PON（Passive Optical Network，无源光网络）是当前 FTTX（Fiber to the X，光纤到某处）宽带接入网络的主流技术体系之一。

从技术演进路线来看，PON 经历了 APON → BPON → EPON / GPON → 10G-EPON / XG-PON / XGS-PON → 50G-PON 的发展过程。随着网络带宽需求不断提升，PON 也正在从千兆接入逐步迈向万兆乃至 50G 时代。

无源光网络技术（PON）

张量积 II

Tensor Product II

张量积 I

Tensor Product I

# §20 Cayley-Hamilton定理
## §20.1 线性变换的矩阵
**定义 20.1** 设 $M$ 是 $F$ 上的向量空间，并确定 $M$ 的一组基 $\vec{v}_{1}, \ldots, \vec{v}_{k}$。（假设 $M$ 是有限维的。）那么，给定任意线性变换
$$
A: M\to M,
$$
$A$ 关于基 $\vec{v}_{1}, \ldots, \vec{v}_{k}$ 的**矩阵(matrix)** 是满足
$$
A\vec{v}_{i}=\sum\limits_{j=1}^{k}A_{ji}\vec{v}_{j}.
$$
的矩阵。



**例 20.1** 
$(A)=\begin{pmatrix}
A_{11}&A_{12}&A_{13}\\
A_{21}&A_{22}&A_{23}\\
A_{31}&A_{32}&A_{33}
\end{pmatrix}$。 则$A\vec{v}_{1}=A_{11}\vec{v}_{1}+A_{21}\vec{v}_{2}+A_{31}\vec{v}_{3}$。

**练习** 写出 $t$ 在以下每个具有指定基的 $F[t]$-模上的乘法矩阵：

(1) $M=F[t]/(t)$，$\vec{v}_{1}=\overline{1}$。

(2) $M=F[t]/(t-\alpha)$，$\vec{v}_{1}=\overline{1}$。

(3) $M=F[t]/(t-\alpha)^{2}$，$\vec{v}_{2}=\overline{1}$，$\vec{v}_{1}=\overline{t-\alpha}$。

(4) $M=F[t]/(t-\alpha)^{3}$，$\vec{v}_{3}=\overline{1}$，$\vec{v}_{2}=\overline{t-\alpha}$，$\vec{v}_{1}=\overline{(t-\alpha)^{2}}$。

解：一般来说，$F[t]$ 在 $F[t]/I$ 上的作用是 $f\cdot\overline{g}=\overline{fg}$。

(2) $t\cdot\overline{1}=\overline{t\cdot1}=\overline{t-\alpha}+\overline{\alpha}=\alpha\overline{1}$。

因此，$t$ 将 $\overline{1}$ 映射为 $\alpha\overline{1}$，即
$$
(A)=(\alpha)。
$$

(3) $t\cdot\overline{1}=\overline{t\cdot1}=\overline{t-\alpha}+\overline{\alpha}=\vec{v}_{1}+\alpha\vec{v}_{2}$。所以 $(A)=\begin{pmatrix}
        ? & 1\\
          & \alpha 
    \end{pmatrix}$。

计算：
$$
\begin{aligned}
    t\cdot\overline{(t-\alpha)} &= (t-\alpha)\cdot\overline{(t-\alpha)} + \alpha\cdot\overline{(t-\alpha)} \\
    &= \overline{(t-\alpha)^2} + \alpha\overline{(t-\alpha)} \\
    &= 0 + \alpha\vec{v}_{1}
    \end{aligned}
$$
因此，$A=\begin{pmatrix}
        \alpha & 1\\
        0 & \alpha
    \end{pmatrix}$。

**命题 20.1** 对于 $M=F[t]/(t-\alpha)^{n}$，在以下基下 $\overline{(t-\alpha)}^{n-1}$，$\overline{(t-\alpha)}^{n-2}$，$\ldots$，$\overline{(t-\alpha)}^{1}$，$\overline{1}$，线性变换
$$
\begin{aligned}
M&\to M\\
\vec{v}&\mapsto t\vec{v}
\end{aligned}
$$
的矩阵为
$$
\begin{pmatrix}
\alpha & 1 & 0 & & 0 & 0\\
0 & \alpha & 1 & & 0 & 0\\
0 & 0 & \alpha & & 0 & 0\\
 &  &  & \ddots &  & \\
0 & 0 & 0 & & \alpha & 1\\
0 & 0 & 0 & & 0 & \alpha
\end{pmatrix}
$$
其中对角线上为 $\alpha$，每个 $\alpha$ 的正上方元素为 $1$，除了最上方的 $\alpha$。

证明：
$$
\begin{aligned}
tv_{i}=t\overline{(t-\alpha)^{n-i}}&=((t-\alpha)+\alpha)\overline{(t-\alpha)}^{n-i}\\
&=(t-\alpha)\overline{(t-\alpha)}^{n-i}+\alpha\overline{(t-\alpha)}^{n-i}\\
&=\overline{(t-\alpha)}^{n-i+1}+\alpha\vec{v}_{i}\\
&=\begin{cases}
\vec{v}_{i-1}+\alpha\vec{v}_{i}, & i\leqslant n-1,\\
\alpha\vec{v}_{n}, & i=n.
\end{cases}
\end{aligned}
$$
$$
~\tag*{$\square$}
$$
**注** 
如果 $M$ 和 $N$ 是有限维的 $F[t]$-模，并且存在基 $\begin{aligned}
v_{1}, \ldots, v_{m}~\text{为}~M~\text{的基}\\
w_{1}, \ldots, w_{n}~\text{为}~N~\text{的基}
\end{aligned}$，使得“$t$ 的乘法”分别由 $\begin{aligned}
\text{矩阵}~A~\text{作用在}~M\\
\text{矩阵}~B~\text{作用在}~N
\end{aligned}$ 给出，那么在 $M \oplus N$ 上，$t$ 的作用由矩阵
$$
\begin{pmatrix}
A & 0\\
0 & B
\end{pmatrix}
$$
给出，即分块对角矩阵。
## §20.2 Jordan标准形
由于任意 $F[t]$-模同构于 $\bigoplus\limits_{i}F[t]/(p_{i}n_{i})$，我们有以下推论：

**推论 20.2** 设 $T: F^{n}\to F^{n}$ 是一个 $F$-线性变换，且 $F$ 是代数闭域。那么，存在 $F^{n}$ 的一组基，使得
$$
\begin{pmatrix}
A_{1} & 0 & \cdots & 0\\
0 & A_{2} & \cdots & 0\\
\vdots & & \ddots & \vdots\\
0 & \cdots & 0 & A_{e}
\end{pmatrix}
$$
其中
$A_{1}=\begin{pmatrix}
    \alpha_{1} & 1 & \cdots & 0\\
    0 & \ddots &  & \vdots\\
    \vdots & & \ddots & 1\\
    0 & \cdots & 0 & \alpha_{1}
\end{pmatrix}$, $A_{2}=\begin{pmatrix}
    \alpha_{2} & 1 & \cdots & 0\\
    0 & \ddots &  & \vdots\\
    \vdots & & \ddots & 1\\
    0 & \cdots & 0 & \alpha_{2}
\end{pmatrix}, \ldots,$ $A_{e}=\begin{pmatrix}
    \alpha_{e} & 1 & \cdots & 0\\
    0 & \ddots &  & \vdots\\
    \vdots & & \ddots & 1\\
    0 & \cdots & 0 & \alpha_{e}
\end{pmatrix}$。

**定义 20.2** 这称为 $T$ 的**Jordan标准形(Jordan normal form)**。
## §20.3 特征多项式
**定义 20.3** 特征多项式
$$
\det(tI - A) \in F[t]
$$

**例 20.2** 如果 $A = \begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{pmatrix}$，则 $tI - A = \begin{pmatrix} t - A_{11} & A_{12} \\ A_{21} & t - A_{22} \end{pmatrix}$。因此
$$
\det(tI - A) = t^{2} - (A_{11} + A_{22})t + (A_{11}A_{22} - A_{12}A_{21}).
$$

**注** 如果 $A$ 是 $k \times k$ 的矩阵，则特征多项式的第 $i$ 项系数是 $A$ 的一个不变量，在共轭变换下保持不变。
$$
\det(B(tI - A)B^{-1}) = \det BB^{-1} \det(tI - A) = \det(tI - A).
$$
**定义 20.4** 任意 $A \in M_{k \times k}(F)$ 确定了一个映射
$$
f: F[t] \to M_{k \times k}(F),
$$
由于 $F[t]$ 是一个主理想整环，$\ker(f) = (p)$，其中 $p \in F[t]$。

取唯一的 $p$，使得：
- $p = t^{d} + a_{d}t^{d-1} + \cdots$，

- $\ker(f) = (p)$，

我们称 $p$ 为 $A$ 的**最小多项式（minimal polynomial）**。
## §20.4 Cayley-Hamilton定理
**定理 20.3 (Cayley-Hamilton)**
任意矩阵 $A$ 都满足它的特征多项式。

**注** 该定理在 $F$ 不是代数闭域的情况下也成立！

证明：在上述给定的基下，
$$
\det(tI - A) = \prod\limits_{i=1}^{e}(t - \alpha_{i})^{n_{i}}.
$$
因此需要证明
$$
\prod\limits_{i=1}^{e}(A - \alpha_{i}I)^{n_{i}} = 0.
$$
但 $\overline{1}_{1}, \ldots, \overline{1}_{e}$ 作为模生成了 $v$，并且
$$
(A - \alpha_{i}I)^{n_{i}}\overline{1}_{j} = 0.
$$
（因为 $(A - \alpha_{i}I)^{n_{i} - 1}(\overline{1}_{i})$ 是特征向量。）
$$
~\tag*{$\square$}
$$

**例 20.3** 如果 $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$，则它的特征多项式为 $t^{2} - t(a + d) + (ad - bc)$。该定理表明
$$
A^{2} - A(a + d) + I(ad - bc) = 0 \in M_{2 \times 2}(F).
$$

**例 20.4** 另一种形式，如果 $A \in M_{k \times k}(F)$，则其特征多项式的形式为
$$
t^{k} + b_{k-1}t^{k-1} + \cdots + b_{1}t + b_{0}.
$$
该定理表明，对于任意 $\vec{v} \in F^{k}$，我们有
$$
A^{k}\vec{v} + b_{k-1}A^{k-1}\vec{v} + \cdots + b_{1}A\vec{v} + b_{0}\vec{v} = 0.
$$

**推论 20.4** $A$ 的最小多项式整除其特征多项式。

**命题 20.5** 设 $A \in M_{k \times k}(F)$。当且仅当 $A$ 的列构成一个基时，$A$ 是可逆的。

证明：设 $T_{A}$ 为由 $A$ 给出的线性变换 $F^{k} \to F^{k}$。我们需要证明 $T_{A}$ 是可逆的，即 $T_{A}$ 是单射且满射。
$$
T_{A}(\vec{e}_{i}) = \vec{v}_{i} \quad \text{其中}~\vec{v}_{i}~\text{是}~A~\text{的列向量。}
$$
$$
\begin{pmatrix}
    \vdots &  & \vdots\\
    \vec{v}_{1} & \cdots & \vec{v}_{k}\\
    \vdots &  & \vdots
\end{pmatrix} \begin{pmatrix}
    0\\ \vdots \\ 1 \\ \vdots \\ 0
\end{pmatrix} =
\begin{pmatrix}
    ~ \\ ~ \\ \vec{v}_{i} \\ ~ \\ ~
\end{pmatrix}
$$
所以 $T_{A}(\sum b_{i}\vec{e}_{i}) = \vec{0} \Leftrightarrow \sum b_{i}\vec{v}_{i} = 0 \Leftrightarrow b_{i} = 0$。

由于 $\ker T_{A} = \{0\}$ 且 $T_{A}$ 是 $F^{k}$ 到 $F^{k}$ 之间的线性映射（维数相同），所以 $T_{A}$ 是可逆的。
$$
~\tag*{$\square$}
$$

**命题 20.6** 设 $A^{\prime}$ 是 $T_{A}$ 在某个基 $v_{1}, \ldots, v_{k}$ 下的矩阵。令 $B = \begin{pmatrix} \vdots & & \vdots \\ \vec{v}_{1} & \cdots & \vec{v}_{k} \\ \vdots & & \vdots \end{pmatrix}$。则
$$
A^{\prime} = BAB^{-1}。
$$

证明：
$$
\begin{aligned}
BAB^{-1}(\vec{v}_{i}) &= BA\vec{e}_{i} \\
&= B\left(\sum A_{ji}\vec{e}_{j}\right) \\
&= \sum A_{ji}\vec{v}_{j} \\
&= A^{\prime}(\vec{v}_{i})。
\end{aligned}
$$
$$
~\tag*{$\square$}
$$

**命题 20.7** 如果 $B$ 是可逆的且 $A^{\prime} = BAB^{-1}$，则
$$
\det(tI - A^{\prime}) = \det(tI - A) \in F[t].
$$

证明：一般来说，
$$
\begin{aligned}
\det(BCB^{-1}) &= \det B \det C \det B^{-1} \\
&= \det B \det B^{-1} \det C \\
&= \det BB^{-1} \det C \\
&= \det I \det C \\
&= \det C.
\end{aligned}
$$
因此
$$
\det (B(tI - A)B^{-1}) = \det(tI - A).
$$
并且
$$
\begin{aligned}
\det(B(tI - A)B^{-1}) &= \det(BtIB^{-1} - BAB^{-1}) \\
&= \det(tI - BAB^{-1})。
\end{aligned}
$$
$$
~\tag*{$\square$}
$$

因此，为了计算任意 $A \in M_{k \times k}(F)$ 的特征多项式？如果 $F$ 是代数闭域，那么
$$
A = B A^{\prime} B^{-1}
$$
其中
$$
A^{\prime} = \begin{pmatrix}
A_{1} & 0 & \cdots & 0\\
0 & A_{2} & \cdots & 0\\
\vdots & \cdots & \ddots & \vdots\\
0 & \cdots & 0 & A_{e}
\end{pmatrix}
$$
是一个分块对角矩阵，其中
$$
A_{i} = \begin{pmatrix}
\alpha_{i} & 1 & 0 & \cdots & 0\\
0 & \alpha_{i} & 1 & \cdots & 0\\
0 & 0 & \alpha_{i} & \cdots & 0\\
\vdots & \cdots & \cdots & \ddots & \vdots\\
0 & \cdots & \cdots & 0 & \alpha_{i}
\end{pmatrix}, \quad \alpha_{i} \in F.
$$
其中 $\alpha_{i}$ 位于对角线上，每个 $\alpha_{i}$ 的右上方是 $1$。因此，我们只需计算
$$
\begin{aligned}
\det(tI - A^{\prime}) &= \det\begin{pmatrix}
t - \alpha_{1} & -1 & \cdots & 0\\
0 & t - \alpha_{1} & \cdots & 0\\
\vdots & \cdots & \ddots & \vdots\\
0 & \cdots & \cdots & -1\\
0 & \cdots & \cdots & t - \alpha_{e}
\end{pmatrix}\\
&= (t - \alpha_{1})^{n_{1}} \cdot (t - \alpha_{2})^{n_{2}} \cdot \cdots \cdot (t - \alpha_{e})^{n_{e}}.
\end{aligned}
$$
因为 $tI - A^{\prime}$ 是上三角矩阵，其行列式为对角线上元素的乘积。

为证明
$$
(A-\alpha_{1})^{n_{1}}\cdot\cdots\cdot(A-\alpha_{e})^{n_{e}}=0
$$
我们只需证明对于任意 $\vec{v}$，有
$$
(A-\alpha_{1})^{n_{1}}\cdot \cdots\cdot(A-\alpha_{e})^{n_{e}}\vec{v}=0.
$$
现在选择一个基
$$
\begin{aligned}
\vec{v}_{1,n_{1}} &= \overline{1}, \ldots, \vec{v}_{1,1} = \overline{(t - \alpha_{1})}^{n_{1} - 1} \in F[t]/(t - \alpha_{1})^{n_{1}} \\
&\vdots \\
\vec{v}_{e,n_{e}} &= \overline{1}, \ldots, \vec{v}_{e,1} = \overline{(t - \alpha_{e})}^{n_{e} - 1} \in F[t]/(t - \alpha_{e})^{n_{e}}
\end{aligned}
$$
则 $\{\vec{v}_{i, n_{j}}\}$ 构成 $A$ 作用的向量空间的基。此外，
$$
\begin{aligned}
&(A - \alpha_{1})^{n_{1}} \cdot \cdots \cdot (A - \alpha_{i})^{n_{i}} \cdot \cdots \cdot (A - \alpha_{e})^{n_{e}} \vec{v}_{i,j} \\
=& (A - \alpha_{1})^{n_{1}} \cdot \cdots \cdot (A - \alpha_{e})^{n_{e}} (A - \alpha_{i})^{n_{i}} \vec{v}_{i,j} \\
=& (A - \alpha_{1})^{n_{1}} \cdot \cdots \cdot (A - \alpha_{e})^{n_{e}} \vec{0} \\
=& \vec{0}
\end{aligned}
$$
第二和第三个等式成立的原因是我们知道（通过选择的基），
$$
A\vec{v}_{i,j} = \vec{v}_{i,j-1} + \alpha_{i} \vec{v}_{i,j}.
$$
另一方面，
$$
A\vec{v}_{i,1} = \alpha_{i}\vec{v}_{i,1}.
$$
因此
$$
(A - \alpha_{i}) \vec{v}_{i,j} = \begin{cases} 
\vec{v}_{i,j-1} & j > 1 \\ 
0 & j = 1 
\end{cases}
$$

这样我们就证明了，如果 $F$ 是代数闭域，那么任何 $A \in M_{k \times k}(F)$ 都满足其特征多项式 $\det(tI - A)$。如果 $F$ 不是代数闭域呢？令 $\overline{F}$ 是一个包含 $F$ 的代数闭域，则我们有包含映射
$$
M_{k \times k}(F) \hookrightarrow M_{k \times k}(\overline{F})
$$
和
$$
F[t] \hookrightarrow \overline{F}[t]。
$$
我们可以将任何以 $F$ 为系数的矩阵视作以 $\overline{F}$ 为系数的矩阵。同样，任何以 $F$ 为系数的多项式也可以视作以 $\overline{F}$ 为系数的多项式。将这些联系组合起来，
<div align=center>
    <img src=https://tex.xumin.net/svgb/U1HhiklKTc_Mqy7JzK5KTqnlclSISSwqyi-PLtJRyE0sKFYoyY-FCaWkpMAFdRSSUvNSFIoy0zNKbE0sYxXUFKo1HHUUYlJSSzRKPBV0FRw1NWuRDFOKKUmtKKl-smfDi3ULa5WQTFdTsFeIieHyja_OVogpycxNLVbIrtVw04RrRjhARyEjPz8bbBkW5RCOW3RJbC1evThtgqvBdE5MfllqUU5mXmq1Wy2yyzBdgqYSIo4siOI-jIDB4kA0E2OwRxKhCEBSaMAVA4w8eJyrqAAA>
</div>

是交换的。因此，如果 $?\to 0$，则表明? 必然为零。简单来说，将矩阵代入其特征多项式的结果是相同的，无论我们将矩阵视作具有 $F$ 还是 $\overline{F}$ 系数。

# §20 Cayley-Hamilton Theorem

## §20.1 Matrices of Linear Transformations

**Definition 20.1** Let $M$ be a vector space over $F$, and fix a basis

$$
\vec{v}_{1},\ldots,\vec{v}_{k}
$$

of $M$.

Assume that $M$ is finite-dimensional.

Given any linear transformation

$$
A:M\to M,
$$

the **matrix** of $A$ with respect to the basis
$\vec{v}*{1},\ldots,\vec{v}*{k}$
is the matrix satisfying

$$
A\vec{v}_{i}
=
\sum_{j=1}^{k}
A_{ji}\vec{v}_{j}.
$$



**Example 20.1**

Let

$$
(A)
=
\begin{pmatrix}
A_{11}&A_{12}&A_{13}\\
A_{21}&A_{22}&A_{23}\\
A_{31}&A_{32}&A_{33}
\end{pmatrix}.
$$

Then

$$
A\vec{v}_{1}
=
A_{11}\vec{v}_{1}
+
A_{21}\vec{v}_{2}
+
A_{31}\vec{v}_{3}.
$$

**Exercise** Write down the matrix of multiplication by $t$ on each of the following $F[t]$-modules with the specified basis.

(1)

$$
M=F[t]/(t),
\qquad
\vec{v}_{1}=\overline{1}.
$$

(2)

$$
M=F[t]/(t-\alpha),
\qquad
\vec{v}_{1}=\overline{1}.
$$

(3)

$$
M=F[t]/(t-\alpha)^2,
\qquad
\vec{v}_{2}=\overline{1},
\qquad
\vec{v}_{1}=\overline{t-\alpha}.
$$

(4)

$$
M=F[t]/(t-\alpha)^3,
\qquad
\vec{v}_{3}=\overline{1},
\qquad
\vec{v}_{2}=\overline{t-\alpha},
\qquad
\vec{v}_{1}=\overline{(t-\alpha)^2}.
$$

Solution: In general, the action of $F[t]$ on

$$
F[t]/I
$$

is

$$
f\cdot\overline{g}
=
\overline{fg}.
$$

(2)

$$
t\cdot\overline{1}
=
\overline{t\cdot1}
=
\overline{t-\alpha}
+
\overline{\alpha}
=
\alpha\overline{1}.
$$

Therefore, $t$ sends
$\overline{1}$
to
$\alpha\overline{1}$, so

$$
(A)
=
(\alpha).
$$

(3)

$$
t\cdot\overline{1}
=
\overline{t\cdot1}
=
\overline{t-\alpha}
+
\overline{\alpha}
=
\vec{v}_{1}
+
\alpha\vec{v}_{2}.
$$

Thus

$$
(A)
=
\begin{pmatrix}
? & 1\\
& \alpha
\end{pmatrix}.
$$

Now compute:

$$
\begin{aligned}
t\cdot\overline{(t-\alpha)}
&=
(t-\alpha)\cdot\overline{(t-\alpha)}
+
\alpha\cdot\overline{(t-\alpha)}\\
&=
\overline{(t-\alpha)^2}
+
\alpha\overline{(t-\alpha)}\\
&=
0+\alpha\vec{v}_{1}.
\end{aligned}
$$

Therefore,

$$
A
=
\begin{pmatrix}
\alpha & 1\\
0 & \alpha
\end{pmatrix}.
$$

**Proposition 20.1** For

$$
M=F[t]/(t-\alpha)^n,
$$

with respect to the basis

$$
\overline{(t-\alpha)}^{n-1},
\overline{(t-\alpha)}^{n-2},
\ldots,
\overline{(t-\alpha)}^{1},
\overline{1},
$$

the matrix of the linear transformation

$$
\begin{aligned}
M&\to M\\
\vec{v}&\mapsto t\vec{v}
\end{aligned}
$$

is

$$
\begin{pmatrix}
\alpha & 1 & 0 & & 0 & 0\\
0 & \alpha & 1 & & 0 & 0\\
0 & 0 & \alpha & & 0 & 0\\
& & & \ddots & &\\
0 & 0 & 0 & & \alpha & 1\\
0 & 0 & 0 & & 0 & \alpha
\end{pmatrix},
$$

where the diagonal entries are $\alpha$, and the entry immediately above each $\alpha$ is $1$, except above the topmost $\alpha$.

Proof:

$$
\begin{aligned}
tv_i
=
t\overline{(t-\alpha)^{n-i}}
&=
((t-\alpha)+\alpha)
\overline{(t-\alpha)}^{n-i}\\
&=
(t-\alpha)
\overline{(t-\alpha)}^{n-i}
+
\alpha
\overline{(t-\alpha)}^{n-i}\\
&=
\overline{(t-\alpha)}^{n-i+1}
+
\alpha\vec{v}_i\\
&=
\begin{cases}
\vec{v}_{i-1}+\alpha\vec{v}_i,
& i\leqslant n-1,\\
\alpha\vec{v}_n,
& i=n.
\end{cases}
\end{aligned}
$$

$$
~\tag*{$\square$}
$$

**Remark**

Suppose $M$ and $N$ are finite-dimensional $F[t]$-modules, and there are bases

$$
\begin{aligned}
v_1,\ldots,v_m
&~\text{for }M,\\
w_1,\ldots,w_n
&~\text{for }N,
\end{aligned}
$$

such that multiplication by $t$ is given respectively by

$$
\begin{aligned}
&\text{the matrix }A~\text{on }M,\\
&\text{the matrix }B~\text{on }N.
\end{aligned}
$$

Then the action of $t$ on

$$
M\oplus N
$$

is given by the matrix

$$
\begin{pmatrix}
A&0\\
0&B
\end{pmatrix},
$$

i.e. a block-diagonal matrix.

## §20.2 Jordan Normal Form

Since every $F[t]$-module is isomorphic to

$$
\bigoplus_iF[t]/(p_in_i),
$$

we obtain the following corollary.

**Corollary 20.2** Let

$$
T:F^n\to F^n
$$

be an $F$-linear transformation, and suppose that $F$ is algebraically closed.

Then there exists a basis of $F^n$ with respect to which the matrix of $T$ is

$$
\begin{pmatrix}
A_1&0&\cdots&0\\
0&A_2&\cdots&0\\
\vdots&&\ddots&\vdots\\
0&\cdots&0&A_e
\end{pmatrix},
$$

where

$$
A_1
=
\begin{pmatrix}
\alpha_1&1&\cdots&0\\
0&\ddots&&\vdots\\
\vdots&&\ddots&1\\
0&\cdots&0&\alpha_1
\end{pmatrix},
$$

$$
A_2
=
\begin{pmatrix}
\alpha_2&1&\cdots&0\\
0&\ddots&&\vdots\\
\vdots&&\ddots&1\\
0&\cdots&0&\alpha_2
\end{pmatrix},
$$

and so on, up to

$$
A_e
=
\begin{pmatrix}
\alpha_e&1&\cdots&0\\
0&\ddots&&\vdots\\
\vdots&&\ddots&1\\
0&\cdots&0&\alpha_e
\end{pmatrix}.
$$

**Definition 20.2** This is called the **Jordan normal form** of $T$.

## §20.3 Characteristic Polynomials

**Definition 20.3** The **characteristic polynomial** is

$$
\det(tI-A)\in F[t].
$$

**Example 20.2** If

$$
A
=
\begin{pmatrix}
A_{11}&A_{12}\\
A_{21}&A_{22}
\end{pmatrix},
$$

then

$$
tI-A
=
\begin{pmatrix}
t-A_{11}&A_{12}\\
A_{21}&t-A_{22}
\end{pmatrix}.
$$

Therefore,

$$
\det(tI-A)
=
t^2-(A_{11}+A_{22})t
+
(A_{11}A_{22}-A_{12}A_{21}).
$$

**Remark** If $A$ is a
$k\times k$
matrix, then the coefficient of each term in the characteristic polynomial is an invariant of $A$, remaining unchanged under conjugation:

$$
\det(B(tI-A)B^{-1})
=
\det BB^{-1}\det(tI-A)
=
\det(tI-A).
$$

**Definition 20.4** Any

$$
A\in M_{k\times k}(F)
$$

determines a map

$$
f:F[t]\to M_{k\times k}(F).
$$

Since
$F[t]$
is a principal ideal domain,

$$
\ker(f)
=
(p),
$$

where

$$
p\in F[t].
$$

Choose the unique $p$ such that

*

$$
p
=
t^d+a_dt^{d-1}+\cdots;
$$

*

$$
\ker(f)=(p).
$$

We call $p$ the **minimal polynomial** of $A$.

## §20.4 Cayley-Hamilton Theorem

**Theorem 20.3 (Cayley-Hamilton)**

Every matrix $A$ satisfies its characteristic polynomial.

**Remark** This theorem is also true when $F$ is not algebraically closed.

Proof: With respect to the basis above,

$$
\det(tI-A)
=
\prod_{i=1}^{e}
(t-\alpha_i)^{n_i}.
$$

Therefore, we need to prove

$$
\prod_{i=1}^{e}
(A-\alpha_iI)^{n_i}
=
0.
$$

But

$$
\overline{1}_1,\ldots,\overline{1}_e
$$

generate $v$ as a module, and

$$
(A-\alpha_iI)^{n_i}\overline{1}_j
=
0.
$$

This is because

$$
(A-\alpha_iI)^{n_i-1}
(\overline{1}_i)
$$

is an eigenvector.

$$
~\tag*{$\square$}
$$

**Example 20.3** If

$$
A
=
\begin{pmatrix}
a&b\\
c&d
\end{pmatrix},
$$

then its characteristic polynomial is

$$
t^2-t(a+d)+(ad-bc).
$$

The theorem says that

$$
A^2-A(a+d)+I(ad-bc)
=
0
\in
M_{2\times2}(F).
$$

**Example 20.4** In another form, if

$$
A\in M_{k\times k}(F),
$$

then its characteristic polynomial has the form

$$
t^k
+
b_{k-1}t^{k-1}
+
\cdots
+
b_1t
+
b_0.
$$

The theorem says that for every

$$
\vec{v}\in F^k,
$$

we have

$$
A^k\vec{v}
+
b_{k-1}A^{k-1}\vec{v}
+
\cdots
+
b_1A\vec{v}
+
b_0\vec{v}
=
0.
$$

**Corollary 20.4** The minimal polynomial of $A$ divides its characteristic polynomial.

**Proposition 20.5** Let

$$
A\in M_{k\times k}(F).
$$

Then $A$ is invertible if and only if its columns form a basis.

Proof: Let

$$
T_A
$$

be the linear transformation

$$
F^k\to F^k
$$

defined by $A$.

We need to prove that
$T_A$
is invertible, i.e. that
$T_A$
is injective and surjective.

$$
T_A(\vec{e}_i)
=
\vec{v}_i,
\quad
\text{where }\vec{v}_i~\text{is the }i\text{th column of }A.
$$

$$
\begin{pmatrix}
\vdots&&\vdots\\
\vec{v}_1&\cdots&\vec{v}_k\\
\vdots&&\vdots
\end{pmatrix}
\begin{pmatrix}
0\\
\vdots\\
1\\
\vdots\\
0
\end{pmatrix}
=
\begin{pmatrix}
~\\
~\\
\vec{v}_i\\
~\\
~
\end{pmatrix}.
$$

Thus

$$
T_A\left(\sum b_i\vec{e}_i\right)
=
\vec{0}
\Longleftrightarrow
\sum b_i\vec{v}_i
=
0
\Longleftrightarrow
b_i=0.
$$

Since

$$
\ker T_A
=
\{0\},
$$

and $T_A$ is a linear map from
$F^k$
to
$F^k$,
which have the same dimension, $T_A$ is invertible.

$$
~\tag*{$\square$}
$$

**Proposition 20.6** Let $A^{\prime}$ be the matrix of $T_A$ with respect to some basis

$$
v_1,\ldots,v_k.
$$

Let

$$
B
=
\begin{pmatrix}
\vdots&&\vdots\\
\vec{v}_1&\cdots&\vec{v}_k\\
\vdots&&\vdots
\end{pmatrix}.
$$

Then

$$
A^{\prime}
=
BAB^{-1}.
$$

Proof:

$$
\begin{aligned}
BAB^{-1}(\vec{v}_i)
&=
BA\vec{e}_i\\
&=
B\left(
\sum A_{ji}\vec{e}_j
\right)\\
&=
\sum A_{ji}\vec{v}_j\\
&=
A^{\prime}(\vec{v}_i).
\end{aligned}
$$

$$
~\tag*{$\square$}
$$

**Proposition 20.7** If $B$ is invertible and

$$
A^{\prime}
=
BAB^{-1},
$$

then

$$
\det(tI-A^{\prime})
=
\det(tI-A)
\in F[t].
$$

Proof: In general,

$$
\begin{aligned}
\det(BCB^{-1})
&=
\det B\det C\det B^{-1}\\
&=
\det B\det B^{-1}\det C\\
&=
\det BB^{-1}\det C\\
&=
\det I\det C\\
&=
\det C.
\end{aligned}
$$

Therefore,

$$
\det(B(tI-A)B^{-1})
=
\det(tI-A).
$$

Also,

$$
\begin{aligned}
\det(B(tI-A)B^{-1})
&=
\det(BtIB^{-1}-BAB^{-1})\\
&=
\det(tI-BAB^{-1}).
\end{aligned}
$$

$$
~\tag*{$\square$}
$$

So how do we compute the characteristic polynomial of an arbitrary

$$
A\in M_{k\times k}(F)?
$$

If $F$ is algebraically closed, then

$$
A
=
BA^{\prime}B^{-1},
$$

where

$$
A^{\prime}
=
\begin{pmatrix}
A_1&0&\cdots&0\\
0&A_2&\cdots&0\\
\vdots&\cdots&\ddots&\vdots\\
0&\cdots&0&A_e
\end{pmatrix}
$$

is a block-diagonal matrix, with

$$
A_i
=
\begin{pmatrix}
\alpha_i&1&0&\cdots&0\\
0&\alpha_i&1&\cdots&0\\
0&0&\alpha_i&\cdots&0\\
\vdots&\cdots&\cdots&\ddots&\vdots\\
0&\cdots&\cdots&0&\alpha_i
\end{pmatrix},
\qquad
\alpha_i\in F.
$$

The entries
$\alpha_i$
lie on the diagonal, and immediately above each $\alpha_i$ is a $1$.

Therefore, we only need to compute

$$
\begin{aligned}
\det(tI-A^{\prime})
&=
\det
\begin{pmatrix}
t-\alpha_1&-1&\cdots&0\\
0&t-\alpha_1&\cdots&0\\
\vdots&\cdots&\ddots&\vdots\\
0&\cdots&\cdots&-1\\
0&\cdots&\cdots&t-\alpha_e
\end{pmatrix}\\
&=
(t-\alpha_1)^{n_1}
\cdot
(t-\alpha_2)^{n_2}
\cdot
\cdots
\cdot
(t-\alpha_e)^{n_e}.
\end{aligned}
$$

This is because
$tI-A^{\prime}$
is upper triangular, so its determinant is the product of its diagonal entries.

To prove

$$
(A-\alpha_1)^{n_1}
\cdot
\cdots
\cdot
(A-\alpha_e)^{n_e}
=
0,
$$

it is enough to prove that for every vector
$\vec{v}$,

$$
(A-\alpha_1)^{n_1}
\cdot
\cdots
\cdot
(A-\alpha_e)^{n_e}
\vec{v}
=
0.
$$

Now choose a basis

$$
\begin{aligned}
\vec{v}_{1,n_1}
&=
\overline{1},
\ldots,
\vec{v}_{1,1}
=
\overline{(t-\alpha_1)}^{n_1-1}
\in
F[t]/(t-\alpha_1)^{n_1}\\
&\vdots\\
\vec{v}_{e,n_e}
&=
\overline{1},
\ldots,
\vec{v}_{e,1}
=
\overline{(t-\alpha_e)}^{n_e-1}
\in
F[t]/(t-\alpha_e)^{n_e}.
\end{aligned}
$$

Then

$$
\{\vec{v}_{i,n_j}\}
$$

forms a basis of the vector space on which $A$ acts.

Moreover,

$$
\begin{aligned}
&
(A-\alpha_1)^{n_1}
\cdot
\cdots
\cdot
(A-\alpha_i)^{n_i}
\cdot
\cdots
\cdot
(A-\alpha_e)^{n_e}
\vec{v}_{i,j}\\
={}&
(A-\alpha_1)^{n_1}
\cdot
\cdots
\cdot
(A-\alpha_e)^{n_e}
(A-\alpha_i)^{n_i}
\vec{v}_{i,j}\\
={}&
(A-\alpha_1)^{n_1}
\cdot
\cdots
\cdot
(A-\alpha_e)^{n_e}
\vec{0}\\
={}&
\vec{0}.
\end{aligned}
$$

The second and third equalities hold because, by our choice of basis,

$$
A\vec{v}_{i,j}
=
\vec{v}_{i,j-1}
+
\alpha_i\vec{v}_{i,j}.
$$

On the other hand,

$$
A\vec{v}_{i,1}
=
\alpha_i\vec{v}_{i,1}.
$$

Therefore,

$$
(A-\alpha_i)\vec{v}_{i,j}
=
\begin{cases}
\vec{v}_{i,j-1},
& j>1,\\
0,
& j=1.
\end{cases}
$$

Thus we have proved that if $F$ is algebraically closed, then every

$$
A\in M_{k\times k}(F)
$$

satisfies its characteristic polynomial

$$
\det(tI-A).
$$

What if $F$ is not algebraically closed?

Let

$$
\overline{F}
$$

be an algebraically closed field containing $F$.

Then we have inclusion maps

$$
M_{k\times k}(F)
\hookrightarrow
M_{k\times k}(\overline{F})
$$

and

$$
F[t]
\hookrightarrow
\overline{F}[t].
$$

We may regard any matrix with coefficients in $F$ as a matrix with coefficients in
$\overline{F}$.

Similarly, any polynomial with coefficients in $F$ may also be regarded as a polynomial with coefficients in
$\overline{F}$.

Putting these observations together, the diagram

<div align=center>
    <img src=https://tex.xumin.net/svgb/U1HhiklKTc_Mqy7JzK5KTqnlclSISSwqyi-PLtJRyE0sKFYoyY-FCaWkpMAFdRSSUvNSFIoy0zNKbE0sYxXUFKo1HHUUYlJSSzRKPBV0FRw1NWuRDFOKKUmtKKl-smfDi3ULa5WQTFdTsFeIieHyja_OVogpycxNLVbIrtVw04RrRjhARyEjPz8bbBkW5RCOW3RJbC1evThtgqvBdE5MfllqUU5mXmq1Wy2yyzBdgqYSIo4siOI-jIDB4kA0E2OwRxKhCEBSaMAVA4w8eJyrqAAA>
</div>

commutes.

Therefore, if

$$
?\to0,
$$

then this shows that

$$
?
$$

must itself be zero.

In simpler terms, substituting a matrix into its characteristic polynomial gives the same result whether we regard the matrix as having coefficients in
$F$
or in
$\overline{F}$.

张量积 II

1. 简介

张量积 I

1. 简介

§20 Cayley-Hamilton定理

§20.1 线性变换的矩阵

§19 主理想整环(PID)

§19.1 多项式环