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characteristic polynomial

Characteristic polynomial and Cayley-Hamilton theorem. See also determinant.

Lemma

Let R be a commutative ring, and let A be an n×n matrix with entries in R. Then there exists an n×n matrix A˜ with entries in R such that AA˜=A˜A=det(A)I n.

Proof

We may as well take R to be the polynomial ring [a ij] 1i,jn, since we are then free to interpret the indeterminates a ij however we like along a ring map [a ij]R. Let A denote the corresponding generic matrix.

Guided by Cramer’s rule, put

A˜ ji=det(a 1,,e i,a n),\tilde{A}_{j i} = \det(a_1, \ldots, e_i, \ldots a_n),

the a i being columns of A and e i, the column vector with 1 in the i th row and 0’s elsewhere, appearing as the j th column. If we pretend A is invertible, then we know AA˜=det(A)I n=A˜A by Cramer’s rule. We claim this holds for general A.

Indeed, we can interpret this as a polynomial equation in [a ij] and check it there. As an equation between polynomial functions on the space of matrices AMat n()=Spec([a ij]), it holds on the dense subset GL n()Mat n(). Therefore, by continuity, it holds on all of Mat n(). But a polynomial function equation with coefficients in implies the corresponding polynomial identity, and the proof is complete.

Theorem

(Cayley-Hamilton) Let V be a finitely generated free module over a commutative ring R, and let f:VV be an R-module map. Let p(t)R[t] be the characteristic polynomial det(t1 Vf) of f, and let ϕ f:R[t]Mod R(V,V) be the unique R-algebra map sending t to f. Then p(f)ϕ f(p) is the zero map 0:VV.

Proof

Via ϕ f, regard V as an R[t]-module, and with regard to some R-basis {v i} 1in of V, represent f by a matrix A. Now consider tI nA as an n×n matrix B(t) with entries in R[t]. By definition of the module structure, this matrix B(t), seen as acting on V n, annihilates the length n column vector c whose i th row entry is v i.

By the previous lemma, there is B˜(t) such that B˜(t)B(t) is det(tI nA) times the identity matrix. It follows that

det(tI nA)c=B˜(t)B(t)c=B˜(t)0=0\det(t \cdot I_n - A) c = \tilde{B}(t) B(t) c = \tilde{B}(t) 0 = 0

i.e., det(tI nA)v i=0 for each i. Since the v i form an R-basis, the R[t]-scalar det(tI nA) annihilates the R[t]-module V, as was to be shown.

The Cayley-Hamilton theorem easily generalizes to finitely generated R-modules (not necessarily free) as follows. Let f:VV be a module endomorphism, and suppose π:R nV is an epimorphism. Since R n is projective, the map fπ can be lifted through π to a map A:R nR n. Let P(t) be the characteristic polynomial of A.

Proposition

P(f)=0.

Proof

Write P(t)= ia it i. We already know P(A)=0. From fπ=πA, it follows that f iπ=πA i for any i0. Hence P(f)π=πP(A)=0. Since π is epic, P(f)=0 follows.

We give an interesting and perhaps surprising consequence of the Cayley-Hamilton theorem below, after establishing a lemma close in spirit to Nakayama's lemma.

Lemma

Suppose V is a finitely generated R-module, and g:VV is a module map such that g(V)IV for some ideal I of R. Then there is a polynomial p(t)=t n+a 1t n1++a n, with all a iI, such that p(g)=0.

Proof

For some finite n0, we have a surjective map R nM, and by hypothesis we have a surjective map I nim(g) in

I n R n M g im(g)\array{ & & & & I^n \\ & & & & \downarrow \\ R^n & \to & M & \stackrel{g}{\to} & im(g) }

By projectivity of R n, we can lift the bottom composite to a map R nI n making the diagram commute. Let A be the R-module map R nI nR n, regarded as a matrix. Then the characteristic polynomial of A satisfies the conclusion, by the Cayley-Hamilton theorem.

Proposition

Let V be a finitely generated module over a commutative ring R, and let f:VV be a surjective module map. Then f is an isomorphism.

Proof

Regard V as a finitely generated R[t]-module via ϕ f:R[t]Mod R(V,V). Since f is assumed surjective, we have IV=V for the ideal I=(t) of R[t]. Now take g=1 V as in the preceding lemma, a module map over the ring R=R[t]. By the lemma, we see that g n+a 1g n1++a n=0 where a i(t), in other words the R[t]-scalar

(1+a 1++a n)1 V=0(1 + a_1 + \ldots + a_n)1_V = 0

as an operator on V. Write a i=b i(t)t for polynomials b i(t)R[t]. Now we may rewrite the previous displayed equation as

1 V(v)=( ib i(t))tv1_V(v) = -(\sum_i b_i(t)) t \cdot v

for all vV, which translates into saying that 1 V= ib i(f)f, i.e., that ib i(f) is a retraction of f. Since f is epic, we now see f is an isomorphism.

Literature

The proof of the Cayley-Hamilton theorem follows the treatment in

  • Serge Lange, Algebra (3 rd edition), Addison-Wesley, 1993.

Revised on April 19, 2013 21:16:14 by Todd Trimble (67.81.93.26)