Mathematics desk
< April 1	<< Mar \| April \| May >>	April 3 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

April 2

is a sum of f(eigenvalue) meaningful?

I just recently got interested in eigenvectors. If $x_{i}$ are the eigenvalues of a given matrix, is $\sum x_{i}$ or $\sum x_{i}^{2}$ or $\prod x_{i}$ equal to anything interesting? —Tamfang 20:47, 2 April 2007 (UTC)[reply]

The sum and product of the eigenvalues (taken with appropriate multiplicities) are related to simple properties of the matrix. There is more information in our article on eigenvalues (although you have to look quite hard to find it). Gandalf61 21:00, 2 April 2007 (UTC)[reply]

Since the trace, or the sum of the elements on the main diagonal of a matrix, is preserved by unitary equivalence, the Jordan normal form tells us that it is equal to the sum of the eigenvalues;
Similarly, because the eigenvalues of a triangular matrix are the entries on the main diagonal, the determinant equals the product of the eigenvalues (counted according to algebraic multiplicity).

From eigenvalues. --Xedi 23:43, 2 April 2007 (UTC)[reply]

Thanks. Why I asked: the application that interests me at the moment is deriving topological coordinates from a graph's adjacency matrix. It occurred to me that the entropy function of the set of (some increasing function of) eigenvalues would give an objective, albeit fuzzy, value for the "natural" number of dimensions, possibly better than the subjective number given by the "scree test". —Tamfang 06:18, 3 April 2007 (UTC)[reply]

If your matrix has finite order, then the sums $\sum x_{i}^{k}$ over all k in fact determine all the eigenvalues. (A fact of basic algebra, not matrices or eigenvalues. This fact is related to the power of character theory in the group representation theory of finite groups. (If this interests you and you can't see how this is relevant, I can elaborate. Respond here or leave me a message.) Tesseran 01:39, 4 April 2007 (UTC)[reply]

Simple groups

Is it true that any group is expressible as the direct product of a number of simple groups? —The preceding unsigned comment was added by 129.78.208.4 (talk) 22:56, 2 April 2007 (UTC).[reply]

~~This is true for any finite group~~ No - any finite group can be constructed as a product of simple groups, but not necessarily as a direct product. A construction that creates a finite group as the ~~direct~~ product of a sequence of simple groups is called a composition series, and every finite group has a composition series (note that this is not true for infinite groups). As an analogy, we can construct the following composition series to show that 12 is the product of a sequence of prime numbers:

1{\xrightarrow {\times 2}}2{\xrightarrow {\times 2}}4{\xrightarrow {\times 3}}12

Of course, this composition series for 12 is not unique - two alternative composition series are:

1{\xrightarrow {\times 2}}2{\xrightarrow {\times 3}}6{\xrightarrow {\times 2}}12

1{\xrightarrow {\times 3}}3{\xrightarrow {\times 2}}6{\xrightarrow {\times 2}}12

We can see that the prime numbers involved in each of these composition series are the same (2,2 and 3) although they appear in different orders. The Jordan-Hölder theorem says that the same property holds for groups - each composition series for a given group contains the same set of simple groups as composition factors, in different orders. In this sense, the "factorisation" of a finite group into simple groups is unique. Gandalf61 08:46, 3 April 2007 (UTC)[reply]

Thanks, I think I might have actually asked that before, but promptly forgot the answer ;) —The preceding unsigned comment was added by 149.135.62.165 (talk) 10:05, 3 April 2007 (UTC).[reply]

I think the answer is false, even for finite groups. – b_jonas 13:49, 3 April 2007 (UTC)[reply]

Ah - yes - because the group products involved in a composition series are not necessarily direct products ? Gandalf61 14:40, 3 April 2007 (UTC)[reply]

Sometimes they're semidirect, but it gets worse than that even. Incidentally, the simplest counterexample to the original question is the cyclic group of order 4. Algebraist 00:19, 4 April 2007 (UTC)[reply]

Right, that counterexample is helpful. Then does that mean that the necessary paragraph on Simple group is wrong? (or if not wrong, needs clarification?) Is there still some known notion of "irreducible" arising from a decomposition of a group into a direct product of smaller groups? (we would consider the cyclic group of order 4 as an "irreducible", then). There is a well-known notion for abelian groups. Restrict to finite groups if needed. 01:20, 4 April 2007 (UTC)

Okay, I have corrected my original response above, to avoid confusion. One more question - why don't the product operations in a composition series have to even be semidirect products ? Since a composition series is a normal series, each subgroup in the series is a normal subgroup of the one above it - so doesn't that mean that the product operations are at least semidirect, even if they are not direct ? Gandalf61 09:24, 4 April 2007 (UTC)[reply]

No. Consider C₄ again. It has a normal subgroup isomorphic to C₂, but no disjoint C₂ subgroup to form a semidirect product with. I think (haven't done enough algebra to be sure) that the Jordan-Hölder theorem is about the best you can do here. Algebraist 19:04, 4 April 2007 (UTC)[reply]