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characteristic polynomial

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

Lemma

Let RR be a commutative ring, and let AA be an n×nn \times n matrix with entries in RR. Then there exists an n×nn \times n matrix A˜\tilde{A} with entries in RR such that AA˜=A˜A=det(A)I nA \tilde{A} = \tilde{A} A = \det(A) \cdot I_n.

Proof

We may as well take RR to be the polynomial ring [a ij] 1i,jn\mathbb{Z}[a_{i j}]_{1 \leq i, j \leq n}, since we are then free to interpret the indeterminates a ija_{i j} however we like along a ring map [a ij]R\mathbb{Z}[a_{i j}] \to R. Let AA 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 ia_i being columns of AA and e ie_i, the column vector with 11 in the i thi^{th} row and 00‘s elsewhere, appearing as the j thj^{th} column. If we pretend AA is invertible, then we know AA˜=det(A)I n=A˜AA \tilde{A} = \det(A) \cdot I_n = \tilde{A} A by Cramer’s rule. We claim this holds for general AA.

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

Theorem

(Cayley-Hamilton) Let VV be a finitely generated free module over a commutative ring RR, and let f:VVf \colon V \to V be an RR-module map. Let p(t)R[t]p(t) \in R[t] be the characteristic polynomial det(t1 Vf)\det(t \cdot 1_V - f) of ff, and let ϕ f:R[t]Mod R(V,V)\phi_f \colon R[t] \to Mod_R(V, V) be the unique RR-algebra map sending tt to ff. Then p(f)ϕ f(p)p(f) \coloneqq \phi_f(p) is the zero map 0:VV0 \colon V \to V.

Proof

Via ϕ f\phi_f, regard VV as an R[t]R[t]-module, and with regard to some RR-basis {v i} 1in\{v_i\}_{1 \leq i \leq n} of VV, represent ff by a matrix AA. Now consider tI nAt \cdot I_n - A as an n×nn \times n matrix B(t)B(t) with entries in R[t]R[t]. By definition of the module structure, this matrix B(t)B(t), seen as acting on V nV^n, annihilates the length nn column vector cc whose i thi^{th} row entry is v iv_i.

By the previous lemma, there is B˜(t)\tilde{B}(t) such that B˜(t)B(t)\tilde{B}(t) B(t) is det(tI nA)\det(t \cdot I_n - A) 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\det(t \cdot I_n - A) \cdot v_i = 0 for each ii. Since the v iv_i form an RR-basis, the R[t]R[t]-scalar det(tI nA)\det(t \cdot I_n - A) annihilates the R[t]R[t]-module VV, as was to be shown.

The Cayley-Hamilton theorem easily generalizes to finitely generated RR-modules (not necessarily free) as follows. Let f:VVf \colon V \to V be a module endomorphism, and suppose π:R nV\pi \colon R^n \to V is an epimorphism. Since R nR^n is projective, the map fπf \circ \pi can be lifted through π\pi to a map A:R nR nA \colon R^n \to R^n. Let P(t)P(t) be the characteristic polynomial of AA.

Proposition

P(f)=0P(f) = 0.

Proof

Write P(t)= ia it iP(t) = \sum_i a_i t^i. We already know P(A)=0P(A) = 0. From fπ=πAf \circ \pi = \pi \circ A, it follows that f iπ=πA if^i \circ \pi = \pi \circ A^i for any i0i \geq 0. Hence P(f)π=πP(A)=0P(f) \circ \pi = \pi \circ P(A) = 0. Since π\pi is epic, P(f)=0P(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 VV is a finitely generated RR-module, and g:VVg \colon V \to V is a module map such that g(V)IVg(V) \subseteq I V for some ideal II of RR. Then there is a polynomial p(t)=t n+a 1t n1++a np(t) = t^n + a_1 t^{n-1} + \ldots + a_n, with all a iIa_i \in I, such that p(g)=0p(g) = 0.

Proof

For some finite n0n \geq 0, we have a surjective map R nMR^n \to M, and by hypothesis we have a surjective map I nim(g)I^n \to im(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 nR^n, we can lift the bottom composite to a map R nI nR^n \to I^n making the diagram commute. Let AA be the RR-module map R nI nR nR^n \to I^n \hookrightarrow R^n, regarded as a matrix. Then the characteristic polynomial of AA satisfies the conclusion, by the Cayley-Hamilton theorem.

Proposition

Let VV be a finitely generated module over a commutative ring RR, and let f:VVf \colon V \to V be a surjective module map. Then ff is an isomorphism.

Proof

Regard VV as a finitely generated R[t]R[t]-module via ϕ f:R[t]Mod R(V,V)\phi_f \colon R[t] \to Mod_R(V, V). Since ff is assumed surjective, we have IV=VI V = V for the ideal I=(t)I = (t) of R[t]R[t]. Now take g=1 Vg = 1_V as in the preceding lemma, a module map over the ring R=R[t]R' = R[t]. By the lemma, we see that g n+a 1g n1++a n=0g^n + a_1 g^{n-1} + \ldots + a_n = 0 where a i(t)a_i \in (t), in other words the R[t]R[t]-scalar

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

as an operator on VV. Write a i=b i(t)ta_i = b_i(t) t for polynomials b i(t)R[t]b_i(t) \in 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 vVv \in V, which translates into saying that 1 V= ib i(f)f1_V = -\sum_i b_i(f) f, i.e., that ib i(f)-\sum_i b_i(f) is a retraction of ff. Since ff is epic, we now see ff is an isomorphism.

Literature

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

  • Serge Lange, Algebra (3 rd3^{rd} edition), Addison-Wesley, 1993.

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