# Help with Cayley-Hamilton determinant trick theorem from Matsumura's Commutative Algebra.

Theorem 2.1. Suppose that $$M$$ is an $$A$$-module generated by $$n$$ elements, and that $$\varphi \in \text{Hom}_A(M,M)$$; let $$I$$ be an ideal of $$A$$ such that $$\varphi(M) \subset IM$$. Then there is a relation of the form: $$\varphi^n + a_1 \varphi^{n-1} + \dots + a_{n-1} \varphi + a_n = 0,$$ with $$a_i \in I$$ for $$1 \leq i \leq n$$ (where both sides are considered as endomorphisms of $$M$$).

Proof. Let $$M = A w_1 + \dots Aw_n$$. By the assumption $$\varphi(M) \subset IM$$ there exist $$a_{ij} \in I$$ such that $$\varphi(w_i) = \sum_{j=1}^n a_{ij} w_j$$. This can be rewritten $$\sum_{j=1}^n (\varphi \delta_{ij} - a_{ij}) w_j = 0, \ i=1..n$$. The coefficients of this systme of linear equations can be viewed as a square matrix $$(\varphi \delta_{ij} - a_{ij})$$ of elements of $$A'[\varphi]$$, the commutative subring of the endomorphisms ring $$E = \text{Hom}_A(M,M)$$ generated by the image $$A'$$ of $$A$$ under $$a \mapsto (x \mapsto ax)$$, together with $$\varphi$$. Let $$b_{ij}$$ denote its $$(i,j)$$th cofactor, and $$d$$ its determinant. Multiply the equation through by $$b_{ik}$$ and sum over $$i$$: $$\sum_{i=1}^n b_{ik}(\sum_{j=1}^n (\varphi \delta_{ij} - a_{ij}) w_j ) = 0$$

For example, if $$B = (\varphi \delta_{ij} - a_{ij})$$ is a $$3 \times 3$$ matrix we have:
$$B = -\begin{pmatrix} a_{11} - \varphi & a_{12} & a_{13} \\ a_{21} & a_{22} -\varphi & a_{23} \\ a_{31} & a_{32} & a_{33} - \varphi \end{pmatrix}$$

Then for example at $$i = 1$$ we have the following cofactors: $$b_{11} = -\det\begin{pmatrix} a_{22} - \varphi & a_{23} \\ a_{32} & a_{33} - \varphi \end{pmatrix}, b_{12} = \det\begin{pmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} - \varphi \end{pmatrix}, b_{13} = -\det\begin{pmatrix} a_{21} & a_{22} - \varphi \\ a_{31} & a_{32} \end{pmatrix}$$ Then the sum:

$$\sum_{i=1}^n b_{ik} \sum_{j=1}^n B_{ij} w_j = 0 \iff \\ \sum_{j=1}^n \sum_{i=1}^n b_{ik} B_{ij} w_j = 0 \iff \\ \sum_{j=1}^n (\det B) w_j = 0 \iff \\ (\det B) \sum_{j=1}^n w_j = 0 \ \ \textbf{( wrong here )}$$

I believe it should be $$(\det B) w_j = 0, \ \forall j=1..n$$ instead, as in this lecture video.

I followed the recipe to multiply by $$b_{ik}$$ and sum over $$i$$ and I get the wrong answer. Please help me find where I've made a mistake.

I shall defer slightly from your notation. $$b_{ij}=(i,j)$$th entry of $$\operatorname{Adj}B$$.
We know that $$\displaystyle{}\sum_{j=1}^n B_{ij}w_j=0 \ \forall \ i=1,2,\dots, n$$. Then as you did , we get $$\displaystyle{}\sum_{i=1}^nb_{ki}\sum_{j=1}^n B_{ij}w_j=0 \ \forall \ k=1,2,\dots, n \\ \implies \displaystyle{}\sum_{j=1}^n\left( \sum_{i=1}^n b_{ki}B_{ij}\right )w_j=0 \ \forall \ k=1,2,\dots, n \\ \implies\displaystyle{ \sum_{j=1}^n(\det B)\delta_{kj}w_j=0 \ \forall \ k=1,2,\dots, n }\\ \implies \displaystyle{(\det B)w_k=0 \ \forall \ k=1,2,\dots, n }$$
Edit: Your mistake was writing $$\displaystyle {\sum_{i}b_{ik}B_{ij}=\det B}$$ instead of $$(\det B)\delta_{jk}$$. After all, you should expect the sum to depend on $$k$$ in some way since you are not summing over $$k$$.
• I'm not understanding how you can introduce $\delta_{kj}$ on the third step. Please explain this step. Oct 17, 2020 at 19:09
• $\operatorname{Adj}B\cdot B=(\det B)I$. In other words $$\sum_{i=1}^n b_{ki}B_{ij}=(k,j)^{\text{th}}\text{ entry of }(\det B)I=(\det B)\delta_{jk}$$ Oct 17, 2020 at 20:00