2. choose a basis of R^n that contains the 2 vectors {v,Av}
3. let B be a matrix that takes v to v, i.e. Bv=v, and takes Av to 0, B(Av)=0.
4. then verify that ABv \neq BAv and so AB \neq BA.
I haven't justified each step rigorously , but these are just direct applications of the basic theorems in LA. except for step 1, can't think of a simple reason for that.


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if g is an element of G, then f(g) is an element of Aut(G). that is, f(g) is an automorphism. remember, automorphisms are functions.
the formula f(g)(h)=ghg^-1 is telling you that
f(g) is the automorphism on G that takes a h in G to

if the notation is confusing (as it often does when you have a function that outputs another function)then try writing ("to every there is an associated automorphism that acts on G according the following rule:

dont get confused, there are 2  'levels' of morphisms here, the first level is the fact that the map is a homomorphism , and the 2nd one is the fact that the map  is an automorphism

[kamil9876]

Another way you could prove that is to write and then use the binomial theorem to expand that.

does the binomial theorem hold for matrix multiplication? u need commutativity to prove it.

also for fun:

are and (with addition only) isomorphic? I'll give 10 quid to anyone who proves it without the continuum hypothesis.

[zzdfa]

yes, because I commutes with everything

also, nfi how to approach your question. hints?

[kamil9876]

ahh yes sorry I was high on English beer. I made the problem up myself, I tried to prove in vain that they are not isomorphic ( I first proved that Z and Z^2, Q and Q^2 are not isomorphic). When I come back home on saturday I will post in more detail. I'm not even sure if it is decidable without the continumm hypothesis or axiom of choice. A clue is to think of R and R^2 as vector spaces over Q.

[zzdfa]

yeah I thought about it a little more:
The Z->Z^2 case follows from that fact that p(z)=zp(1)
The Q->Q^2 case follows from the fact that any homomorphism is in fact a linear transformation (by fiddling around with the defn of homomorphism), and we know that there is no linear isomorphism Q->Q^2 (different dimensions)


For the R->R^2 case,  assuming that bases for R and R^2 exist and have the same cardinality, find a bijection between the bases and we're done.
And it seems like you need AC to find the basis and CH to prove that they have the same cardinality.

[Ahmad]

An idea for showing R and R^2 as Q vectorspaces have dimension of the continuum: If R had a countable basis then we can express any element of R uniquely as a finite linear combination of elements of the basis B which means R is the union of all elements expressible by a linear combination of 1 basis element, and those expressible as a linear combination of 2 basis elements, and 3 basis elements and so on. This is a countable union of countable sets (why are they countable?), which is countable but R is uncountable.

So the basis has size more than aleph_0 and at most the continuum (R is a spanning set for itself, of course), now you need CH. Same goes for R^2. Intriguing result.

[kamil9876]

yep good

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Let be a cyclic group of order . Can someone please explain how if are coprime then is isomorphic to ? I arrived late to one of the lectures and must have missed the explanation. Thanks