Suppose $A\in R^{n\times n}$, where $R$ is a commutative ring. Let $p\_i \in R$ be the coefficients of the characteristic polynomial of $A$: $\operatorname{det}(A-xI) = p\_0 + p\_1x + \dots + p\_n x^n$.


I am looking for a proof that
$-\operatorname{adj}(A) = p\_1 I + p\_2 A + \dots + p\_n A^{n-1}$.


In the case where $\operatorname{det}(A)$ is a unit, $A$ is invertible, and the proof follows from the Cayley–Hamilton theorem. But what about the case where $A$ is not invertible?



Expressing $-\operatorname{adj}(A)$ as a polynomial in $A$?
Here is a direct proof along the lines of the standard proof of the Cayley–Hamilton theorem. [*This works universally, i.e. over the commutative ring $R=\mathbb{Z}[a\_{ij}]$ generated by the entries of a generic matrix $A$.*]


The following lemma combining Abel's summation and Bezout's polynomial remainder theorem is immediate.


**Lemma** Let $A(\lambda)$ and $B(\lambda)$ be matrix polynomials over a (noncommutative) ring $S.$ Then $A(\lambda)B(\lambda)-A(0)B(0)=\lambda q(\lambda)$ for a polynomial $q(\lambda)\in S[\lambda]$ that can be expressed as


$$q(\lambda)=A(\lambda)\frac{B(\lambda)-B(0)}{\lambda}+\frac{A(\lambda)-A(0)}{\lambda}B(0)=A(\lambda)b(\lambda)+a(\lambda)B(0) \qquad (\*)$$


with $a(\lambda),b(\lambda)\in S[\lambda].$


  

Let $A(\lambda)=A-\lambda I\_n$ and $B(\lambda)=\operatorname{adj} A(\lambda)$ [*viewed as elements of $S[\lambda]$ with $S=M\_n(R)$*], then 


$$A(\lambda)B(\lambda)=\det A(\lambda)=p\_A(\lambda)=p\_0+p\_1\lambda+\ldots+p\_n\lambda^n$$ is the characteristic polynomial of $A$ and $$A(0)B(0)=p\_0 \text{ and } q(\lambda)=p\_1+\ldots+p\_n\lambda^{n-1}$$ 


Applying $(\*),$ we get


$$q(\lambda)=(A-\lambda I)b(\lambda)-\operatorname{adj} A \qquad (\*\*) $$


for some matrix polynomial $b(\lambda)$ commuting with $A.$ Specializing $\lambda$ to $A$ in $(\*\*),$ we conclude that


$$q(A)=-\operatorname{adj} A\qquad \square$$


