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proof of CayleyHamilton theorem by formal substitutions
Let $A$ be a $n\times n$ matrix with entries in a commutative ring with identity, and let $p(\lambda)=c_{0}+c_{1}\lambda+\dots+c_{n}\lambda^{n}$ be its characteristic polynomial. We will prove that $p(A):=c_{0}I+c_{1}A+\dots+c_{n}A^{n}=0$.
Proof$\,$ (Popular fake proof):
In the expression
$\displaystyle p(t)=\det(AtI)=c_{0}+c_{1}t+\cdots+c_{n}t^{n}\,,$ 
substitute $t=A$; then $p(A)=\det(AAI)=\det(0)=0$. $\square$
$\,$
It is clear why the argument is faulty. But interestingly, there is a way to rescue it using clever formal substitution arguments. For the moment, we assume that the matrix $A$ is over the complex field.
Since the notation $p(\lambda)$ and $p(A)$ can be confusing at first sight, as one expression takes scalar values and the other matrix values, we will change it. From now on, we will use the notation $\widetilde{p}(t)$ when applying the polymial $p$ to a matrix $t$. In this sense $p(\cdot)$ is a function of $\mathbb{C}$ and $\widetilde{p}(\cdot)$ is function of matrices. Also, by definition,
$\displaystyle\widetilde{p}(t):=c_{0}I+c_{1}t+\dots+c_{n}t^{n}\,,\qquad\qquad% \text{where $t$ is a matrix}$ 
Of course, we intend to prove that $\widetilde{p}(A)=0$.
Proof $\,$ (Proof of the complex case):
For each $\lambda\in\mathbb{C}$ let $B(\lambda)$ be the classical adjoint to $A\lambda I$. We have
$B(\lambda)(A\lambda I)=\det(A\lambda I)I=p(\lambda)I=c_{0}I+\lambda(c_{1}I)+% \cdots+\lambda^{n}(c_{n}I)\,.$  (1) 
From the definition of the classical adjoint, it is clear that $B(\lambda)$ can be written as a polynomial of $\lambda$ having degree $\leq n1$, whose coefficients are matrices. That is,
$\displaystyle B(\lambda)=D_{0}+D_{1}\lambda+\dots+D_{{n1}}\lambda^{{n1}}\,,$ 
for some constant coefficient matrices $D_{i}$.
Now, we would like $B$ to be defined for matrices (just like from the polynomial $p$ we have considered $\widetilde{p}$), so we define for every matrix $t$
$\displaystyle\widetilde{B}(t):=D_{0}+D_{1}t+\dots+D_{{n1}}t^{{n1}}\,.$ 
Now consider the following function of matrices:
$\displaystyle Q(t):=(D_{0}Ac_{0}I)+(D_{1}AD_{0}c_{1}I)t+\dots+(D_{{n1}}AD% _{{n2}}c_{{n1}}I)t^{{n1}}+(D_{{n1}}c_{n}I)t^{n}$  (2) 
The above expression may look strange, but if we think that the matrix $t$ commutes with all $D_{0},\dots,D_{{n1}}$ and $A$, then expression (2) is easily seen to be equal to
$\displaystyle(D_{0}+D_{1}t+\dots+D_{{n1}}t^{{n1}})(AIt)c_{0}Ic_{1}t\dots% c_{n}t^{n}$ 
This means that
$\displaystyle Q(t)=\widetilde{B}(t)(AIt)\widetilde{p}(t)\;\;\;\;\;\;\;\text{% whenever}\;t\text{commutes with}D_{0},\dots,D_{{n1}},A$  (3) 
The reason for not defining $Q(t)$ by the expression in (3) is that we want $Q(t)$ to be some kind of ”polynomial” in $t$ (with matrix coefficients on the left of each $t^{k}$).
We now state some properties that can be easily checked by straightforward calculation:

$\widetilde{p}(\lambda I)=p(\lambda)I$

$\widetilde{B}(\lambda I)=B(\lambda)$
Notice that matrices of the form $\lambda I$ with $\lambda\in\mathbb{C}$ commute with every other matrix, so that
$\displaystyle Q(\lambda I)=\widetilde{B}(\lambda I)(A\lambda I)\widetilde{p}% (\lambda I)=B(\lambda)(A\lambda I)p(\lambda)I=0$ 
Now $Q(\lambda I)$ is also a matrix whose entries $q_{{ij}}(\lambda)$ are polynomials in $\lambda$. Since $Q(\lambda I)=0$ we must have $q_{{ij}}(\lambda)=0$ for all $\lambda\in\mathbb{C}$. This means that $q_{{ij}}(t)$ is the zero polynomial and, since this occurs for all $i,j$, it follows that the matrix coefficients of $t^{k}$ occurring in (2) are all zero, i.e. $Q(t)$ is the zero matrix for all matrices $t$.
Taking $t=A$ we can also see that
$\displaystyle Q(A)$  $\displaystyle=$  $\displaystyle(D_{0}Ac_{0}I)+(D_{1}AD_{0}c_{1}I)A+\dots+(D_{{n1}}AD_{{n2}% }c_{{n1}}I)A^{{n1}}+(D_{{n1}}c_{n}I)A^{n}$  
$\displaystyle=$  $\displaystylec_{0}Ic_{1}A\dotsc_{{n1}}A^{{n1}}c_{n}A^{n}$  
$\displaystyle=$  $\displaystyle\widetilde{p}(A)$ 
Hence $\widetilde{p}(A)=0$, which finishes the proof. $\square$
$,$
Actually, $\mathbb{C}$ could have been substituted with $\mathbb{R}$ or $\mathbb{Q}$ in the proof above. The only property of $\mathbb{C}$ that was used is that it is an infinite integral domain.
Proof $\,$ (Proof for an arbitrary commutative ring with identity):
Let $A=(a_{{ij}})$, where the entries $a_{{ij}}$ are in commutative ring with identity $R$. First notice that, since $c_{0}+c_{1}\lambda+\dots+c_{n}\lambda^{n}=p(\lambda)=\det(A\lambda I)$, where $\lambda\in R$, we have that the coefficients $c_{0},\dots,c_{n}$ are polynomials in $\{a_{{ij}}\}$.
Hence $\widetilde{p}(A):=c_{0}I+c_{1}A+\dots+c_{n}A^{n}$ is a matrix whose entries are also polynomials in $\{a_{{ij}}\}$. These polynomials vanish for every assignment of $\{a_{{ij}}\}$ to numbers in $\mathbb{C}$, because the complex case of the theorem has already been proven (this would be the same as substituting the matrix $A$ by a matrix with complex entries). Therefore these polynomials are zero polynomials and we conclude that $\widetilde{p}(A)=0$ as we inteded to prove.$\square$
Comments on other proofs
Yet another proof of the CayleyHamilton Theorem is to establish it for diagonalizable matrices, and then by a density argument (i.e. every matrix can be approximated by diagonalizable ones in an algebraically closed field), we conclude that $p(A)=0$ is an identity for all matrices over a field. This kind of proof is also presented as an exercise in [1].
The “standard approach” can be found in [3]. It proves the result for matrices over any field, but requires no “abstract algebra” (no algebraic closure, Zariski topology or formal substitutions).
The two other proofs just mentioned can be extended to matrices over an arbitrary commutative ring simply by repeating the last argument in our proof.
References
 1 Michael Artin. Algebra. PrenticeHall, 1991.
 2 Martin Braun. Differential equations and their applications: an introduction to applied mathematics, 3rd edition. SpringerVerlag, 1983.
 3 Friedberg, Insel, Spence. Linear Algebra, 3rd edition. PrenticeHall, 1997.
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Comments
proof needs correction
The proof on this page is not quite correct. I found your page by a link from the wikipedia (http://en.wikipedia.org/wiki/CayleyHamilton_theorem), where I recently corrected the also erroneous proof that use to be there, and provided ample discussion of how and why, so please look there for details. When you say "we think of the formal symbol $t$ as representing an $n \times n$ matrix" you ignore the fact that already it has been treated as commuting with matrices, for instance by placing them arbitrarily at the left of the terms in the expansion of $B(t)$. So you simply cannot just substitute a matrix for $t$ and expect to get a valid identity, even if (like you do) you care about left and right _after_ the substitution.
A valid argument can be given by doing some noncommutative theory first, and carefully distinguishing leftevaluation from rightevaluation. For instance another proof I found on PlanteMath, http://planetmath.org/encyclopedia/ProofOfCayleyHamiltonTheoremInACommut... is basically correct, provided one supplies proper definitions of left factor and left hand value, and proves the result used (which however is about as hard as proving CayleyHamilton in the first place...).
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Re: proof needs correction
Hello van Leeuwen,
I think Steve's proof is some incomplete, but I cannot see mistakes there. My opinion is that one may circumvent commutativity issue. On the ring of matrices (an algebra) is valid X^{1}X=XX^{1}=I. The definition of matrix adjugate B(\lambda) leads to
(\lambda IA)B(\lambda)=\lambda IAI, (1)
and
B(\lambda)(\lambda IA)=\lambda IAI. (2)
In both equations the RHS as well as B(\lambda) on the LHS are polynomial matrices in \lambda. One see in those eq., that \lambda IAI us divisible without remainder. So by applying generalized Bezout theorem,
http://planetmath.org/encyclopedia/GeneralizedBezoutTheoremOnMatrices.html
CayleyHamilton may be proved.
perucho