# What is going on in this proof of the Cayley-Hamilton theorem?

I am reading a proof of the Cayley-Hamilton theorem here. For a rough outline of the proof, let $$A$$ be a matrix representing the endomorphism $$\phi$$ over finitely-generated module $$M$$ with generators $$m_1,...,m_n$$. Now, we can regard $$M$$ as an $$R[x]$$-module by letting $$x$$ act as $$\phi$$.

This next part is where I am confused. They let $$\mathfrak{m}$$ the column vector whose entries are the $$m_j$$. Then, we get $$(xI-A)\mathfrak{m}=0$$, which I guess works by letting matrix multiplication be letting the elements in the matrix with coefficients in $$R[x]$$ act on the elements of $$\mathfrak{m}$$.

The next step multiplies both sides by the adjugate matrix to get $$[\det(xI-A)]I\cdot\mathfrak{m}=0$$, which then completes the proof, as we $$p(\phi)=0$$, where $$p(x):=\det(xI-A)$$.

I guess my real question is: what is actually going on? I've never seen matrices used this way; is this just a formal manipulation? But then it feels like multiplying by the adjugate matrix is "wrong". How do I know that the manipulation in $$R[x]$$ preserves the module action structure? I'm sorry for phrasing this poorly but I just have a feeling that something is off and I may not have it articulated it that well.

• As you say, $M$ is an $R[x]$-module. What us happening here is that $M^n$ (considered as a space of column vectors) is a left-module for the ring $M_n(R[x])$ of $n$ by $n$ matrices over $R[x]$. Both $xI-A$ and its adjugate are elements of $M_n(R[x])$. Sep 14, 2020 at 3:36
• @AnginaSeng It seems like this is a mistake in the proof: I don't think we want to look at $M^n,$ but rather think of elements of $M$ itself as coming from a surjection $R^n\to M$ (i.e., choosing a generating set $\{m_i\}$ of $M$). Otherwise, $\phi$ is not necessarily representable as a matrix, and moreover even if it were, it would be diagonal and there would be little reason to consider it as a matrix in the first place (we would simply be looking at multiplication by a scalar $r\in R$ on $M$). Sep 14, 2020 at 4:48

My previous answer made a false claim -- that we wanted to view $$M$$ as an $$M_{n\times n}(R[x])$$-module. In fact, this will not work in general: while given an endomorphism $$\phi\in\operatorname{End}_R(M)$$ and a generating set $$\{m_1,\dots, m_n\}$$ of $$M$$ we may produce a matrix $$A_\phi\in M_{n\times n}(R)$$ such that $$\require{AMScd} \begin{CD} R^n @>A_\phi>> R^n \\ @V\pi VV @VV\pi V\\ M @>>\phi > M \end{CD}$$ commutes, it is not the case that in general a matrix $$B\in M_{n\times n}(R)$$ induces a well-defined endomorphism of $$M.$$ However, this doesn't mean we can't use the main idea that $$(xI - A)^{\textrm{adj}}(xI -A) = \det(xI - A)I\in M_{n\times n}(R[x]),$$ we just need to be careful.
First, let's choose our generating set $$\{m_1,\dots, m_n\}$$ of $$M$$ and our matrix representation $$A_\phi$$ of $$\phi$$ with respect to this generating set. Explicitly, we have some collection of constants $$r_{ij}\in R$$ such that $$\phi(m_i) = \sum_{j=1}^n r_{ij} m_j.$$ If we let $$\delta_{ij} = \begin{cases} 1,\quad i = j\\ 0,\quad i\neq j\end{cases}$$ and we consider $$M$$ as an $$R[x]$$-module where $$x$$ acts on $$M$$ by $$xm = \phi(m),$$ then the previous equation is equivalent to $$\sum_{j}(x\delta_{ij} - a_{ij})m_j = 0.$$
Observe that if we assemble the coefficients of the $$m_j$$ as we range over all $$j$$ and all $$i$$ into a matrix, we obtain $$(x\delta_{ij} - a_{ij})_{ij} = xI - A_\phi.$$ Now we apply the adjugate trick. Write $$(xI - A_\phi)^{\textrm{adj}} = (b_{ij})_{ij}.$$ Then the fact that $$(xI - A_\phi)^{\textrm{adj}}(xI - A_\phi) = \det(xI - A_\phi) I$$ means that $$\sum_{k=1}^n b_{ik}(x\delta_{kj} - a_{kj}) = \det(xI - A_\phi)\delta_{ij}.$$ Taking our equation $$0 = \sum_{j}(x\delta_{kj} - a_{kj})m_j$$ and multiplying by $$b_{ik},$$ we have $$0 = \sum_j b_{ik}(x\delta_{kj} - a_{kj})m_j.$$ Next we sum these equations over $$k$$: \begin{align*} 0 &= \sum_{k=1}^n\sum_{j=1}^n b_{ik}(x\delta_{kj} - a_{kj})m_j\\ &=\sum_{j=1}^n\sum_{k=1}^n b_{ik}(x\delta_{kj} - a_{kj})m_j\\ &= \sum_{j=1}^n\det(xI - A_\phi)\delta_{ij} m_j\\ &= \det(xI - A_\phi)m_i. \end{align*} This holds for any $$i,$$ so that $$\det(xI - A_\phi) = p(x)$$ acts on $$M$$ identically as zero; i.e., $$p(\phi) : M\to M$$ is zero.
• @Vasting: The short answer is that each entry of $B\mathbf{r}$ is an $R$-linear combination of the components of $\mathbf{r}.$ Since $K$ is an $R$-submodule of $R^n$ and each component of $\mathbf{r}$ is in $K,$ the result follows. More explicitly, write $B = (b_{ij})$ with all $b_{ij}\in R.$ If $\mathbf{r} = (r_1,r_2,\dots, r_n)^T\in K^n$ then the $k$th component of $B\mathbf{r}\in R^n$ is $\sum_{i = 1}^nb_{ki}r_i.$ Then each $b_{ki}r_i\in K$ and the sum of elements in $K$ is again in $K,$ as $K$ is an $R$-submodule of $M.$ So each entry of $B\mathbf{r}$ is in $K,$ which proves the result. Sep 14, 2020 at 5:43
• @Vasting Ah, you're right: in fact, it doesn't make sense to say each $r_i\in K,$ as $K\subseteq R^n,$ not $R.$ Something is off here, but I'm not 100% sure how to fix it at the moment. I'll try to return later and update this with a proper explanation! Sep 14, 2020 at 20:48
• @Vasting I'm not sure that the original approach of considering $M$ as an $M_{n\times n}(R[x])$-module is salvageable, so I've explained how to apply the adjugate trick without actually treating $M$ as a module over the matrix ring. Sep 15, 2020 at 20:04