---- > [!theorem] Theorem. ([[Cayley-Hamilton Theorem]]) > Let $R$ be a [[commutative ring|commutative]] [[ring]] and let $M$ be a [[submodule generated by a subset|finitely generated]] $R$-[[module]]. Let $f:M \to M$ be an $R$-[[linear map|linear map]]. Let $\mathfrak{a}$ be an [[ideal]] of $R$ such that $f(M) \subset \mathfrak{a}M$.[^3] Then there is $n \geq 1$ and $a_{1},\dots,a_{n} \in \mathfrak{a}$ such that $f^{n}+a_{1}f^{n-1}+\dots+a_{n}f^{0}=0 \text{ in } \text{End }M.$ > [!proposition] The Stabilization Corollary. > Let $M$ be a [[submodule generated by a subset|finitely generated]] $R$-[[module]] and $\mathfrak{a}$ an [[ideal]] of $R$ such that $\mathfrak{a}M=M$. Then there is a 'stabilizing' $a \in \mathfrak{a}$ such that $am=m$ for all $m \in M$. ^proposition > [!proof] Proof of Corollary. > Put $f=\id_{M}$. Note that $\id_{M}^{0}=1_{\text{End }M}=\id_{M}$. Since $f(M)=M=\mathfrak{a}M$, the C-H hypothesis is satisfied and we have $a_{1},\dots,a_{n}$ such that $\id_{M}^{n}+a_{1}\id_{M}^{n-1}+\dots+a_{n} \id_{M}^{0}=0$, i.e., $(1+a_{1}+\dots+a_{n}) \id_{M}=0.$ Now set $a=-(a_{1}+\dots+a_{n})$. ^proof > [!proof]+ Proof. ([[Cayley-Hamilton Theorem]]) > Take a generating set $\{ m_{1},\dots ,m_{n} \}$ for the $R$-[[module]] $M$. Each element of $\mathfrak{a}M$ is an $\mathfrak{a}$-linear combination of $m_{1},\dots,m_{n}$. Since $f(M) \subset \mathfrak{a}M$, we have $\begin{bmatrix} > f(m_{1}) \\ > \vdots \\ > f(m_{n}) > \end{bmatrix}=P \begin{bmatrix} > m_{1} \\ > \vdots \\ > m_{n} > \end{bmatrix} \ \ (*)$ > for a [[matrix]] $P \in M_{n \times n}(R)$ *with entries in $\mathfrak{a} \subset R$*. > > The structure [[ring homomorphism]] $\rho:R \to \text{End }M$ that defines $M$ also turns it into a (noncommutative) $R$-[[algebra]]. The [[universal property of polynomial algebras]] tells us that an $R$-[[algebra]] [[algebra homomorphism|homomorphism]] $R[T]\to\text{End }M$ is determined by where it sends $T$ and choice of image of $T$ extends to an $R$-[[algebra]] homomorphism.[^1] We send $T \mapsto f$. Thus $M$ becomes an $R[T]$-[[module]] via $Tm :=f(m)$ for $m \in M$. > > Then $(*)$ can be written in the form: $\underbrace{T\begin{bmatrix} > m_{1} \\ \vdots \\ m_{n} > \end{bmatrix}}_{\text{scalar-vector mult.}}=\underbrace{P\begin{bmatrix} > m_{1} \\ \vdots \\ m_{n} > \end{bmatrix}}_{\text{matrix-vector mult.}}$ > so that the matrix $Q: TI_{n}-P=\text{diag}(T, \dots, T)-P \in M_{n \times n}(R[T])$[^2] satisfies $Q \begin{bmatrix} > m_{1} \\ \vdots \\ m_{n} > \end{bmatrix}=0.$ > Multiplying both sides by the [[matrix adjugate]] $\text{adj }Q$ on the left, we have $(\det Q) \begin{bmatrix}m_{1} \\ \vdots \\ m_{n}\end{bmatrix}=0$. Since $m_{1},\dots,m_{n}$ generate $M$, $\underbrace{(\det Q)}_{\in R[T]}m=0$ for all $m \in M$. Recalling how $R[T]$ acts on $M$, this means that the image of $\det Q$ in $\text{End }M$, obtaining by substituting $f$ for $T$, is the zero endomorphism. Since $\det Q$ is a monic degree-$n$ [[polynomial 4|polynomial]], where the coefficients of $T^{0},\dots,T^{n-1}$ are in $\mathfrak{a}$ (since the entries of $P$ are in $\mathfrak{a}$), this completes the proof. [^1]: We don't have to worry that $\text{End }M$ is noncommutative because our polynomials are only in a single variable. [^2]: Viewing the entries of $P$ (elements of $\mathfrak{a}$) as polynomials of degree zero. > [!theorem] Theorem. ([[Cayley-Hamilton Theorem]]) > $\mathbb{F}$ denotes $\mathbb{R}$ or $\mathbb{C}$. Let $V$ be a [[vector space]] over $\mathbb{F}$.Let $T \in$[[vector space of operators on a vector space]] have [[characteristic polynomial]] $q$. Then $q(T)=0$. [^3]: $\mathfrak{a}M$ is the [[submodule]] generated by $\{ am: a \in \mathfrak{a} , m \in M\}$. ---- #### ----- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as Tag > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```