# 张量积 II

## 1. 简介

继续研究张量乘积，我们将看到如何把两个线性映射 $M \rightarrow M^{\prime}$ 和 $N \rightarrow N^{\prime}$ 结合成一个线性映射$M \otimes_{R} N \rightarrow M^{\prime} \otimes_{R} N^{\prime}$。这就引出了平坦模块和基扩张之间的线性映射。然后，我们将研究向量空间的张量积（包括缩并）的特殊性，$R$-代数的张量积以及最后的$R$-模的张量代数。



## 2. 线性映射中的张量积

## 3. 平坦模

## 4. 线性映射和基扩张的张量积

## 5. 向量空间

## 6. 张量缩并

## 7. $R$-代数的张量积

## 8. $R$-模的张量代数

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
$\overline{F}$.

§ DS-20 Caylay-Hamilton定理

§ DS-20 Caylay-Hamilton Theorem

§ DS-19 主理想整环

§ DS-19 Principal Ideal Domain

# §18 向量空间
## §18.1 张成、线性无关与基
**定义 18.1** 固定 $x_{1},\ldots,x_{n}\in M$。

(1) 如果映射 $X: R^{n}\to M$ 是满射，那么称该集合**张成(span)** $M$。

(2) 如果映射 $X: R^{n}\to M$ 是单射，那么称该集合在 $M$ 中是**线性无关(linearly independent)** 的。

(3) 如果 $X$ 同时是单射和满射，那么称该集合是 $M$ 的**基(basis)**。



你会在线性代数中认出这些术语。用方程来表达，这些定义正是你所预想的：

**命题 18.1** 设 $M$ 是一个左 $R$-模，$x_{1},\ldots,x_{n}\in M$ 是一个有序集合。

(1) 如果对每个 $y\in M$，存在一个集合 $a_{1},\ldots,a_{n}\in R$ 使得
$$
y=a_{1}x_{1}+\cdots+a_{n}x_{n}，
$$
那么该集合张成 $M$。

(2) 如果方程
$$
0=a_{1} x_{1}+\cdots+a_{n} x_{n}
$$
仅有唯一解 $\left(a_{1}, \ldots, a_{n}\right)=(0, \ldots, 0)$，那么该集合是线性无关的。

(3) 如果对任何 $y \in M$，方程
$$
y=a_1 x_1+\ldots+a_n x_n
$$
仅有唯一的解 $\left(a_{1}, \ldots, a_{n}\right)$，那么该集合是基。

证明：第一条是满射的定义。后一条结论成立，因为同态映射是单射当且仅当核是平凡的，而 $(0, \ldots, 0) \in R^{n}$ 是 $R^n$ 的加法单位元。最后一条是双射的定义。
$$
~\tag*{$\square$}
$$

**定义 18.2** 如果存在某个整数 $n \in \mathbb{Z}_{\geqslant 0}$ 和一个满 $R$-模同态 $R^{n} \rightarrow M$，称一个模是**有限生成的(finitely generated)**。

这与群的情况类似。一个群 $G$ 是有限生成的当且仅当存在有限个元素 $g_{i}$ 使得所有其他元素都可以表示为 $g_{i}$ 和它们的逆元的乘积。同样地，如果存在有限个 $x_{i}$，使得 $M$ 的每个元素都可以通过 $x_{i}$ 的线性组合得到，那么 $M$ 是有限生成的。

**例 18.1** 并非所有的 $R$-模都承认基。这与向量空间不同。例如，若 $R=\mathbb{Z}$ 且 $M=\mathbb{Z}/n\mathbb{Z}$，则对于任何 $x\in M$，方程
$$
a x=0
$$
有许多解——$a$ 可以是 $n, 2n, \ldots$。

**要点**: 并非每个有限生成的 $R$-模都承认基。

## §18.2 向量空间与子空间
**定义 18.3** 如果 $R-\{0\}$ 在乘法下构成群，则称一个交换环是**域(field)**。

**定义 18.4** 设 $F$ 是一个域。$F$ 上的模称为 $F$ 上的**向量空间(vector space)**。

**定义 18.5** 设 $V$ 是一个向量空间。那么 $V$ 的一个子模称为 $V$ 的**线性子空间(linear subspace)**。

## §18.3 张成集大于无关集
以下是域与环的最重要区别之一：

**定理 18.2** 设 $F$ 是一个域，$M$ 是一个 $F$ 上的向量空间。如果 $v_{1}, \ldots, v_{n}$ 张成 $M$，而 $w_{1}, \ldots, w_{m}$ 是线性无关的，那么 $n \geqslant m$。

证明：设 $y_{1}, \ldots, y_{m}$ 是线性无关的，$v_{1}, \ldots, v_{n}$ 是张成的。若有必要，可以对 $v_{i}$ 重新排序，以使
$$
y_{1}=a_{1} v_{1}+\ldots+a_{n} v_{n}
$$
且 $a_{1} \neq 0$。于是 $y_{1}, v_{2}, \ldots, v_{n}$ 也张成 $M$，因为我们可以通过 $y_{1}$ 和 $v_{i}$ 的线性组合得到 $v_{1}$——只需将上述方程除以 $a_{1} \neq 0$ 并重新排列即可。

让 $M_{1} \subset M$ 是由 $y_{1}$ 生成的子模——即 $R \rightarrow M$ 的像，由 $1 \mapsto y_{1}$ 定义——并考虑商模
$$
M/M_{1}
$$
（可以证明这是一个 $R$-模——即，一个向量空间。）那么 $\overline{y_{2}}, \ldots, \overline{y_{m}}$ 仍然是线性无关的，因为它们的线性组合等于零当且仅当
$$
a_{1} y_{1}=a_{2} y_{2}+\ldots+a_{m} y_{m}
$$
对于某个 $a_{1} \in F$ 成立，而此类方程仅当所有 $a_{i}=0$ 时成立，因为假定 $y_{i}$ 是线性无关的。注意 $\overline{y_{1}}=0, \overline{v_{2}}, \ldots, \overline{v_{n}}$ 仍是张成的，因此 $\overline{v_{2}}, \ldots, \overline{v_{n}}$ 是张成的。所以我们有 $n-1$ 个向量张成 $M/M_{1}$，且其中有 $m-1$ 个线性无关的向量。

通过重复上述技巧，如果我们在一个向量空间中有 $m$ 个线性无关的元素且有 $n$ 个元素张成它，我们可以在商向量空间中得到 $m-k$ 个线性无关的元素和 $n-k$ 个张成它的元素。那么这些数哪个会先到 0 呢？如果 $n-k=0$ 先到零，那么我们位于一个由 $0$ 个元素张成的商向量空间——即零向量空间——因此我们必须得出 $m-k=0$，因为在零向量空间中没有线性无关的向量。在这种情况下，$m=n$。如果 $m-k$ 在 $n-k$ 之前到零，那么 $m \leqslant n$。
$$
~\tag*{$\square$}
$$

## §18.4 维数
**推论 18.3** 如果 $M$ 是一个有限生成的向量空间，那么 $M$ 的任意两个基的元素数相同。

证明：如果 $\left\{v_{i}\right\}$ 和 $\left\{w_{i}\right\}$ 均是张成且线性无关的，我们有 $n \geqslant m$ 且 $m \geqslant n$。因此 $m=n$。
$$
~\tag*{$\square$}
$$

**定义 18.6** 设 $V$ 是一个有限生成的 $F$-模——即有限生成的向量空间。称此类 $V$ 为**有限维**向量空间，并定义 $V$ 的**维数(dimension)**
$$
\dim_{F} V
$$
为 $V$ 的任意基中的元素数。

**注** 这是线性代数中最重要的事实：我们有维数的概念。人类花了几千年才理解什么是 $n$ 维空间，因此不要轻视这一概念！

**例 18.2** $0$ 维向量空间是由平凡Abel群 $M=\{0\}$ 给出的模。

**推论 18.4** 如果 $M$ 是一个有限生成的向量空间，那么任意线性无关集合 $w_{1}, \ldots, w_{m}$ 可以扩展为一个基——也就是说，我们可以找到 $w_{m+1}, \ldots, w_{n}$ 使得所得集合 $w_{1}, \ldots, w_{n}$ 同时线性无关且张成。

证明：由于 $M$ 是有限生成的，存在某个 $N$ 使得我们有一个满射 $F^{N} \rightarrow M$。因此，任意线性无关向量集的大小必须 $\leqslant N$。如果 $X_{m}: F^{m} \rightarrow M$ 是由 $w_{1}, \ldots, w_{m}$ 决定的映射，且 $X_{m}$ 不是满射，则选择一个不在 $\mathrm{im}\left(X_{m}\right)$ 中的元素 $w_{m+1}$。注意所得集合 $w_{1}, \ldots, w_{m+1}$ 仍然是线性无关的，因为如果
$$
a_{1} w_{1}+\ldots+a_{m+1} w_{m+1}=0
$$
则
$$
a_{1} w_{1}+\ldots+a_{m} w_{m}=-a_{m+1} w_{m+1}
$$
如果 $a_{m+1}=0$，根据 $w_{i}$ 的线性无关性，我们知道所有 $a_{i}=0$。另一方面，如果 $a_{m+1} \neq 0$，我们通过除法得出矛盾：
$$
\frac{a_{1}}{-a_{m+1}} w_{1}+\ldots+\frac{a_{m}}{-a_{m+1}} w_{m}=w_{m+1}
$$
左边在 $X_{m}$ 的像中，但 $w_{m+1}$ 被选择为不在其中。

因此我们有一个单射同态 $X_{m+1}: F^{m+1} \rightarrow M$。如果 $X_{m+1}$ 不是满射，我们重复上述论证。根据定理，$X_{m+k}$ 在某个 $m+k \leqslant N$ 处必须成为满射。让 $k$ 是 $X_{m+k}$ 首次成为满射的整数。根据上述论证，它仍然是单射，因此我们有一个由生成元 $w_{1}, \ldots, w_{m+k}$ 决定的基。
$$
~\tag*{$\square$}
$$

## §18.5 若干推论
我们将要做什么？你以前已经研究过元素为实数的矩阵。你对它们进行了多种操作——相乘、相加，以及判断它们何时是可逆的。我声称几乎所有关于实数矩阵的操作都可以在任何域的矩阵上完成。

**推论 18.5** 任意有限生成的域 $F$ 上的模与 $F^{n}$ 对某个 $n$ 同构。

证明：从线性无关集合 0 开始扩展为基。一个基定义了从 $F^{n}$ 到你的模的同构。
$$
~\tag*{$\square$}
$$

**注** 如果 $R$ 不是域，这一结论对于 $R$-模不成立——毕竟，任何有限Abel群都是一个 $\mathbb{Z}$-模，但任何自由 $\mathbb{Z}$-模要么是零模，要么是无限模。

**推论 18.6** 如果 $V^{\prime}\subset V$ 是子空间，
$$
\dim V^{\prime}=\

dim V\quad\Leftrightarrow\quad V=V^{\prime}.
$$

证明：一方向显然成立。对于另一方向，令 $y_{1},\ldots,y_{n}$ 是 $V^{\prime}$ 的基。由于这些向量是线性无关的，它们可以通过上述推论之一扩展为 $V$ 的基。但根据维数的定义，该基必须恰好有 $n$ 个元素——换句话说，$y_{i}$ 已经是一个基。
$$
~\tag*{$\square$}
$$

**推论 18.7** 设 $V^{\prime} \subset V$ 是一个子空间。则有 $\dim V^{\prime}+\dim V/V^{\prime}=\dim V$。

证明：设 $v_{1}, \ldots, v_{\dim V^{\prime}}$ 是 $V^{\prime}$ 的基。设 $\overline{u_{1}}, \ldots, \overline{u_{\dim V/V^{\prime}}}$ 是 $V/V^{\prime}$ 的基。选择代表 $u_{i}$ 作为 $\overline{u_{i}}$，则集合
$$
v_{1}, \ldots, v_{\dim V^{\prime}}, u_{1}, \ldots, u_{\dim V/V^{\prime}}
$$
是 $V$ 的基。

它显然是张成的，因为对于每个 $a \in V$，$\overline{a}$ 是 $\overline{u_{i}}$ 的线性组合，因此 $a$ 位于某个线性组合的 $V^{\prime}$-轨道内。它也是线性无关的，因为如果
$$
0 = a_{1} v_{1} + \cdots + a_{\dim V^{\prime}} v_{\dim V^{\prime}} + b_{1} u_{1} + \cdots + b_{\dim V/V^{\prime}} u_{\dim V/V^{\prime}}
$$
那么
$$
\overline{0} = a_{1} \overline{v_{1}} + \cdots + a_{\dim V^{\prime}} \overline{v_{\dim V^{\prime}}} + b_{1} \overline{u_{1}} + \cdots + b_{\dim V/V^{\prime}} \overline{u_{\dim V/V^{\prime}}}
$$
因为 $\overline{v_{i}} = 0$，$a_{i}$ 项消失，因此我们得到一个等式，表示 $\overline{u_{i}}$ 的线性组合等于零。由于 $\overline{u_{i}}$ 是线性无关的，得出每个 $b_{i}$ 必须为零。原等式表明 $0 = \sum a_{i} v_{i}$，因此根据 $v_{i}$ 的线性无关性，所有 $a_{i}$ 也必须为零。
$$
~\tag*{$\square$}
$$

**推论 18.8 (秩-零度定理)**
设 $f: V \rightarrow W$ 是一个 $F$-模之间的映射，且假设 $V$ 是有限生成的。那么有 $\dim \ker f + \dim \mathrm{im} f = \dim V$。

证明：根据同态基本定理，我们知道存在一个群同构 $V/\ker f \cong \mathrm{im} f$。但这个同态映射也是一个 $F$-模映射，可以手工验证。因此 $\mathrm{im} f \cong V/\ker f$。
$$
~\tag*{$\square$}
$$

**推论 18.9 (同构的判别准则)**
设 $f: V \rightarrow W$ 是有限维向量空间之间的线性映射。那么 $f$ 是同构当且仅当 $f$ 是单射且 $\dim V = \dim W$。

证明：根据秩-零度定理，$f$ 的像的维数等于 $V$ 的维数，因为 $f$ 是单射。
$$
~\tag*{$\square$}
$$

## §18.6 总结
从上述所有内容中得到的总结是维数概念的强大。无论你的域是 $\mathbb{R}$ 这样熟悉的，还是类似 $\mathbb{Z}/p\mathbb{Z}$ 这样陌生的；无论线性映射是熟悉的矩阵，还是像多项式函数的值这样的线性映射，我们都有一种强大的方法来研究线性映射。

## §18.7 行列式
我们从线性代数中得到的另一个强大工具是行列式的概念。行列式只需要一个关于乘以 $-1$（取加法逆元）、矩阵元素的乘法和加法的概念。因此我们应该能够定义任何系数在环 $R$ 上的矩阵的行列式。

事实证明，如果环 $R$ 不是交换的，有些公式可能不成立——因为乘法的顺序很重要——因此我们将限制自己使用交换环。

**定义 18.7** 设 $R$ 是一个交换环。一个 $k \times k$ 的**矩阵(matrix)** 是元素的集合
$$
A_{ij} \in R
$$
其中 $i \in 1, \ldots, k$，$j \in 1, \ldots, k$。我们用符号表示矩阵
$$
A = \left(A_{ij}\right).
$$

**例 18.3** 一个 $3 \times 3$ 的 $R$ 上的矩阵可以用通常的方式表示：
$$
\begin{pmatrix}
A_{11} & A_{12} & A_{13} \\
A_{21} & A_{22} & A_{23} \\
A_{31} & A_{32} & A_{33}
\end{pmatrix}
$$

**定义 18.8** $R$ 中 $k \times k$ 矩阵的**环(ring)**，记作 $M_{k \times k}(R)$，其加法定义为
$$
\left(A_{i j}\right)+\left(B_{i j}\right)=\left(A_{i j}+B_{i j}\right)
$$
且乘法定义为
$$
\left(A_{i j}\right)\left(B_{i j}\right)=\left(\sum_{l=1}^k A_{i l} B_{l j}\right)
$$
即加法是逐项相加，而乘法中第 $i,j$ 项是 $B$ 的第 $j$ 列与 $A$ 的第 $i$ 行的配对。

**定义 18.9 (余子式矩阵)**
设 $A$ 是一个 $k \times k$ 矩阵。$A$ 的第 $(i,j)$ 项的**余子式矩阵(cofactor matrix)** 是删除 $A$ 的第 $i$ 行和第 $j$ 列所得的矩阵。当 $A$ 是已知时，我们记作
$$
C_{i, j}
$$
表示由 $A$ 的第 $(i, j)$ 项余子式得到的 $(k-1) \times (k-1)$ 矩阵。

**定义 18.10** $R$ 中 $1 \times 1$ 矩阵的**行列式(determinant)** 是矩阵的唯一元素 $A_{11}$。

递归定义：设 $A$ 是一个 $k \times k$ 矩阵。则 $A$ 的行列式定义为以下求和：
$$
\det A = A_{11} \det C_{1,1} - A_{21} \det C_{2,1} + \cdots + (-1)^{1+k} A_{k 1} \det C_{k, 1}。
$$
使用求和符号表示为：
$$
\det A := \sum_{i=1}^k (-1)^{i+1} A_{i 1} \det C_{i, 1}。
$$
这定义了一个函数
$$
\det: M_{k \times k}(R) \rightarrow R。
$$

**例 18.4** 如果 $A$ 是一个 $2 \times 2$ 矩阵，
$$
\det(A) = A_{11} A_{22} - A_{12} A_{21}。
$$

我们不证明以下定理，但你对实数进行的证明同样适用于一般情形：

**定理 18.10** 设 $A$ 和 $B$ 是 $k \times k$ 矩阵。那么
$$
\det(A) \det(B) = \det(AB)
$$
且
$$
\det\left(A^\mathrm{T}\right) = \det(A).
$$

**定理 18.11** 设 $\mathrm{adj}(A)$ 是一个 $k \times k$ 矩阵，其第 $(i, j)$ 项由
$$
(-1)^{i+j} \det C_{j, i}
$$
给出。那么
$$
A \cdot (\mathrm{adj} A) = (\mathrm{adj} A) \cdot A = \det A \cdot I
$$
其中 $\det A \cdot I$ 是对角矩阵，对角元素由元素 $\det A \in R$ 给出。

**注** 如果你之前没有见过这个定理的最后一条说明，我会简要说明一下证明思路。第一乘法的 $(i, j)$ 项由以下给出：
$$
\sum_{l=1}^k A_{i l} (\mathrm{adj} A)_{l j} = \sum_{l=1}^k A_{i l} (-1)^{j+l} \det C_{j, l}.
$$
例如，$(1,1)$ 项就是 $A$ 的行列式的定义。通过使用关于交换行只改变行列式符号的性质，可以证明每个对角线项都是 $A$ 的行列式。

对于非对角线项，注意到上述求和式变为具有两行相等的矩阵的行列式，因此为零。

**推论 18.12** 设 $A \in M_{k \times k}(R)$。则 $A$ 是可逆矩阵当且仅当 $\det A \in R$ 有乘法逆元。

证明：令 $B = \det A^{-1} \mathrm{adj} A$。则有
$$
B A = \det A^{-1} \mathrm{adj} A \cdot A = \det A^{-1} \det A \cdot I = I.
$$
同样可以证明 $BA = I$。
$$
~\tag*{$\square$}
$$

**例 18.5** 如果 $A$ 是一个仅包含整数元素的矩阵，那么当且仅当 $\det A = \pm 1$ 时存在一个仅包含整数元素的逆矩阵。

**例 18.6** 设 $A$ 是 $\mathbb{Z}/n\mathbb{Z}$ 上的一个矩阵。则当且仅当其行列式与 $n$ 互素时，$A$ 是可逆的。

# §18 Vector Spaces

## §18.1 Spanning Sets, Linear Independence, and Bases

**Definition 18.1** Fix

$$
x_{1},\ldots,x_{n}\in M.
$$

(1) If the map

$$
X:R^{n}\to M
$$

is surjective, then we say that the set **spans** $M$.

(2) If the map

$$
X:R^{n}\to M
$$

is injective, then we say that the set is **linearly independent** in $M$.

(3) If $X$ is both injective and surjective, then we say that the set is a **basis** of $M$.



You will recognize these terms from linear algebra. Expressed in terms of equations, these definitions are exactly what you would expect.

**Proposition 18.1** Let $M$ be a left $R$-module, and let

$$
x_{1},\ldots,x_{n}\in M
$$

be an ordered set.

(1) If for every

$$
y\in M
$$

there exist

$$
a_{1},\ldots,a_{n}\in R
$$

such that

$$
y=a_{1}x_{1}+\cdots+a_{n}x_{n},
$$

then the set spans $M$.

(2) If the equation

$$
0=a_{1}x_{1}+\cdots+a_{n}x_{n}
$$

has the unique solution

$$
(a_{1},\ldots,a_{n})=(0,\ldots,0),
$$

then the set is linearly independent.

(3) If for every

$$
y\in M,
$$

the equation

$$
y=a_{1}x_{1}+\ldots+a_{n}x_{n}
$$

has a unique solution

$$
(a_{1},\ldots,a_{n}),
$$

then the set is a basis.

Proof: The first statement is simply the definition of surjectivity.

The second follows because a homomorphism is injective if and only if its kernel is trivial, and

$$
(0,\ldots,0)\in R^{n}
$$

is the additive identity of
$R^n$.

The final statement is the definition of a bijection.

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

**Definition 18.2** A module is called **finitely generated** if there exists some integer

$$
n\in\mathbb{Z}_{\geqslant0}
$$

and a surjective $R$-module homomorphism

$$
R^{n}\to M.
$$

This is analogous to the case of groups.

A group $G$ is finitely generated if and only if there exist finitely many elements
$g_i$
such that every other element can be expressed as a product of the
$g_i$
and their inverses.

Similarly, a module $M$ is finitely generated if there exist finitely many
$x_i$
such that every element of $M$ can be obtained as a linear combination of the
$x_i$.

**Example 18.1** Not every $R$-module admits a basis.

This is different from the situation for vector spaces.

For example, let

$$
R=\mathbb{Z}
$$

and

$$
M=\mathbb{Z}/n\mathbb{Z}.
$$

Then for any

$$
x\in M,
$$

the equation

$$
ax=0
$$

has many solutions: $a$ may be

$$
n,2n,\ldots.
$$

**Key Point:** Not every finitely generated $R$-module admits a basis.

## §18.2 Vector Spaces and Subspaces

**Definition 18.3** A commutative ring is called a **field** if

$$
R-\{0\}
$$

forms a group under multiplication.

**Definition 18.4** Let $F$ be a field. A module over $F$ is called a **vector space over $F$**.

**Definition 18.5** Let $V$ be a vector space. A submodule of $V$ is called a **linear subspace** of $V$.

## §18.3 A Spanning Set Is at Least as Large as an Independent Set

The following is one of the most important differences between fields and general rings.

**Theorem 18.2** Let $F$ be a field and let $M$ be a vector space over $F$.

If

$$
v_{1},\ldots,v_{n}
$$

span $M$, and

$$
w_{1},\ldots,w_{m}
$$

are linearly independent, then

$$
n\geqslant m.
$$

Proof: Let

$$
y_{1},\ldots,y_{m}
$$

be linearly independent, and let

$$
v_{1},\ldots,v_{n}
$$

span.

If necessary, reorder the $v_i$ so that

$$
y_{1}
=
a_{1}v_{1}
+\ldots+
a_{n}v_{n}
$$

with

$$
a_{1}\neq0.
$$

Then

$$
y_{1},v_{2},\ldots,v_{n}
$$

also span $M$, because we can express $v_1$ as a linear combination of
$y_1$
and the other $v_i$ simply by dividing the equation above by
$a_1\neq0$
and rearranging.

Let

$$
M_{1}\subset M
$$

be the submodule generated by
$y_1$,
i.e. the image of

$$
R\to M
$$

defined by

$$
1\mapsto y_1,
$$

and consider the quotient module

$$
M/M_1.
$$

One can prove that this is an $R$-module, hence a vector space.

Then

$$
\overline{y_2},\ldots,\overline{y_m}
$$

remain linearly independent, because a linear combination of them is zero if and only if

$$
a_1y_1
=
a_2y_2+\ldots+a_my_m
$$

for some

$$
a_1\in F,
$$

and such an equation can hold only when all the
$a_i$
are zero, since the
$y_i$
are assumed to be linearly independent.

Notice that

$$
\overline{y_1}=0,
\qquad
\overline{v_2},\ldots,\overline{v_n}
$$

still span.

Therefore,

$$
\overline{v_2},\ldots,\overline{v_n}
$$

span

$$
M/M_1.
$$

Thus we have
$n-1$
vectors spanning
$M/M_1$
and
$m-1$
linearly independent vectors in it.

Repeating this argument, if a vector space has $m$ linearly independent elements and is spanned by $n$ elements, then in a quotient vector space we obtain

$$
m-k
$$

linearly independent elements and

$$
n-k
$$

spanning elements.

Which of these numbers reaches $0$ first?

If

$$
n-k=0
$$

first, then we are in a quotient vector space spanned by $0$ elements, i.e. the zero vector space.

Therefore,

$$
m-k=0,
$$

because there are no linearly independent vectors in the zero vector space.

In this case,

$$
m=n.
$$

If

$$
m-k
$$

reaches zero before
$n-k$,
then

$$
m\leqslant n.
$$

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

## §18.4 Dimension

**Corollary 18.3** If $M$ is a finitely generated vector space, then any two bases of $M$ contain the same number of elements.

Proof: If

$$
\{v_i\}
$$

and

$$
\{w_i\}
$$

are both spanning and linearly independent, then

$$
n\geqslant m
$$

and

$$
m\geqslant n.
$$

Therefore,

$$
m=n.
$$

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

**Definition 18.6** Let $V$ be a finitely generated $F$-module, i.e. a finitely generated vector space.

Such a $V$ is called a **finite-dimensional** vector space, and the **dimension** of $V$ is defined by

$$
\dim_F V
$$

to be the number of elements in any basis of $V$.

**Remark** This is one of the most important facts in linear algebra: we have a notion of dimension.

It took humanity thousands of years to understand what an $n$-dimensional space is, so do not take this concept lightly!

**Example 18.2** The $0$-dimensional vector space is the module given by the trivial Abelian group

$$
M=\{0\}.
$$

**Corollary 18.4** If $M$ is a finitely generated vector space, then any linearly independent set

$$
w_{1},\ldots,w_{m}
$$

can be extended to a basis.

That is, we can find

$$
w_{m+1},\ldots,w_{n}
$$

such that

$$
w_{1},\ldots,w_{n}
$$

is both linearly independent and spanning.

Proof: Since $M$ is finitely generated, there exists some $N$ such that we have a surjection

$$
F^N\to M.
$$

Therefore, the size of any linearly independent set of vectors must be

$$
\leqslant N.
$$

If

$$
X_m:F^m\to M
$$

is the map determined by

$$
w_1,\ldots,w_m
$$

and $X_m$ is not surjective, choose an element

$$
w_{m+1}
$$

not lying in

$$
\mathrm{im}(X_m).
$$

Notice that the resulting set

$$
w_1,\ldots,w_{m+1}
$$

is still linearly independent.

Indeed, if

$$
a_1w_1+\ldots+a_{m+1}w_{m+1}
=
0,
$$

then

$$
a_1w_1+\ldots+a_mw_m
=
-a_{m+1}w_{m+1}.
$$

If

$$
a_{m+1}=0,
$$

then the linear independence of the $w_i$ implies that all
$a_i=0$.

On the other hand, if

$$
a_{m+1}\neq0,
$$

then dividing gives the contradiction

$$
\frac{a_1}{-a_{m+1}}w_1
+\ldots+
\frac{a_m}{-a_{m+1}}w_m
=
w_{m+1}.
$$

The left-hand side lies in the image of
$X_m$,
while $w_{m+1}$ was chosen not to lie in it.

Therefore, we obtain an injective homomorphism

$$
X_{m+1}:F^{m+1}\to M.
$$

If $X_{m+1}$ is not surjective, repeat the argument.

By the theorem, $X_{m+k}$ must become surjective for some

$$
m+k\leqslant N.
$$

Let $k$ be the first integer for which
$X_{m+k}$
is surjective.

By the argument above, it is still injective.

Therefore, the generators

$$
w_1,\ldots,w_{m+k}
$$

determine a basis.

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

## §18.5 Some Corollaries

What are we going to do with this?

You have already studied matrices with real entries.

You performed many operations on them: multiplication, addition, and determining when they are invertible.

I claim that almost everything you do with real matrices can also be done with matrices over any field.

**Corollary 18.5** Every finitely generated module over a field $F$ is isomorphic to

$$
F^n
$$

for some $n$.

Proof: Starting with the linearly independent set containing $0$ elements, extend it to a basis.

A basis defines an isomorphism from

$$
F^n
$$

to your module.

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

**Remark** If $R$ is not a field, this statement fails for $R$-modules.

After all, every finite Abelian group is a
$\mathbb{Z}$-module,
but every free
$\mathbb{Z}$-module
is either the zero module or infinite.

**Corollary 18.6** If

$$
V^{\prime}\subset V
$$

is a subspace, then

$$
\dim V^{\prime}
=
\dim V
\quad\Leftrightarrow\quad
V=V^{\prime}.
$$

Proof: One direction is obvious.

For the other direction, let

$$
y_1,\ldots,y_n
$$

be a basis of
$V^{\prime}$.

Since these vectors are linearly independent, one of the previous corollaries allows us to extend them to a basis of $V$.

But by the definition of dimension, this basis must have exactly
$n$
elements.

In other words, the
$y_i$
already form a basis of $V$.

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

**Corollary 18.7** Let

$$
V^{\prime}\subset V
$$

be a subspace.

Then

$$
\dim V^{\prime}
+
\dim V/V^{\prime}
=
\dim V.
$$

Proof: Let

$$
v_1,\ldots,v_{\dim V^{\prime}}
$$

be a basis of
$V^{\prime}$.

Let

$$
\overline{u_1},
\ldots,
\overline{u_{\dim V/V^{\prime}}}
$$

be a basis of

$$
V/V^{\prime}.
$$

Choose representatives
$u_i$
for the
$\overline{u_i}$.

Then the set

$$
v_1,\ldots,v_{\dim V^{\prime}},
u_1,\ldots,u_{\dim V/V^{\prime}}
$$

is a basis of $V$.

It clearly spans, because for every

$$
a\in V,
$$

the class

$$
\overline{a}
$$

is a linear combination of the
$\overline{u_i}$.

Therefore, $a$ lies in the
$V^{\prime}$-orbit of some such linear combination.

It is also linearly independent.

Indeed, if

$$
0
=
a_1v_1
+\cdots+
a_{\dim V^{\prime}}v_{\dim V^{\prime}}
+
b_1u_1
+\cdots+
b_{\dim V/V^{\prime}}u_{\dim V/V^{\prime}},
$$

then

$$
\overline{0}
=
a_1\overline{v_1}
+\cdots+
a_{\dim V^{\prime}}\overline{v_{\dim V^{\prime}}}
+
b_1\overline{u_1}
+\cdots+
b_{\dim V/V^{\prime}}\overline{u_{\dim V/V^{\prime}}}.
$$

Since

$$
\overline{v_i}=0,
$$

the terms involving the
$a_i$
vanish.

Thus we obtain an equation saying that a linear combination of the
$\overline{u_i}$
is zero.

Since the
$\overline{u_i}$
are linearly independent, every
$b_i$
must be zero.

The original equation then becomes

$$
0
=
\sum a_iv_i.
$$

Since the
$v_i$
are linearly independent, all
$a_i$
must also be zero.

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

**Corollary 18.8 (Rank-Nullity Theorem)**

Let

$$
f:V\to W
$$

be a map of $F$-modules, and assume that $V$ is finitely generated.

Then

$$
\dim\ker f
+
\dim\mathrm{im}f
=
\dim V.
$$

Proof: By the fundamental theorem of homomorphisms, we know that there is a group isomorphism

$$
V/\ker f
\cong
\mathrm{im}f.
$$

But this homomorphism is also an $F$-module map, as can be verified directly.

Therefore,

$$
\mathrm{im}f
\cong
V/\ker f.
$$

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

**Corollary 18.9 (Criterion for Isomorphisms)**

Let

$$
f:V\to W
$$

be a linear map between finite-dimensional vector spaces.

Then $f$ is an isomorphism if and only if $f$ is injective and

$$
\dim V
=
\dim W.
$$

Proof: By the Rank-Nullity Theorem, the dimension of the image of $f$ equals the dimension of $V$ because $f$ is injective.

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

## §18.6 Summary

The main lesson from everything above is the power of the concept of dimension.

Whether your field is something familiar like
$\mathbb{R}$
or something more unfamiliar like
$\mathbb{Z}/p\mathbb{Z}$;
whether your linear map is a familiar matrix or something like evaluation of polynomial functions, we have a powerful method for studying linear maps.

## §18.7 Determinants

Another powerful tool from linear algebra is the concept of the determinant.

The determinant requires only notions of multiplication by
$-1$,
i.e. taking additive inverses, multiplication of matrix entries, and addition.

Therefore, we should be able to define the determinant of a matrix whose entries lie in any ring $R$.

It turns out that if the ring $R$ is not commutative, some formulas may fail because the order of multiplication matters.

Therefore, we restrict ourselves to commutative rings.

**Definition 18.7** Let $R$ be a commutative ring.

A
$k\times k$
**matrix** is a collection of elements

$$
A_{ij}\in R
$$

where

$$
i\in1,\ldots,k,
\qquad
j\in1,\ldots,k.
$$

We denote the matrix by

$$
A=(A_{ij}).
$$

**Example 18.3** A
$3\times3$
matrix over $R$ may be written in the usual way:

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

**Definition 18.8** The **ring** of
$k\times k$
matrices over $R$, denoted

$$
M_{k\times k}(R),
$$

has addition defined by

$$
(A_{ij})+(B_{ij})
=
(A_{ij}+B_{ij}),
$$

and multiplication defined by

$$
(A_{ij})(B_{ij})
=
\left(
\sum_{l=1}^{k}
A_{il}B_{lj}
\right).
$$

Thus addition is performed entrywise, while the
$(i,j)$
entry of the product is the pairing of the $i$th row of $A$ with the $j$th column of $B$.

**Definition 18.9 (Cofactor Matrix)**

Let $A$ be a
$k\times k$
matrix.

The **cofactor matrix** associated with the
$(i,j)$
entry of $A$ is the matrix obtained by deleting the $i$th row and the $j$th column of $A$.

When $A$ is understood, we write

$$
C_{i,j}
$$

for the
$(k-1)\times(k-1)$
matrix obtained from the cofactor associated with the
$(i,j)$
entry of $A$.

**Definition 18.10** The **determinant** of a
$1\times1$
matrix over $R$ is its unique entry

$$
A_{11}.
$$

Recursively, let $A$ be a
$k\times k$
matrix.

The determinant of $A$ is defined by the sum

$$
\det A
=
A_{11}\det C_{1,1}
-
A_{21}\det C_{2,1}
+
\cdots
+
(-1)^{1+k}
A_{k1}\det C_{k,1}.
$$

Using summation notation,

$$
\det A
:=
\sum_{i=1}^{k}
(-1)^{i+1}
A_{i1}
\det C_{i,1}.
$$

This defines a function

$$
\det:
M_{k\times k}(R)
\to
R.
$$

**Example 18.4** If $A$ is a
$2\times2$
matrix, then

$$
\det(A)
=
A_{11}A_{22}
-
A_{12}A_{21}.
$$

We will not prove the following theorem, but the proofs you know over the real numbers work equally well in this general setting.

**Theorem 18.10** Let $A$ and $B$ be
$k\times k$
matrices.

Then

$$
\det(A)\det(B)
=
\det(AB),
$$

and

$$
\det(A^{\mathrm{T}})
=
\det(A).
$$

**Theorem 18.11** Let

$$
\mathrm{adj}(A)
$$

be the
$k\times k$
matrix whose
$(i,j)$
entry is

$$
(-1)^{i+j}
\det C_{j,i}.
$$

Then

$$
A\cdot(\mathrm{adj}A)
=
(\mathrm{adj}A)\cdot A
=
\det A\cdot I,
$$

where

$$
\det A\cdot I
$$

is the diagonal matrix whose diagonal entries are the element

$$
\det A\in R.
$$

**Remark** If you have not seen the final statement of this theorem before, here is a brief sketch of the proof.

The
$(i,j)$
entry of the first product is

$$
\sum_{l=1}^{k}
A_{il}(\mathrm{adj}A)_{lj}
=
\sum_{l=1}^{k}
A_{il}
(-1)^{j+l}
\det C_{j,l}.
$$

For example, the
$(1,1)$
entry is exactly the definition of the determinant of $A$.

Using the fact that exchanging two rows only changes the sign of the determinant, one can prove that every diagonal entry is equal to
$\det A$.

For the off-diagonal entries, observe that the sum above becomes the determinant of a matrix having two identical rows, and is therefore zero.

**Corollary 18.12** Let

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

Then $A$ is invertible if and only if

$$
\det A\in R
$$

has a multiplicative inverse.

Proof: Let

$$
B
=
\det A^{-1}
\mathrm{adj}A.
$$

Then

$$
BA
=
\det A^{-1}
\mathrm{adj}A\cdot A
=
\det A^{-1}
\det A\cdot I
=
I.
$$

Similarly, one can prove

$$
BA=I.
$$

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

**Example 18.5** If $A$ is a matrix with integer entries, then it has an inverse with integer entries if and only if

$$
\det A=\pm1.
$$

**Example 18.6** Let $A$ be a matrix over

$$
\mathbb{Z}/n\mathbb{Z}.
$$

Then $A$ is invertible if and only if its determinant is relatively prime to $n$.

张量积 II

1. 简介

张量积 I

1. 简介

§20 Cayley-Hamilton Theorem

§20.1 Matrices of Linear Transformations

§19 Principal Ideal Domains (PIDs)

§19.1 Polynomial Rings

§18 Vector Spaces

§18.1 Spanning Sets, Linear Independence, and Bases