Posts Tagged ‘Cayley-Hamilton theorem’

Prp1. Let K be a field and S=M_n(K). The characteristic polynomial and the reduced characteristic polynomial of an element of S are equal.

Proof. Well, f: S \otimes_K K \longrightarrow S = M_n(K) defined by f(a \otimes_K \alpha) = \alpha a, \ a \in S, \alpha \in K, is a K-algebra isomorphism. Thus \text{Prd}_S(a,x) = \det(xI - f(a \otimes_K 1)) = \det(xI - a). \ \Box

Prp2. (Cayley-Hamilton) Let A be a finite dimensional central simple k-algebra and a \in A. Then \text{Prd}_A(a,a)=0.

Proof. Let K be a splitting field of A with a K-algebra isomorphism f: A \otimes_k K \longrightarrow M_n(K). We have \text{Prd}_A(a,x)=\det(xI-f(a \otimes_k 1))=x^n + \alpha_{n-1}x^{n-1} + \ldots + \alpha_1 x + \alpha_0 \in k[x]. Since \text{Prd}_A(a,x) is just the characteristic polynomial of f(a \otimes_k 1) \in M_n(K), we may apply the Cayley-Hamilton theorem from linear algebra to get (f(a \otimes_k 1))^n + \alpha_{n-1}(f(a \otimes_k 1))^{n-1} + \ldots + \alpha_1 f(a \otimes_k 1) + \alpha_0=0. Thus, since f is a K-algebra homomorphism, we get

f((a^n + \alpha_{n-1}a^{n-1} + \ldots + \alpha_1 a + \alpha_0) \otimes_k 1)=0.

Hence, since f is injective, we must have (a^n + \alpha_{n-1}a^{n-1} + \ldots + \alpha_1 a + \alpha_0) \otimes_k 1 = 0 which implies

\text{Prd}_A(a,a)=a^n + \alpha_{n-1}a^{n-1} + \ldots + \alpha_1 a + \alpha_0 = 0. \ \Box

Prp3. Reduced characteristic polynomials are invariant under extension of scalars.

Proof. First let’s understand the question! We have a finite dimensional central simple k-algebra A with a \in A. We are asked to prove that if K/k is a field extension and B= A \otimes_k K, then \text{Prd}_B(a \otimes_k 1, x) = \text{Prd}_A(a,x). Note that B is a central simple K-algebra and \deg A = \deg B = n. Now, let L be a splitting field of B with an L-algebra isomorphism f : B \otimes_K L \longrightarrow M_n(L). But

B \otimes_K L =(A \otimes_k K) \otimes_K L \cong A \otimes_k (K \otimes_K L) \cong A \otimes_k L.

So we also have an isomorphism g : A \otimes_k L \longrightarrow M_n(L). Clearly f((a \otimes_k 1) \otimes_K 1)=g(a \otimes_k 1) and hence \text{Prd}_B(a \otimes_k 1, x) = \text{Prd}_A(a,x). \ \Box

Prp4. Reduced characteristic polynomials are invariant under isomorphism of algebras.

Proof. So A_1,A_2 are finite dimensional central simple k-algebras and f : A_1 \longrightarrow A_2 is a k-algebra isomorphism. We want to prove that \text{Prd}_{A_1}(a_1,x) = \text{Prd}_{A_2}(f(a_1),x) for all a_1 \in A_1. Let K be a splitting field of A_2 with a K-algebra isomorphism g : A_2 \otimes_k K \longrightarrow M_n(K). The map f \otimes \text{id}_K: A_1 \otimes_k K \longrightarrow A_2 \otimes_k K is also a K-algebra isomorphism. Thus we have a K-algebra isomorphism h=g \circ (f \otimes \text{id}_K) : A_1 \otimes_k K \longrightarrow M_n(K). So if a_1 \in A_1, then

\text{Prd}_{A_2}(f(a_1),x)=\det (xI - g(f(a_1) \otimes_k 1))=\det (xI - h(a_1 \otimes_k 1)) = \text{Prd}_{A_1}(a_1,x). \ \Box


Problem. Let A and B be n \times n matrices with entries from some field k. Let c_1, \cdots , c_{n+1} be n+1 distinct elements  in k such that A+c_1B, \cdots , A+c_{n+1}B are all nilpotent. Prove that A and B are nilpotent.

Solution. Since A + c_1B, \cdots , A+c_{n+1}B are nilpotent, their eigenvalues are all zero and hence, by the Cayley-Hamilton theorem, (A+c_iB)^n=0, for all 1 \leq i \leq n+1. Let x be an indeterminate. So the equation

(A+xB)^n = 0 \ \ \ \ \ \ \ \ \ \ \ \ (1)

has at least n+1 roots x =c_1, \cdots , c_{n+1} in k. We will show that this is not possible unless A^n=B^n=0, which will complete the solution.  To do so, let’s expand the left hand side in (1) to get

B^nx^n + D_1x^{n-1} + \cdots + D_{n-1}x + A^n = 0, \ \ \ \ \ \ \ \ \ \ \ \ \ (2),

where each D_i is in the k-algebra generated by A and B. Let 1 \leq i,j \leq n and let a_{ij} and b_{ij} be the (i,j)-entries of A^n and B^n respectively. Then the (i,j)-entry of the matrix on the left hand side of (2) is

b_{ij}x^n + d_1x^{n-1} + \cdots + d_{n-1}x + a_{ij}=0, \ \ \ \ \ \ \ \ \ \ \ \ \ (3)

where d_1, \cdots , d_{n-1} are the (i,j)-entries of D_1, \cdots , D_{n-1}. So (3) is telling us that the polynomial p(x) = b_{ij}x^n + d_1x^{n-1} + \cdots + d_{n-1}x + a_{ij} \in k[x], which has degree at most n, has at least n+1 roots in k. This is not possible unless p(x)=0. Thus a_{ij}=b_{ij}=0. \ \Box