Talk:Cayley–Hamilton theorem

message: The proof using polynomials with matrix coefficients (3.5) can be significantly simplified. The remark that evaluating such a polynomial by replacing the “unknown” (I would prefer the term “indeterminate”) by a polynomial is not a morphism of rings is essential.
But for a given product of such polynomials, it is not necessary for the matrix in whith the evaluation is done to be in the center of the ring generated by the coefficients of the two polynomials. It is sufficient that this matrix commutes with the coefficients of the left factor if evaluation is done with coefficients on the right – or with the right factors if evaluation is done with coefficients on the left.
To put it in formulas, we can write this lemma:
Let M(t) and N(t) be two polynomials in t with matrix coefficients
Let us use evaluation with coefficients on the right, that is M(P)\,:={\displaystyle \sum_{i}P\cdot M_{i}}
Then, if P commutes with all the coefficients M_{i}, (M\cdot N)(P)={\displaystyle M(P)\cdot N(P)}
The proof of this lemma is quite obvious: it is based on the equalities
P^{i+j}\cdot M_{i}\cdot N_{j}=P^{i}\cdot M_{i}\cdot P^{j}\cdot N_{j}
which hold when P commutes with all the M_{j}
One can apply this lemma to the identity used for the proof, {\displaystyle (tI_{n}-A)\cdot B=p(t)\cdot I_{n}}
Obiously, A commutes with the left factor (tI_{n}-A)
It is then legitimate to replace t by A in this identity, and we are done.
It is true that A commutes also with the matrix coefficients in the right factor, B, but this is not necessary for the proof.
I let it to the author to double-check the reasoning above and to simplify the demonstration acccordingly.
With this simplification, this proof using polynomials with matrix coefficients seems to me the nicest and simplest proof.
Silvain.dupertuis (talk) 14:27, 9 March 2018 (UTC)
thread_title:  Proof can be simplified 

message: 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)
thread_title:  Accessibility 

message: 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)
thread_title:  Undefined term 

message: 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)
thread_title:  Fun with page names 

message: given here is actually valid for matrices over any field, not just the complex numbers, since over any Algebraically closed field, the diagonalizable matrices always form a dense subset in the zariski topology. Liransh Talk 15:30, 10 April 2009 (UTC)
thread_title:  The topological proof 

message: Sheldon Axler, in his (very nice, it won an award) American Mathematical Monthly article "Down with Determinants!", and presumably also in his later book Linear Algebra Done Right, gives a short proof of the Cayley-Hamilton theorem using a definition of the characteristic polynomial that avoids determinants entirely. Is it equivalent to one of the proofs here? If not, I wonder (but doubt) whether it's possible to incorporate the proof somehow, without having to re-prove all the five pages of theorems beforehand. Shreevatsa (talk) 08:41, 29 September 2009 (UTC)
replies: text: I've just rapidly read the article. While the paper is interesting, I cannot consider the result it calls "Cayley-Hamilton theorem" to be the same as the one this wikipedia article is about. That is because the notion of characteristic polynomial it uses, being defined without mentioning determinants, is very hard to match with the formal algebraic object defined here to be the characteristic polynomial. For one thing the article depends quite essentially on the complex numbers. It would be rather hard to deduce from its approach what the characteristic polynomial of a matrix with entries in for instance the a finite field (or in the integers, or in a commutative ring with zero divisors) would have to be, and why it takes coefficient in the same field (or commutative ring) as the matrix; all this is immediately obvious from the "determinant" definition. Also one has to read all the way to the end of the article, where the usual definition of determinants is finally "deduced", to understand why the coefficients of the characteristic polynomial depend in a polynomial (or even continuous) fashion on the coefficients of the matrix. In fact this is the fundamental reason why the characteristic polynomial, rather than the minimal polynomial (which the article naturally comes upon first), is of crucial importance for the theory. In the paper the characteristic polynomial comes across as a somewhat artificially "blown up" version of the minimal polynomial (by replacing the numbers \alpha_i by the dimensions of generalized eigenspaces), making the fact that it is divisible by the minimal polynomial (i.e., the "Cayley-Hamilton theorem") more or less built into the definition. So in short, the answers to your questions are are no and no (even if one does re-prove those five pages). Marc van Leeuwen (talk) 21:02, 29 September 2009 (UTC)
replies: text:  Hmm, good point. Elsewhere, he says: Another of the comments on this blog asks how to deal with linear operators on vector spaces over non-complete fields. There are two ways to do this: (1) Embed the non-complete field in its algebraic completion, as I do in the paper or (2) Give up on factoring a polynomial into linear factors and work directly with the non-complete field; the techniques for doing this are illustrated in Chapter 9 (titled “Operators on Real Vector Spaces”) of my book. So it may be possible to extend the approach, but it doesn't seem utopian anymore. :-) Shreevatsa (talk) 02:28, 30 September 2009 (UTC)
thread_title:  "Determinant-free" proof 

message: One proof that is easy to follow is based on the similarity of any complex matrix to an upper-triangular one.
This is the one I would choose to lecture engineers or scientists on this subject.
Thanks!
Gold gerry (talk) 18:53, 18 January 2010 (UTC)
thread_title:  one more proof? 

message: In the Preliminaries section, it is said that "for matrices with real entries the diagonizable ones do not form a dense set". This is wrong. If M=PJP^{-1} is a Jordan decomposition of a real square matrix M, one can always perturb J by D=\textrm{diag}(\epsilon, \epsilon^2, ..., \epsilon^n) for an arbitrarily small \epsilon>0 to obtain a matrix J+D with distinct eigenvalues, and obtain in turn a diagonalizable real square matrix P(J+D)P^{-1}.The suffocated (talk) 14:09, 22 April 2010 (UTC)
replies: text: This is not true. Rotation of R2 by ninety degrees gives an example of a real matrix that cannot be approximated by real diagonalizable matrices. Algebraist 16:03, 22 April 2010 (UTC)
replies: text: But it is complex-diagonalisable. I mean, the original text is confusing because the set of complex-diagonalisable real matrices is indeed dense in GL(n,R).  You just cannot perform diagonalisation over the real field, but the proof can still work without modification.  Even if you insists doing arithmetics on R, you can still follow the same idea and approximate the matrix by a direct sum of real scalars and 2x2 blocks (modulo a similarity transform). So you just need an additional step to prove the theorem for the 2x2 case.The suffocated (talk) 19:31, 22 April 2010 (UTC)
replies: text: That's what that line in the article says, isn't it? You can prove Cayley–Hamilton for real matrices by extending to the complexes and diagonalizing, but it's a rather unsatisfactory proof. Algebraist 20:03, 22 April 2010 (UTC)
replies: text: What I'm saying are: (1) the line is imprecise as one may (actually I did) mistake it as saying "the subset of all complex-diagonalisable matrices in GL(n,R) is not dense in GL(n,R)"; (2) the proof is unsatisfactory, but not because it does not work on R.  It does work on R without extending to the complex field.  All you need is to add one small step to verify the theorem for scalar multiples of 2x2 rotation matrices (which is easy). On a second thought, I also think that dismissing the proof for complex matrices as "strange" is ... strange. I′ll bet most people will find this proof more natural and intuitive than the others, because it reduces the whole theorem to the (trivial) scalar case. Also, continuity arguments are very useful and common tricks in matrix analysis. Transforming one problem into another easier one is also a usual practice in mathematics. If one solves a differential equation by using Fourier transform, shall we say that working on the frequency domain is strange? ;-D The suffocated (talk) 21:57, 22 April 2010 (UTC)
thread_title:  Diagonalisable matrices *are* dense in GL(n,R) 

message: What is this ludicrous thing doing in the article?  —Preceding unsigned comment added by 95.181.12.52 (talk) 09:29, 22 May 2010 (UTC)
replies: text: I am not sure if `ludicrous' refers to the phrase `non-proof' or the wrong proof itself.  If you are asking what purpose does the wrong proof serve, I believe it is here because its validity is the most confusing part for novices.  See, for example, the discussion thread ″I_may_be_stupid_but...″ in the above.The suffocated (talk) 13:09, 26 May 2010 (UTC)
thread_title:  A `Non-proof'? 

message: There seem to be people who have already spent a lot of time on this, so I won't change anything and set off an edit war, but the explanation of the bogus "proof" seems to be needlessly obfuscated. This is really a simple issue with a simple solution: people don't understand the notation p(A), so we should write out p(A) explicitly. As currently written, this section never points out what p(A) actually is. This seems absurd to me. Note that p(A) is currently the fifth displayed line of this section, but the accompanying text gives the impression that this line is some sort of error. (More precisely, that line would be p(A), if the given text didn't re-interpret the matrix as a 4x4 matrix instead of a 2x2 matrix with entries in R[A].) Some of the other stuff here is worthwhile -- I like pointing out that the bogus proof tries to compare a scalar to a matrix, since I've seen too many of my students try to equate things that aren't even the same type of object -- but we should lead with the simple explanation "p(A) is THIS, not THAT."  165.123.215.110 (talk) 04:36, 10 February 2013 (UTC)
replies: text: I suppose it is always hard to clearly steer the reader to take the received wrong steps correctly! The correct 2x2 matrix definition of p(A) is already in the last displayed eqn of the Example, section 1, at the top of the article, indeed, the takeout message of the whole article. I would argue against encouraging the confused reader to take the "right two left turns" to get to the proper p(A) here--it might be a good idea to not mix the right with the wrong in the bogus section. If the reader knew how to properly and safely compute and interpret non-scalar-component matrices, as you assume, she/he  need not be here, in the first place! Any discussion of entries in R[A] would compound the confusion. Cuzkatzimhut (talk) 13:12, 10 February 2013 (UTC)
thread_title:  Unclear explanation of the bogus proof 

message: The article claims that:
"(for a matrix to be diagonalizable it suffices for instance that its characteristic polynomial not have multiple roots)"
Here is a simple counter example: A = [[1,1],[0,1]]. A has charpoly (t-1)*(t-1), but it is already in nondiagonal Jordan form and thus is not diagonalizable.
Now, perhaps the author here intended to say that the matrix have distinct roots. Perhaps this could be stated (although not clearly IMO) that the characteristic polynomial not have repeated roots, but the claim as stated is very misleading.
:I don't understand your problem with the cited formulation. In your example (t-1)*(t-1) does have a multiple root, namely 1; are you making a fine semantic distinction between "multiple roots" and "repeated roots" (and if so which)? Maybe you just read "multiple roots" as saying "more than one root" (in which case to "not have multiple roots" would be more simply expressed as "have at most one root", which is entirely different from what is meant). If that is the case, than I will just modify the language to avoid this confusion. Marc van Leeuwen (talk) 13:02, 9 March 2014 (UTC)
thread_title:  Claim about sufficient condition for a matrix to be diagonal is false as stated. 

message: If Hamilton really did prove that the theorem holds for quaternions it is quite a feat, especially if this was done for matrices of arbitrary order. Note that the quaternions constitute a non-commutative ring. This seems to me to strongly suggest that it holds for the complex numbers because they are embedded in the quaternions and all its subrings. It also ought to hold for rings isomorphic to subrings of the quaternions in general.
Something is not right, either in my reasoning above or that Hamilton's quaternion proof was in full generality for them. Is it the case that there are plenty of rings not isomorphic to a subring of the quaternions? In any case, this needs to be cleared up in the article.
(The statement that it holds for quaternions was buried in the reference section before my recent edits. The name of the reference itself clearly indicates that it really is quaternions we are talking about.) YohanN7 (talk) 21:41, 29 January 2015 (UTC)
thread_title:  Quaternions 

message: Unless somebody comes up with a convincing rationale for keeping the "bogus proof", I will delete it. You could construe the same sort of nonsense proof for whatever theorem, so why report this one? -- Julia Abril (talk) 13:54, 5 February 2015 (UTC)
replies: text: Please do not! It is obviously not for you. The C-H theorem is exceptionally prone to such nonsense proofs, and they must be addressed. The section has a long history, and controversy, and lots of people do not like it; but this is a particularly pervasive and recurring question and this is the place to put it to rest. If you can improve it, cf. items 4,14,15 above, go ahead. But this is an educational section for the novice and the teacher dealing with him---a thinking  expert need not be here! Recall this is not a treatise, but an educational resource.  Dismissing an obvious common canard efficiently is not  pointless. Besides, as you can see, there is a tangent on which it can be actually be salvaged. In any case, if you delete it, sooner or later, somebody will perceive a gap and reinsert something like it. If it offends you, and you cannot improve it, skip it in reading.  Cuzkatzimhut (talk) 15:28, 5 February 2015 (UTC)
replies: text: Not convinced, but since you insist that boldly, I will certainly not touch the section. -- Julia Abril (talk) 20:09, 13 February 2015 (UTC)
thread_title:  "Bogus proof" should be deleted 

message: CAYLEY-HAMILTON THEOREM FOR MATRICES OVER AN ARBITRARY RING
by Jeno Szigeti
Serdica Math. J. 32 (2006), 269–276.
Communicated by V. Drensky
Abstract. For an n × n matrix A over an arbitrary unitary ring R, we
obtain the following Cayley-Hamilton identity with right matrix coefficients:
(λ_0I + C_0) + A(λ_1I + C_1) + · · · + A^(n−1)(λ_(n−1)I + C_n−1) + A^n(n!I+C_n) = 0,
where λ_0 +λ_1x+· · ·+λ_n−1x^(n−1)+n!x^n is the symmetric characteristic polynomial
of A in R[x], I ∈ M_n(R) is the identity matrix and the entries of the n × n
matrices C_i, 0 ≤ i ≤ n are in [R, R]. If R is commutative, then
C_0 = C_1 = · · · = C_(n−1) = C_n = 0
and our identity gives the n! times scalar multiple of the classical Cayley-Hamilton
identity for A.
:We can probably list it in the references with an inline citation. Does the publication have a DOI (Digital object identifier)? (It is accessible via the journal site.) YohanN7 (talk) 17:17, 1 February 2016 (UTC)
replies: text: This seems okay too. Why don't you get an account by the way? You can also edit yourself (unless perhaps you are J. Szigeti in which case it is prunent to ask someone else to do it). YohanN7 (talk) 13:00, 6 February 2016 (UTC)
thread_title:  CAYLEY-HAMILTON THEOREM FOR MATRICES OVER AN ARBITRARY RING 

message: The exposition in the section on power series via Cayley-Hamilton can certainly be improved. The way it stands, it feels like a physicist's dump. Manoguru (talk) 03:23, 9 March 2016 (UTC)
replies: text: Of all categories of talk page comments, the stupidest of all may be this article can be improved. YohanN7 (talk) 08:33, 9 March 2016 (UTC)
thread_title:  Power series 

message: It seems to me that there is an easier proof than the ones given. Since I don't have a source for it, I want to get confirmation before adding it to the article.
Say we are working over a field . The proof is by induction on . The base case  (or ) is straightforward. Suppose we know the theorem for  matrices. Let  be an eigenvalue of our  matrix . Because the characteristic polynomial remains unchanged when we change the basis, we may choose a basis  such that  is an eigenvector of  with eigenvalue . Then  looks like
:A=\begin{pmatrix}&&&0\\&B&&\vdots\\&&&0\\*&\cdots&*&\lambda\end{pmatrix}
where  is an  matrix. By cofactor expansion, we have . Moreover, by the induction hypothesis, .
If we view  as a subspace of  with basis , then  is just the restriction of  to a linear transformation . Thus for any vector ,
:p_B(A)w=p_B(B)w=0w=0.
For any vector , we see from the above form of  that . Applying the above equality with , we get
:p_A(A)v=p_B(A)(A-\lambda I_n)v=0.
Since this holds for all , we conclude that .
Enigmatologist (talk) 23:47, 29 May 2019 (UTC)
replies: text: – this is generally false; the theorem still holds for non-diagonalizable matrices. The proof for the diagonalizable case is trivial and doesn't even require induction; the proof given does not assume the existence of an eigenbasis (or indeed, whether any eigenvectors exist at all; we may not even have eigenvalues in the ring in question).--Jasper Deng (talk) 00:37, 30 May 2019 (UTC)
replies: text: I think you misunderstood. I'm not saying that every basis vector is an eigenvector, just the last one. This can be done as long as there is an eigenvalue. Though I should have assumed that the field is algebraically closed, as there might not be an eigenvalue otherwise. Enigmatologist (talk) 23:23, 31 May 2019 (UTC)
replies: text: I did not misunderstand. To be clear, that last part is the problem. The ring in question need not even be a field, let alone an algebraically closed one. Rings that aren't integral domains cannot have a field of fractions so a proof for the case where R is a field obtained by extending F to its algebraic closure is not powerful enough for the general case. In the algebraically closed case, the Jordan normal form suffices without induction.--Jasper Deng (talk) 23:31, 31 May 2019 (UTC)
thread_title:  Simpler proof 

message: "For such cases, for an eigenvalue λ with multiplicity m, the first m – 1 derivative of p(x) vanishes at the eigenvalues."
What? "The first m - 1 derivative of p(x) vanishes at the eigenvalues."???
If you don't know how to express plurals, and their accompanying verbs, in English, then it would be better to learn how to do this, instead of confusing people on Wikipedia with ridiculous sentences like the one above.  Especially in mathematics.
I'm guessing that the intended sentence is 
"For such cases, for an eigenvalue λ with multiplicity m, the first m – 1 derivatives of p(x) vanish at the eigenvalues." (The boldface is mine, for emphasis.)
But it's hard to be sure.50.205.142.50 (talk) 16:45, 13 March 2020 (UTC)
replies: text: Grammar fixed to salutary approximant. Cuzkatzimhut (talk) 01:00, 14 March 2020 (UTC)
thread_title:  Ridicdulously unclear sentence 

message: I have rewritten the lede to try to make it clearer what exactly the C-H theorem says. I think this makes the "bogus proof" section significantly less necessary, since (to my mind) it stems from an easy misinterpretation of simply defining the analogous matrix equation as simply "consisting of the replacement of the scalar variable λ with the matrix A" from the characteristic polynomial det(λI_n - A). That said, I won't immediately take the initiative to remove it from this page. I also completely removed the sentence "The powers of A, obtained by substitution from powers of λ, are defined by repeated matrix multiplication; the constant term of p(λ) gives a multiple of the power A^0, which is defined as the identity matrix.", because it seems bizarrely elementary for the intended audience of this article, who is presumably already familiar with algebraic operations on matrices.
A Lesbian (talk) 17:01, 30 October 2020 (UTC)
thread_title:  Rewrote the lede, extra notes 

message: This proof section begins with "the Cayley–Hamilton theorem is an immediate consequence of the existence of the Jordan normal form for matrices over algebraically closed fields"  without further clarification.  This is the proof that I would be most interested in.  Does it mean that A=V J V^-1 so that P(lambda) = |A - I lambda| = |J - I lambda|, ie the characteristic equation is invariant, and thus P(A) = (j1 I - J)(j2 I - J)..(jn I - J), which has a 0 at every diagonal position in at least one factor, and thus equals the 0 matrix?  I think a fleshed out version of this would be great to include right after that statement, especially about the invariance of the characteristic equation, and therefore det, tr, and everything in between.  And a reference to Invariants of tensors. Chris2crawford (talk) 03:59, 9 February 2022 (UTC)
thread_title:  immediate consequence of Jordan normal form...