Talk:Cayley–Hamilton theorem

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[edit] Request

Can somebody add a proof outline for the commutative ring case? The proofs I know work only over fields (or integral domains). AxelBoldt 00:52 Mar 30, 2003 (UTC)

[edit] General case

I was going to add this to the article, but it may be a bit too technical to include there. Furthermore, it is poorly written. (PLEASE correct and include in the article.)

I don't think there's such a thing as "too technical", as long as the more technical stuff is at the end of an article. The first few sentences should be written for a high school student, and then progressively more advanced. PAR 03:21, 4 May 2005 (UTC)
I agree. One should not try to knock the heck out of college students by starting with a terribly technical thing. But an article can (and should in many circumstances) gradually develop the subject and end up being quite advanced. Oleg Alexandrov 03:27, 4 May 2005 (UTC)
Okay, I'll move it into the article. It still may need some corrections, though.
I've corrected it (I'm a professional mathematician): the proof was incomplete, so I have filled in the gaps, and I tried to reflect the development from the elementary to the advanced.Geometry guy 12:51, 7 February 2007 (UTC)

Can somebody check on the corecctnes of the example? i think that when you substract the matrix {[t,0][0,t]} from the original one, that you DO NOT negate the second and third number!

[edit] I do not think the given proof works

I will buy the Adj(A-tI)(A-tI) = det(A-tI)I = p(t)I
This is proved using matrices over B = R[t].
However it is an equation in S = all such matrices.
You then apply this to m in M. However you have given no definition of such an action.
Also you have given no evaluation homomorphism for such a thing.
~reader


[edit] problem

The action definition is not given. You have to assign some meaning to Am. If you want to imitate the proof in Atiyah-MacDonald you need to operate on a vector with components from m. You can then drop all reference to the evaluation operation.

~reader

[edit] I may be stupid but...

I don't understand why this theorem is not a triviality...

p(A) = det(A − AI) = det(A − A) = det(0) = 0

Isn't it?

I think the introduction to this article should show why this is not so. 131.220.68.177 09:47, 9 September 2005 (UTC)

Your "proof" only holds for 1×1 matrices. If you read carefully example 1, the idea is that you start with p(t) where t is a number, and then do the calculation of det(A-tI) according to the properties of the determinant. Then, plug in formally instead of t the matrix A, and you will get zero.
The error in your proof is the following. The quantity tI is supposed to be interpretted as
\begin{bmatrix}t & 0 & 0 \\ 0 & t & 0\\ 0 & 0& t \end{bmatrix}.

If you plug right in here instead of the number t the matrix A, you get a matrix of size n2. But then the subtraction A-AI can't be done in your proof, as the sizes don't match. Oleg Alexandrov 16:58, 9 September 2005 (UTC)

Yes I understood that already. In fact that's not that easy. I really see the point. Wouldn't it be possible to make this point clear from the beginning of the article. So that nobody fall in this trap again.131.220.44.10 14:49, 12 September 2005 (UTC)

I don't see a good way of writing that while still maintaining the nice style of the article. I think things are most clear if you look at the first example, they show how to interpret that t. Oleg Alexandrov 15:56, 12 September 2005 (UTC)

I patched up the proof for you. It is just a matter of defining an evaluation map between modules over the polynomial ring. I also thought it would be nicer (given the idea to proceed from the novice to the advanced) to give the proof for matrices before discussing the generalizations. This provides an opportunity to discuss the famous incorrect proof as well.Geometry guy 12:52, 7 February 2007 (UTC)

[edit] Accessibility

I fixed up this article about a month ago, but I would like to edit it again to make the proof (for matrices) more accessible (for instance to undergraduate math students). If anyone has comments or suggestions, please let me know. Geometry guy 21:19, 21 February 2007 (UTC)

[edit] Undefined term

Hi, can the term "N-partitioned matrices" be explained or be linked to an appropriate article? Thanks. Randomblue (talk) 20:43, 17 January 2008 (UTC)

Link introduced. User:PhilDWraight 11:30, 11 February 2008 (UTC)

[edit] Fun with page names

Hmm, this is neat: http://en.wikipedia.org/wiki/Cayley-Hamilton_theorem vs. http://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem But maybe you'll like to change it... —Preceding unsigned comment added by 141.84.69.20 (talk) 00:25, 30 January 2008 (UTC)

Yes, I see the former redirects to the latter one, which has a bizarre page name. I'm not aware of when or how such a thing happened, or how it can be undone, but obviously it should be. Marc van Leeuwen (talk) 10:55, 30 January 2008 (UTC)

Nope, by now I found found that this "bizarre" page name just contained an "en dash", and that the wikipedia manual of style says this is the right symbol to use here, since Cayley and Hamilton are separate persons (the manual gives the Michelson–Morley experiment as an example of correct use of an "en dash"). Marc van Leeuwen (talk) 11:34, 30 January 2008 (UTC)

[edit] What's that about N-partitioned matrices?

The lead says the C-H theorem also holds for N-partitioned matrices. Can anybody explain what that means and why it is any different from the usual case? Or is somebody claiming the theorem makes sense (and holds) for matrices over another ring of matrices? If not, I'm tempted to delete that mention. Marc van Leeuwen (talk) 10:47, 7 March 2008 (UTC)

C-H is important in assessing system stability - re: position of eigenvalues etc - in multidimensional systems this becomes a complex issue. Partitioning the A/system matrix can make the problem much simpler - essentially looking at the problem as sub-systems - particularly useful for on-line analysis. Proof the C-H theorem holds for such matrices is vital. User:PhilDWraight 10.22, 22 April 2008 (UTC)

I understand (somewhat) the concern, but I still don't understand what the statement that C-H holds for partitioned/block matrices means. Suppose for concreteness that I have a block matrix

M=\begin{pmatrix}A&B\\C&D\end{pmatrix}

where A, B, C, D are 10×10 matrices, say. First question: is there a characteristic polynomial of this matrix that is not of degree 20. The usual characteristic polynomial of a 2×2 matrix would be

P = X2 − (A + C)X + AD − BC = X2 − (A + C)X + AD − CB,

but as polynomials with matrix coefficients, the two versions are different (and in any case the unwritten coefficient 1 of X2 should be interpreteed as a 10×10 identity matrix). So what exactly would the Cayley–Hamilton theorem say for this block matrix? Marc van Leeuwen (talk) 07:33, 25 April 2008 (UTC)