# 凱萊–哈密頓定理

（重定向自凱萊-哈密頓定理

${\displaystyle p(\lambda )=\det(\lambda I_{n}-A)}$

${\displaystyle p(A)=O}$

## 例子

${\displaystyle A={\begin{bmatrix}1&2\\3&4\end{bmatrix}}}$

${\displaystyle p(\lambda )={\begin{vmatrix}\lambda -1&-2\\-3&\lambda -4\end{vmatrix}}=(\lambda -1)(\lambda -4)-2\cdot 3=\lambda ^{2}-5\lambda -2}$

${\displaystyle A^{2}-5A-2I_{2}=O}$

${\displaystyle A^{2}-5A-2I_{2}=O}$
${\displaystyle A^{2}=5A+2I_{2}}$

${\displaystyle A^{3}=(5A+2I_{2})A=5A^{2}+2A=5(5A+2I_{2})+2A=27A+10I_{2}}$
${\displaystyle A^{4}=A^{3}A=(27A+10I_{2})A=27A^{2}+10A=27(5A+2I_{2})+10A}$
${\displaystyle A^{4}=145A+54I_{2}}$

${\displaystyle A^{n}=aA+bI}$

${\displaystyle \lambda _{1}^{n}=a\lambda _{1}+b}$
${\displaystyle \lambda _{2}^{n}=a\lambda _{2}+b}$

：一般而言，若${\displaystyle n\times n}$ 矩陣${\displaystyle A}$ 可逆（即：${\displaystyle \det(A)\neq 0}$ ），則${\displaystyle A^{-1}}$ 可以寫成${\displaystyle A}$ 的冪次和：特徵多項式有如下形式

${\displaystyle p(\lambda )=\lambda ^{n}-\operatorname {tr} (A)\lambda ^{n-1}+\cdots +(-1)^{n}\det(A)}$

${\displaystyle A^{-1}={\frac {(-1)^{n-1}}{\det(A)}}(A^{n-1}-\operatorname {tr} (A)A^{n-2}+\cdots )}$

## 定理證明

${\displaystyle S\operatorname {adj} (S)=\det(S)I_{n}}$

${\displaystyle S=tI_{n}-A}$ ，便得到${\displaystyle (tI_{n}-A)\operatorname {adj} (tI_{n}-A)=p_{A}(t)I_{n}}$ 。此式對所有${\displaystyle t}$ 皆成立，由於實數複數域有無窮多元素，上式等式在多項式環${\displaystyle k[t]}$ 內成立。

${\displaystyle M:=k^{n}}$ ，矩陣${\displaystyle A}$ 賦予${\displaystyle M}$ 一個${\displaystyle k[t]}$ -結構：${\displaystyle f(t)\cdot m=f(A)m}$ 。考慮${\displaystyle k[t]}$ -模${\displaystyle M[t]:=M\otimes _{k}k[t]}$ ，我們有${\displaystyle k[t]}$ -模之間的「求值態射」：

${\displaystyle e_{A}:M[t]\to M,\qquad M\otimes t^{i}\mapsto A^{i}m}$

${\displaystyle (tI_{n}-A)\operatorname {adj} (tI_{n}-A)\,m=p_{A}(t)m}$

${\displaystyle B={\mbox{adj}}(tI_{n}-A)}$

${\displaystyle S\operatorname {adj} (S)=\det(S)I_{n}}$

${\displaystyle (tI_{n}-A)B=\det(tI_{n}-A)I_{n}=p(t)I_{n}}$
{\displaystyle {\begin{aligned}p(t)I_{n}&=(tI_{n}-A)B\\&=(tI_{n}-A)\sum _{i=0}^{n-1}t^{i}B_{i}\\&=\sum _{i=0}^{n-1}tI_{n}\cdot t^{i}B_{i}-\sum _{i=0}^{n-1}A\cdot t^{i}B_{i}\\&=\sum _{i=0}^{n-1}t^{i+1}B_{i}-\sum _{i=0}^{n-1}t^{i}AB_{i}\\&=t^{n}B_{n-1}+\sum _{i=1}^{n-1}t^{i}(B_{i-1}-AB_{i})-AB_{0}\end{aligned}}}
${\displaystyle p(t)I_{n}=\det(tI_{n}-A)I_{n}=t^{n}I_{n}+t^{n-1}c_{n-1}I_{n}+\cdots +tc_{1}I_{n}+c_{0}I_{n}}$

${\displaystyle B_{n-1}=I_{n},\qquad B_{i-1}-AB_{i}=c_{i}I_{n}\quad {\text{for }}1\leq i\leq n-1,\qquad -AB_{0}=c_{0}I_{n}~}$

${\displaystyle O=A^{n}+c_{n-1}A^{n-1}+\cdots +c_{1}A+c_{0}I_{n}=p(A)}$