proof of Cayley-Hamilton theorem by formal substitutions



Let A be a n×n matrix with entries in a commutative ring with identityPlanetmathPlanetmath, and let p(λ)=c0+c1λ++cnλn be its characteristic polynomialMathworldPlanetmathPlanetmath. We will prove that p(A):=c0I+c1A++cnAn=0.

Proof (Popular fake proof):

In the expression


substitute t=A; then p(A)=det(A-AI)=det(0)=0.

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(λ) 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 p~(t) when applying the polymial p to a matrix t. In this sense p() is a function of and p~() is function of matrices. Also, by definition,

p~(t):=c0I+c1t++cntn,where t is a matrix

Of course, we intend to prove that p~(A)=0.

Proof (Proof of the complex case):

For each λ let B(λ) be the classical adjoint to A-λI. We have

B(λ)(A-λI)=det(A-λI)I=p(λ)I=c0I+λ(c1I)++λn(cnI). (1)

From the definition of the classical adjoint, it is clear that B(λ) can be written as a polynomialMathworldPlanetmathPlanetmathPlanetmath of λ having degree n-1, whose coefficients are matrices. That is,


for some constant coefficient matrices Di.

Now, we would like B to be defined for matrices (just like from the polynomial p we have considered p~), so we define for every matrix t


Now consider the following function of matrices:

Q(t):=(D0A-c0I)+(D1A-D0-c1I)t++(Dn-1A-Dn-2-cn-1I)tn-1+(-Dn-1-cnI)tn (2)

The above expression may look strange, but if we think that the matrix t commutes with all D0,,Dn-1 and A, then expression (2) is easily seen to be equal to


This means that

Q(t)=B~(t)(A-It)-p~(t)   whenevertcommutes withD0,,Dn-1,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 tk).

We now state some properties that can be easily checked by straightforward calculation:

  • p~(λI)=p(λ)I

  • B~(λI)=B(λ)

Notice that matrices of the form λI with λ commute with every other matrix, so that


Now Q(λI) is also a matrix whose entries qij(λ) are polynomials in λ. Since Q(λI)=0 we must have qij(λ)=0 for all λ. This means that qij(t) is the zero polynomialMathworldPlanetmathPlanetmath and, since this occurs for all i,j, it follows that the matrix coefficients of tk occurring in (2) are all zero, i.e. Q(t) is the zero matrixMathworldPlanetmath for all matrices t.

Taking t=A we can also see that

Q(A) = (D0A-c0I)+(D1A-D0-c1I)A++(Dn-1A-Dn-2-cn-1I)An-1+(-Dn-1-cnI)An
= -c0I-c1A--cn-1An-1-cnAn
= -p~(A)

Hence p~(A)=0, which finishes the proof.


Actually, could have been substituted with or in the proof above. The only property of that was used is that it is an infinite integral domain.

Proof (Proof for an arbitrary commutative ring with identity):

Let A=(aij), where the entries aij are in commutative ring with identity R. First notice that, since c0+c1λ++cnλn=p(λ)=det(A-λI), where λR, we have that the coefficients c0,,cn are polynomials in {aij}.

Hence p~(A):=c0I+c1A++cnAn is a matrix whose entries are also polynomials in {aij}. These polynomials vanish for every assignment of {aij} to numbers in , 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 p~(A)=0 as we inteded to prove.

Comments on other proofs

Yet another proof of the Cayley-Hamilton TheoremMathworldPlanetmath ( is to establish it for diagonalizable matricesMathworldPlanetmath, 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 closureMathworldPlanetmath, 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).


  • 1 Michael Artin. AlgebraMathworldPlanetmathPlanetmathPlanetmath. 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 AlgebraMathworldPlanetmath, 3rd edition. Prentice-Hall, 1997.
Title proof of Cayley-Hamilton theorem by formal substitutions
Canonical name ProofOfCayleyHamiltonTheoremByFormalSubstitutions
Date of creation 2013-03-22 15:27:28
Last modified on 2013-03-22 15:27:28
Owner asteroid (17536)
Last modified by asteroid (17536)
Numerical id 19
Author asteroid (17536)
Entry type Proof
Classification msc 15A18
Classification msc 15A15