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Here is a statement of a standard theorem in commutative algebra (see page 60 of this book):

Theorem. ("Determinant Trick") Suppose that $R$ is a commutative ring with $1$. Let $M$ be a finitely generated $R$-module, generated by $m_{1}, …, m_{n}\in M$. Let $\varphi: M\to M$ is an endomorphism, and assume that $\varphi(m_{i})=\sum_{j=1}^{n} a_{ij} m_{j}$ for each $i=1, 2, …, n$. Then if we let $A=(a_{ij})$, then $p_{A}(\varphi)=0$ where $p_{A}$ is the characteristic polynomial of the matrix $A$.

Now authors (in the linked book above) claim that the usual Cayley-Hamilton theorem (which states that $p_{A}(A)=0$) follows directly from the Determinant Trick by letting $M$ to be generated by standard basis vectors $e_{i}=(0, …, 1, … 0)$, and letting $\varphi: M\to M$ to be $\varphi(m)=Am$ for each $m\in M$. But I think there is one more step that needs to be justified here:

But if $A=(a_{ij})$, then $Ae_{i}=\sum_{j=1}^{n} a_{ji} e_{j}$ and not $Ae_{i}=\sum_{j=1}^{n} a_{ij} e_{j}$ which is the form the determinant trick applies. I think authors are implicitly using the fact that characteristic polynomials of $A$ and $A^{T}$ (transpose) are equal. Am I correct?

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  • $\begingroup$ Yes.$~~~~~~~~~$ $\endgroup$ – Sammy Black Aug 14 '14 at 7:03
  • $\begingroup$ The formula $\varphi(m_{i})=\sum_{j=1}^{n} a_{ij} m_{j}$ makes sense for endomorphisms acting on the right, in which case you probably ought to write something like $(m_i)\varphi$. $\endgroup$ – Sammy Black Aug 14 '14 at 7:07
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    $\begingroup$ @SammyBlack: Hmmm, why does it make sense only for those endomorphisms acting on the right? $\endgroup$ – Prism Aug 14 '14 at 7:22

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