State and prove the Cayley–Hamilton Theorem for an \(n \times n\) matrix \(A\) over a commutative ring \(R\).
Solution
Let \[ p(t)=\det(tI-A)=t^n+c_{n-1}t^{n-1}+\cdots+c_0. \] Over the commutative ring \(R[t]\), the adjugate identity gives \[ (tI-A)\operatorname{adj}(tI-A)=p(t)I. \] Write \[ \operatorname{adj}(tI-A)=B_{n-1}t^{n-1}+\cdots+B_0, \qquad B_i\in M_n(R). \] Comparing coefficients of powers of \(t\) yields \[ B_{n-1}=I, \] \[ B_{k-1}-AB_k=c_kI\qquad(1\le k\le n-1), \] and \[ -AB_0=c_0I. \] Multiply these identities successively by the appropriate powers of \(A\) and add. All terms involving the \(B_i\) telescope, leaving \[ A^n+c_{n-1}A^{n-1}+\cdots+c_0I=0. \] Thus \[ \boxed{p(A)=0}, \] which is the Cayley–Hamilton theorem over any commutative ring.