3. Cayley-Hamilton and McCoy Theorems

A little excursion into linear algebra.

For R a commutative ring, let M_{n}(R) be the R-algebra of n × n matrices with entries in R. We denote the identity matrix by I.

3.1 Lemma. Let R be a commutative ring. For A = (a_{ij}) ∈ M_{n}(R), define the adjoint of A as adj(A) = B = (b_{ij}) ∈ M_{n}(R) with b_{ij} = (−1)^{i+j} det(A_{j}^{i}), where A_{j}^{i} ∈ M_{n−1}(R) results from A by removing the i-th column and the j-th row.

Then

AB = BA = (det A)I.

Proof. Routine verification. □

The adjoint of a matrix can be used to give a short proof of Cayley-Hamilton’s theorem. It uses the natural R-algebra homomorphism between M_{n}(R[x]) and M_{n}(R)[x], which maps the matrix whose (i, j)-th entry is ∑_{k} a_{k}^{(i,j)}x^{k} ∈ R[x], to the polynomial ∑_{k} A_{k}x^{k}, where A_{k} ∈ M_{n}(R) is the matrix whose (i, j)-th entry is a_{k}^{(i,j)}.

If S is a non-commutative ring, such as M_{n}(R), and we want to substitute a ring element s for the variable in f(x) = ∑ a_{k}x^{k} ∈ S[x], we must specify whether substitution happens on the right or on the left of the coefficients, i.e., if f(s) means ∑ a_{k}s^{k} (right substitution) or ∑ s^{k}a_{k} (left substitution).

In any case, it is important to remember that substitution (whether right or left) is in general not a homomorphism: if f, g ∈ S[x] and s ∈ S, it is in general not true that (f · g)(s) = f(s) · g(s) (unless s commutes with the coefficients of f and g). One thing that does work similarly to polynomials over commutative rings is the correspondence between zeros and linear factors:

3.2 Lemma. Let S be a (possibly non-commutative) ring and f = ∑ a_{k}x^{k} ∈ S[x]. For s ∈ S define f(s) = ∑ a_{k}s^{k}. Then f(s) = 0 if and only if f(x) = g(x)(x − s) for some g ∈ S[x].

Proof. If f(s) = 0 then f(x) = f(x)−f(s) = ∑_{k} a_{k}x^{k} −∑_{k} a_{k}s^{k} = ∑_{k} a_{k}(x^{k} −s^{k}) = ∑_{k} a_{k}(∑_{j=0}^{k−1} s^{j}x^{k−j−1})(x − s) = g(x)(x − s).

If, on the other hand, f(x) = g(x)(x − s) with g(x) = ∑_{k=0}^{n} b_{k}x^{k} then f(x) = ∑_{k=0}^{n} b_{k}x^{k+1} − ∑_{k=0}^{n} b_{k}sx^{k} = ∑_{k=0}^{n+1}(b_{k−1} − b_{k}s)x^{k} and right substitution yields f(s) = 0. □

3.3 Theorem (Cayley-Hamilton). Let C be a n × n matrix with entries in a commutative ring R, and χ(x) = det(xI − C) its characteristic polynomial. Then χ(C) = 0.

Proof. Let B ∈ M_{n}(R[x]) be the adjoint of xI − C then

B · (xI − C) = χ(x)I.

Applying the R-algebra isomorphism M_{n}(R[x]) ≃ M_{n}(R)[x], we get

B(x)(x − C) = χ(x),

where B(x) ∈ M_{n}(R)[x] is the polynomial corresponding to the matrix B ∈ M_{n}(R[x]). Now χ(x) has a factor x − C in M_{n}(R)[x], and therefore χ(C) = 0. □

With the argument we used to prove Cayley-Hamilton (3.3), one can actually show a stronger result, characterizing the ideal of those polynomials in R[x], which have a given matrix C ∈ M_{n}(R) as a zero:

3.4 Theorem (McCoy). Let C be a n × n matrix with entries in a commutative ring R. For 1 ≤ k ≤ n let J_{k}(xI − C) be the ideal of R[x] generated by the k × k minors of xI −C. Then the ideal of all polynomials f ∈ R[x] with f(C) = 0 is equal to the ideal quotient (J_{n}(xI −C) : J_{n−1}(xI −C)) = (χ_{C}(x)R[x] :_{R[x]} J_{n−1}(xI −C)).

Proof. Since the generators of J_{n−1}(xI − C), the (n − 1) × (n − 1)-minors of (xI − C), are (up to sign) the entries of adj(xI − C)),

g(x) ∈ (χ(x)R[x] :_{R[x]} J_{n−1}(xI − C))

is equivalent to

g(x)adj(xI − C) ∈ χ(x)M_{n}(R[x]).

Multiplying by (xI − C), which, being monic, is certainly not a zero-divisor in M_{n}(R)[x], we again get an equivalent statement:

g(x)χ(x)I ∈ χ(x)M_{n}(R[x])(xI − C).

We may cancel the scalar χ(x) ∈ R[x] (a monic polynomial - not a zero-divisor in R[x]) to arrive at the equivalent statement

g(x)I ∈ M_{n}(R[x])(xI − C),

and apply the isomorphism between M_{n}(R[x]) and M_{n}(R)[x] to get

g(x) ∈ M_{n}(R)[x](x − C),

which is (by 3.2) equivalent to C being a zero of g(x). □

From Cayley-Hamilton theorem for matrices it is possible to derive a version of Cayley-Hamilton theorem for linear operators, by associating matrices to linear operators:

Suppose M is an R-module generated by m_{1}, …, m_{n}. Then every b ∈ M is representable as b = b_{1}m_{1} + … + b_{n}m_{n} with (in general non-unique) b_{1}, …, b_{n} in R. Let φ ∈ End_{R}(M). We say that a matrix C ∈ M_{n}(R) represents φ with respect to the system of generators m_{1}, …, m_{n}, if for every b ∈ M, b = b_{1}m_{1} + … + b_{n}m_{n} implies φ(b) = b′_{1}m_{1} + … + b′_{n}m_{n} with C[b_{1}, …, b_{n}] = [b′_{1}, …, b′_{n}]. (We use square brackets to denote column vectors: [b_{1}, …, b_{n}] means the transpose of (b_{1}, …, b_{n}). It is easy to see that C represents φ if and only if the k-th column of C is a coordinate vector of φ(m_{k}), i.e., if φ(m_{k}) = c_{1k}m_{1} + c_{2k}m_{2} + … + c_{nk}m_{n}.

If the generators m_{1}, …, m_{n} are not R-linearly independent, then the matrix associated to a linear operator φ ∈ End_{R}(M) is non-unique, and not every matrix in M_{n}(R) represents a linear operator. If C represents φ and D represents ψ, however, then C + D and CD represent φ + ψ and φ ∘ ψ, respectively, and, for r ∈ R, rC represents rφ, so the set E ⊆ M_{n}(R) of matrices representing R-endomorphisms of M is a R-subalgebra of M_{n}(R), and there is a surjective R-algebra homomorphism π: E → End_{R}(M), mapping every matrix in E to the endomorphism it represents. In particular, for every matrix C represening some φ∈End_{R}(M), restriction of π to R[C] gives a surjective R-algebra homomorphism π: R[C] → R[φ], mapping f(C) ∈ R[C] to f(φ) ∈ R[φ] for every f ∈ R[x].

3.5 Lemma. Let M be an R-module generated by n elements and φ ∈ End_{R}(M). If C ∈ M_{n}(R) is a matrix associated to φ (with respect to a system of generators m_{1}, …, m_{n} ∈ M) and χ(x) the characteristic polynomial of C then χ(φ) = 0.

Proof. By Cayley-Hamilton (3.3), χ(C) = 0, where χ = det(xI − C) is the characteristic polynomial of C. Applying the R-algebra homomorphism π: R[C]→ R[φ] mentioned above, we see that 0 = π(χ(C)) = χ(φ). □

3.6 Corollary. Let M be an R-module generated by n elements and φ∈End_{R}(M), such that φ(M) ⊆ JM for some ideal J ⊴ R. Then φ is a zero of a monic polynomial in R[x] of degree n, f(x) = x^{n} + a_{n−1}x^{n−1} + … + a_{1}x + a_{0} with a_{k} ∈ J^{n−k} for 0 ≤ k < n.

Proof. If m_{1}, …, m_{n} generate M as a R-module and φ(M) ⊆ JM then there exist c_{ij} ∈ J with φ(m_{k}) = c_{1k}m_{1} +c_{2k}m_{2} +…+c_{nk}m_{n}. Now χ(x) = det(xI −C), the characteristic polynomial of C = (c_{ij}) is a monic polynomial of degree n with χ(φ) = 0 and the property that the coefficient of x^{k} is in J^{n−k}. □