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 $p(A) := c_0I + c_1 A + \dots + c_n A^n = 0$ .

Proof$\,$ (Popular fake proof):

In the expression

$\,$

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,

where is a matrix

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

(1)

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 $t$

(2)

whenever   commutes with

(3)

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

Taking $t = A$ we can also see that \begin{eqnarray*} Q(A) & = & (D_0A - c_0I)+ (D_1A-D_0-c_1I)A + \dots + (D_{n-1}A- D_{n-2} - c_{n-1}I)A^{n-1}+(-D_{n-1} - c_nI)A^n\\ & = & -c_0I - c_1A - \dots - c_{n-1}A^{n-1} - c_nA^n\\ & = & - \widetilde{p}(A) \end{eqnarray*} Hence $\widetilde{p}(A) = 0$ , which finishes the proof. $\square$

$,$

Actually, $\complex$ could have been substituted with $\real$ or $\rat$ in the proof above. The only property of $\complex$ 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 $\widetilde{p}(A):= c_0I + c_1A+ \dots +c_nA^n$ 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 Cayley-Hamilton 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.

In [2] there is a proof similar to the one presented here (although it has some errors).

Bibliography

1

Michael Artin. Algebra. Prentice-Hall, 1991.

2

Martin Braun. Differential equations and their applications: an introduction to applied mathematics, 3rd edition. Springer-Verlag, 1983.

3

Friedberg, Insel, Spence. Linear Algebra, 3rd edition. Prentice-Hall, 1997.