Talk:Cayley–Hamilton theorem

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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)

Contents 1 General case 2 I do not think the given proof works 3 problem 4 I may be stupid but...

[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.

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

.

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)