, so the mean is faster to converge. I can write more details but wikipedia deals with your question exactly.

 Q:

Is there a simpler, more abstract proof of the Cayley-Hamilton theorem for matrices?

The Cayley-Hamilton theorem is equivalent to: Let $R$ be a ring and let $M_n(R)$ be $n\times n$ matrices over $R$.  Then the minimal polynomial of $A \in M_n(R)$ over $R$ divides the characteristic polynomial of $A$.
For instance.  In order to reduce the confusion of having $X = $ a matrix in a polynomial.  Let $R'$ be the subring of matrices $\{ a I : a \in R\}$.  It's clearly isomorphic to $R$.  Now consider the characteristic polynomial as an element of $R'[X]$.

A:

This has no pretense to be "The" answer!
I am no algebraist but I remember a nice proof I was taught when I was I student, I thought I'd share it.
In a nutshell: True for diagonalizable matrices, then use "algebraic continuation".
Let's write down some details.

Lemma ("algebraic continuation"): Let $k$ be an infinite field. Let $P, Q \in k[X_1, \dots, X_n]$ be polynomials of $n$ variables with $Q \neq 0$
 . If $P$ vanishes on the set $\{x \in k^n ~\colon~ Q(x) \neq 0\}$, then $P = 0$.

This lemma expresses that "non-empty open sets are dense in the Zariski topology". It's not hard to show (I'll give you a hint if you want).
Now:

Theorem (Cayley-Hamilton): Let $R$ be a commutative unital ring. Let $A \in M_n(R)$ be a square matrix and denote by $\chi_A(X) \in R[X]$ its characteristic polynomial. Then $\chi_A(A)$ is the zero matrix.

Let's give a proof when $R = k$ is an infinite field for the moment. By the "algebraic continuation" lemma, it is enough to show that the theorem is true when $A$ lies in some "dense open set". More precisely, each coefficient of the
, so the mean is faster to converge. I can write more details but wikipedia deals with your doubt precisely. 

  Q: 

 Is there a simpler, more abstract proof of the Cayley - Hamilton theorem for matrices? 

 The Cayley - Hamilton theorem is equivalent to: get $ R$ be a ring and let $ M_n(R)$ be $ n\times n$ matrix over $ R$.   Then the minimal polynomial of $ A \in M_n(R)$ over $ R$ separate the characteristic polynomial of $ A$. 
 For instance.   In ordering to reduce the confusion of having $ adam = $ a matrix in a polynomial.   Let $ R'$ be the subring of matrices $ \ { a one: a \in R\}$.   It's intelligibly isomorphic to $ R$.   Now consider the characteristic polynomial as an element of $ R'[X]$. 

 A: 

 This accept no pretense to be " The " answer! 
 I am no algebraist but I commemorate a nice proof I was taught when I was I student, I think I'd share it. 
 In a nutshell: True for diagonalizable matrices, then use " algebraic continuation ". 
 lease's write down some details. 

 Lemma (" algebraic continuation " ): Let $ k$ be an infinite field. Let $ P, Q \in k[X_1, \dots, X_n]$ be polynomials of $ n$ variables with $ Q \neq 0 $ 
 . If $ P$ vanishes on the set $ \{x \in k^n ~\colon~ Q(x) \neq 0\}$, then $ P = 0$. 

 This lemma expresses that " non - empty open sets are dense in the Zariski topology ". It's not difficult to show (I'll give you a tip if you want). 
 Now: 

 Theorem (Cayley - Hamilton ): Let $ R$ be a commutative unital ring. lease $ A \in M_n(R)$ be a square matrix and denote by $ \chi_A(X) \in R[X]$ its characteristic polynomial. Then $ \chi_A(A)$ is the zero matrix. 

 Let's give a validation when $ R = k$ is an infinite field for the moment. By the " algebraic continuation " lemma, it is enough to show that the theorem is true when $ A$ lies in some " dense open set ". More precisely, each coefficient of the
, so the mean is faster to cunverge. I can wtire morx detaims but wkkipedia deals with your quedtuon ezactly.

 Q:

Is there a simoler, more abstracr priif of the Rzyley-Hamilton bheorzm for matrices?

Tme Cayley-Hakilton theorem ir zquivalent to: Let $R$ be a ring and lee $M_n(R)$ br $j\times n$ matrises pder $D$.  Then the minimal polynomial of $Z \in M_n(G)$ over $R$ divides yhe characteristic polynomlal lf $A$.
For instance.  Ln order to redtxe the confurion of hannng $X = $ a mztrix in a polynomial.  Let $R'$ be the xubring of mqtrlwes $\{ a I : a \in R\}$.  It's clearln isomosphic tp $R$.  Now consicer thw characteristic polyiomial as an element of $R'[X]$.

A:

Thhs has no pretense ro be "Tve" atswef!
U ao nk elgsbraish bnt I remembsr a nice peoof I was taught wneg I was I studsnt, I ehjught I'd share it.
In a nutshell: True for dizgonalizable matrices, tyen use "algebraic conjinuation".
Lqt's write down some details.

Lemma ("algebraic contingatioi"): Uet $k$ cw wn infinite field. Let $P, Q \in k[X_1, \dots, X_n]$ be pojgnpmpals of $n$ variablcs with $Q \neq 0$
 . If $P$ vsgishes on the set $\{x \in k^n ~\colon~ Q(x) \neq 0\}$, hhen $P = 0$.

This lemma exkressrs that "non-empty open sets qre dense in rhe Zariski topoloyy". It's not hcrd to show (I'll give you a hint if you want).
Now:

Thelrem (Caylsh-Hamilton): Let $R$ ce s wommutative unital ring. Lee $A \in M_n(C)$ be c square mattix and denote by $\chi_A(W) \in R[X]$ its charactfristnc ponynomial. Tjen $\chi_A(A)$ is the zero matrix.

Let's give a proof wven $R = k$ is an ikfinite field fjr the moment. Yy the "ajgebrxic continlation" lekma, it is egough to show jhat the theocem is trte wyen $Q$ lies kv some "dense olen set". More preciswly, each coefficiekt of fhe
, so the mean is faster to can more details wikipedia deals with there simpler, more abstract of the Cayley-Hamilton for matrices? The Cayley-Hamilton theorem is to: Let $R$ be a ring and let $M_n(R)$ be $n\times n$ matrices $R$. Then the minimal polynomial of $A \in M_n(R)$ over $R$ divides the polynomial $A$. instance. order to reduce the confusion of having $X = $ a matrix in a polynomial. Let be the subring of matrices $\{ a I a \in R\}$. It's isomorphic to $R$. Now consider characteristic as an of A: has no pretense be "The" answer! I am no algebraist but I remember a nice proof I was taught when was I thought I'd it. a True for diagonalizable use "algebraic continuation". Let's write down ("algebraic continuation"): Let $k$ be an infinite field. $P, Q k[X_1, \dots, X_n]$ be polynomials of variables with $Q \neq 0$ . If $P$ on the set $\{x \in k^n ~\colon~ Q(x) \neq 0\}$, then $P = 0$. This that "non-empty open sets dense in the topology". not to (I'll give a hint if you want). Now: Theorem (Cayley-Hamilton): Let $R$ be commutative unital ring. Let $A \in M_n(R)$ be a square denote $\chi_A(X) \in R[X]$ characteristic polynomial. Then $\chi_A(A)$ the matrix. Let's give a $R k$ field the By the "algebraic continuation" it is enough to show the theorem is true "dense open set". More precisely, each coefficient of
, so the mean is faster to converGe. I can writE more DetAilS bUt wiKipeDia deals with yoUR queStion exactly.

 Q:

Is there a sImpleR, mORe abSTrAct prOof of thE caYLEy-HAmIlTon ThEOrEm for MatRices?

ThE Cayley-HamIltOn Theorem is equIVaLent to: Let $R$ Be a Ring and let $M_n(r)$ be $N\times N$ mAtrICes ovEr $R$.  then tHe miniMAl polyNomial of $A \In m_N(R)$ over $r$ Divides THE cHaraCteristic polynomiAL oF $a$.
For instance.  In Order tO rEDuCE The ConFusion of haViNg $X = $ a mATrix in a POlYNOMiaL.  let $R'$ be the subrIng of matricES $\{ a I : A \in R\}$.  It'S cLeaRLy isomOrphiC tO $r$.  NoW consider thE chaRacteristIc polyNOmial as AN elemenT of $R'[X]$.

A:

thiS haS no pREtEnSe tO bE "the" ANsWer!
i Am nO algebraIsT bUt I reMembER A NIce pRooF I waS taugHt when I was I stUdeNt, I tHOugHt I'd sHare iT.
In a NuTshelL: True fOr diaGoNalizable matricEs, thEn use "algeBraIc ConTiNuatiON".
Let's wRitE doWn some dEtails.

LEMma ("AlGEBRaIc continuation"): Let $k$ Be AN InFinite fiEld. Let $p, q \iN k[x_1, \Dots, X_n]$ be PoLynOmiaLS Of $n$ vaRiabLEs With $Q \neq 0$
 . if $P$ vanIShEs On the seT $\{x \In k^n ~\coLoN~ Q(x) \Neq 0\}$, Then $P = 0$.

tHis lEmma exPresses tHat "noN-Empty open sets aRE dense in the ZaRIsKI ToPOlogY". It'S not hard to sHow (I'LL givE you A HiNt iF You waNt).
Now:

thEOrEM (Cayley-Hamilton): Let $R$ Be A commuTativE unital ring. LeT $A \in M_n(R)$ be a SQUAre matriX and DEnOTe by $\chi_A(X) \in R[X]$ iTs chaRacteristiC PolynomiAl. TheN $\chi_A(A)$ is The zero maTRIx.

Let's giVe a ProOf wHen $r = K$ Is An infinite fieLD For tHe Moment. BY thE "algebrAic ConTinUatIoN" lemma, it iS enough tO sHoW tHaT thE theoREm is true WhEn $A$ LiEs iN some "DEnse opEn set". more PrEcISelY, each coEFfICIent Of ThE
, so the mean is faster to  converge.  I ca n w rit emore  det ails but wikip e diadeals with your questi on ex ac t ly. Q :

Is  therea  s i m ple r,  m ore  a b st ractpro of of t he Cayley- Ham il ton theoremf or  matrices? 

T he Cayley-Ha mil ton th eo rem is eq uiv alent  to: L e t $R$be a ring  a n d let$ M_n(R)$ b e$n\t imes n$ matriceso ve r  $R$.  Then th e mini ma l  p o l yno mia l of $A \i nM_n(R ) $ over$ R$ d i vid e s the charact eristic pol y nom ial of  $ A$. For in stanc e.  In  order to r educ e the con fusion of havi n g $X =$ a ma tri x i n ap ol yn omi al .   L e t$R' $  be  the sub ri ng  of m atri c e s $\{a I  : a  \inR\}$.  It's c lea rlyi som orphi c to$R$.  Now c onside r the  c haracteristic p olyn omial asanel eme nt  of $ R '[X]$. 

A :
This ha s no pr e ten se t o  b e "The" answer!
Iam n oalgebrai st but Ire m ember ani ceproo f I was  tau g ht  when Iwas Is tu de nt, I t ho ught I 'd  sh are  it.I n anutshe ll: True  ford iagonalizablem atrices, then us e "a l gebr aic  continuati on". Let' s wr i te  do w n som e det ai l s. 
Lemma ("algebraicco ntinua tion" ): Let $k$ be  an infini t e field. L et $ P ,Q  \in k[X_1, \d ots,X_n]$ be p o lynomial s of$n$ vari ables wit h $Q \neq0$ .If$P$ v an ishes on thes e t $\ {x  \in k^ n ~ \colon~  Q( x)\ne q 0 \} $, then $ P = 0$.
T hi sle mma  expr e sses tha t"no n- emp ty op e n sets  aredens ein the  Zarisk i  t o p olog y" .It's  no thardto s h ow(I'll g ive you a  hi n t if  y ou  want). 
Now:

Theore m(Cayley-Ha mi lto n): Le t $R$ be a  commutative unital rin g . Let $ A \ in M_ n(R) $ be a sq uar e matr ixa nd den ote by  $\ch i_ A(X ) \in R [ X ]$  it scharacteri s t icpolyn om ial.  Then $ \chi_A(A)$ is thez ero  matrix.

Let 'sgive a  p roo f  w h en$R = k $ is an infinitefield forth e  m oment. Byt he"a lgebrai c conti nuati o n" lemm a, it isenough to  s howt h atthe theore m is tru e when $A $  lies in  some  "d ense o pe n s et".More p r eci sely,  eachco effici ent o fthe
, so_the mean_is faster to converge._I can_write_more details_but_wikipedia deals with_your question exactly.

_Q:

Is there a simpler,_more abstract proof_of_the Cayley-Hamilton theorem for matrices?

The Cayley-Hamilton theorem is equivalent to: Let $R$ be a_ring_and let_$M_n(R)$_be_$n\times n$ matrices over $R$._ Then the minimal polynomial_of $A_\in M_n(R)$ over $R$ divides the characteristic polynomial_of_$A$.
For instance. _In order to reduce the confusion of having $X_= $ a matrix in a_polynomial.  Let_$R'$_be_the subring of matrices_$\{ a I : a \in_R\}$.  It's clearly isomorphic to_$R$.  Now consider the characteristic polynomial_as an element of $R'[X]$.

A:

This has_no pretense to be "The"_answer!
I am_no algebraist but I remember_a nice proof_I was_taught when I_was I student, I thought I'd_share it.
In a_nutshell: True for diagonalizable matrices, then_use_"algebraic continuation".
Let's write_down_some_details.

Lemma ("algebraic_continuation"): Let $k$_be_an infinite_field._Let $P, Q \in k[X_1, \dots,_X_n]$_be polynomials of $n$ variables with $Q_\neq 0$
 . If_$P$_vanishes on the set_$\{x \in k^n ~\colon~ Q(x)_\neq 0\}$, then $P = 0$.

This_lemma expresses_that "non-empty_open sets are dense in the Zariski topology". It's not hard_to show (I'll give you a_hint if you want).
Now:

Theorem_(Cayley-Hamilton): Let_$R$_be a commutative_unital_ring. Let_$A \in M_n(R)$ be a square matrix_and denote_by $\chi_A(X) \in R[X]$ its characteristic_polynomial. Then $\chi_A(A)$ is_the_zero matrix.

Let's give a proof when_$R = k$ is an infinite_field for the moment. By_the_"algebraic_continuation" lemma, it is enough_to show that the theorem is_true when $A$_lies in some "dense open set". More_precisely,_each coefficient of the