Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 4 / 11E (Groups, Rings and Modules)

Let $S=R[\varphi ]$ be the commutative subring of endomorphisms generated by scalar multiplication and $\varphi$. Commutativity follows because the scalar maps commute with the $R$-linear map $\varphi$, and every element of $S$ is a polynomial in $\varphi$ with scalar coefficients. Thus determinants and the adjugate identity apply to matrices over $S$.

With the convention for the coefficients in the question, the column of generators satisfiesMultiply by $\mathrm{adj}(C)$ over $S$. Since $\mathrm{adj}(C)C=(\det C)I_n$, the endomorphism $\det C$ kills every generator $m_i$. It therefore kills every element $\sum r_i{m}_i$ of the module, by $R$-linearity. Hence it is the zero endomorphism, andThe argument does not require the generating set to be a basis; relations among the generators cause no difficulty. This is the determinant trick.

A finite basis is a generating set, so the preceding determinant trick applies with $R=k$ and $M=V$. If $p(t)=\det (A-tI)$, the map $k[t]\to k[\varphi ]$ sending $t$ to $\varphi$ is a ring homomorphism. Since a determinant is a polynomial expression in its entries, it sends $p(t)$ to $\det (A-\varphi I)=0$. ThusWith the more usual monic characteristic polynomial ${\chi }_A(t)=\det (tI-A)=(-1{)}^{\dim V}p(t)$, this reads ${\chi }_A(\varphi )=0$, the Cayley-Hamilton theorem. If the matrix convention uses column coordinates instead of the row convention above, transpose $A$; its characteristic polynomial is unchanged.