1、直和
给定两个向量空间$V,W$，维数分别是$n,m$。$V$里有一组向量基底$\left\{\vec{e}_{1}, \vec{e}_{2}, \cdots, \vec{e}_{n}\right\}$，同样$W$里也有一组基底$\left\{\vec{f}_{1}, \vec{f}_{2}, \cdots, \vec{f}_{m}\right\}$.我们现在定义$n+m$个向量构成基组$\left\{\vec{e}_{1}, \cdots, \vec{e}_{n}, \vec{f}_{1}, \cdots, \vec{f}_{m}\right\}$. 由这些基组生成的线性空间叫做$V$与$W$生成的直和空间$V\oplus W$，线性空间维数为$n+m$.
任意向量$\vec{v} \in V$都可以写成基底线性叠加的形式$\vec{v}=v_{1} \vec{e}_{1}+v_{2} \vec{e}_{2}+\cdots+v_{n} \vec{e}_{n}$. 线性代数里我们可以用这些叠加系数构成的列向量来代表这个向量。
\begin{equation}
\left.\vec{v}=\left(\begin{array}{c}
v_{1} \\
v_{2} \\
\vdots \\
v_{n}
\end{array}\right)\right\} n
\end{equation}值得注意，这个列向量是线性空间$V$的一个元素，列向量有$n$个分量。特别地，向量基底可以写成
\begin{equation}
\vec{e}_{1}=\left(\begin{array}{c}
1 \\
0 \\
\vdots \\
0
\end{array}\right), \quad \vec{e}_{2}=\left(\begin{array}{c}
0 \\
1 \\
\vdots \\
0
\end{array}\right)_{V}, \quad \cdots, \quad \vec{e}_{n}=\left(\begin{array}{c}
0 \\
0 \\
\vdots \\
1
\end{array}\right)_{V}
\end{equation}若基组是正交组，$\vec{e}_{i} \cdot \vec{e}_{j}=\delta_{i j}$，则有$v_i=\vec{e}_i\cdot\vec{v}$.同样对于$w$中的任意向量也可以有
\begin{equation}
\left.\vec{w}=\left(\begin{array}{c}
w_{1} \\
w_{2} \\
\vdots \\
w_{n}
\end{array}\right)\right\} m
\end{equation}基向量为
\begin{equation}
\vec{f}_{1}=\left(\begin{array}{c}
1 \\
0 \\
\vdots \\
0
\end{array}\right)_{W}, \quad \vec{f}_{2}=\left(\begin{array}{c}
0 \\
1 \\
\vdots \\
0
\end{array}\right)_{W}, \quad \cdots, \quad \vec{f}_{m}=\left(\begin{array}{c}
0 \\
0 \\
\vdots \\
1
\end{array}\right)_{W}
\end{equation}自然地，对于元素$\vec{v}\in V$和$\vec{w}\in W$在直和空间的列向量形式可以推广成
\begin{equation}
\left.\left.\vec{v}=\left(\begin{array}{c}
v_{1} \\
\vdots \\
v_{n} \\
\hline 0 \\
\vdots \\
0
\end{array}\right)\right\} n+m, \quad \vec{w}=\left(\begin{array}{c}
0 \\
\vdots \\
0 \\
w_{1} \\
\vdots \\
w_{m}
\end{array}\right)\right\} n+m
\end{equation}以看到，只是把其余部分用0填充。
我们也可以定义两个非零向量的直和
\begin{equation}
\left.\vec{v} \oplus \vec{w}=\left(\begin{array}{c}
v_{1} \\
\vdots \\
v_{n} \\
\hline w_{1} \\
\vdots \\
w_{m}
\end{array}\right)=\left(\frac{\vec{v}}{\vec{w}}\right)\right\} n+m
\end{equation}数学上矩阵表示一个线性空间到另一个线性空间的映射。这里我们只讨论一个$V\mapsto V$的特殊情况，因为量子力学里大部分情况便是如此。力学量$A:V\mapsto V$
\begin{equation}
\vec{v}\mapsto A\vec{v}
\end{equation}\begin{equation}
\left(\begin{array}{c}
v_{1} \\
v_{2} \\
\vdots \\
v_{n}
\end{array}\right) \longmapsto\left(\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1 n} \\
a_{22} & a_{22} & \cdots & a_{2 n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n}
\end{array}\right)_{V}\left(\begin{array}{c}
v_{1} \\
v_{2} \\
\vdots \\
v_{n}
\end{array}\right)_{V}
\end{equation}类似的有矩阵$B:W\mapsto W,\vec{w}\mapsto B\vec{w}$
在直和空间里，我们希望矩阵$A,B$经过某种运算后作用在向量上保持$\vec{v}\mapsto A\vec{v}$和$\vec{w}\mapsto B\vec{w}$。这样的运算记成$A\oplus B$. 运算结果可以利用分块对角矩阵构造
\begin{equation}
A \oplus B=\left(\begin{array}{c|c}
A & 0_{n \times m} \\
\hline 0_{m \times n} & B
\end{array}\right)
\end{equation}例如$n=2,m=3$的情况
\begin{equation}
A=\left(\begin{array}{cc}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{array}\right)_{V}, \quad B=\left(\begin{array}{ccc}
b_{11} & b_{12} & b_{13} \\
b_{21} & b_{22} & b_{23} \\
b_{31} & b_{32} & b_{33}
\end{array}\right)_{W}
\end{equation}他们的直和为
\begin{equation}
A \oplus B=\left(\begin{array}{cc|ccc}
a_{11} & a_{12} & 0 & 0 & 0 \\
a_{21} & a_{22} & 0 & 0 & 0 \\
\hline 0 & 0 & b_{11} & b_{12} & b_{13} \\
0 & 0 & b_{21} & b_{22} & b_{23} \\
0 & 0 & b_{31} & b_{32} & b_{33}
\end{array}\right)
\end{equation}显然$A,B$分别作用在不同的子空间里。显然直和矩阵$A\oplus B$作用在直和向量$\vec{v} \oplus \vec{w}$为
\begin{equation}
(A \oplus B)(\vec{v} \oplus \vec{w})=\left(\begin{array}{c|c}
A & 0_{n \times m} \\
\hline 0_{m \times n} & B
\end{array}\right)\left(\frac{\vec{v}}{\vec{w}}\right)=\left(\frac{A \vec{v}}{B \vec{w}}\right)=(A \vec{v}) \oplus(B \vec{w})
\end{equation}若有两个直和矩阵相乘，利用对角矩阵运算显然得到
\begin{equation}
\left(A_{1} \oplus B_{1}\right)\left(A_{2} \oplus B_{2}\right)=\left(A_{1} A_{2}\right) \oplus\left(B_{1} B_{2}\right)
\end{equation}直和空间$V\oplus W$一共有$(n+m)^2$个线性无关矩阵，而子空间$V$和$W$分别只有$n^2,m^2$个线性无关矩阵，这说明并不是所有在直和空间中的矩阵都能表示成直和形式。
利用分块矩阵行列式规则显然得到
\begin{equation}
\operatorname{det}(A \oplus B)=(\operatorname{det} A)(\operatorname{det} B)
\end{equation}分块矩阵求迹得到的是各个子块的迹之和
\begin{equation}
\operatorname{Tr}(A \oplus B)=(\operatorname{Tr} A)+(\operatorname{Tr} B)
\end{equation}2、直积
直积这个词在有些书上叫张量积、Kronecker积
和之前一样，给定两个线性空间$V,W$,分别$n,m$维. $V$中有一组基底$\left\{\vec{e}_{1}, \vec{e}_{2}, \cdots, \vec{e}_{n}\right\}$，$W$中也有一组基底$\left\{\vec{f}_{1}, \vec{f}_{2}, \cdots, \vec{f}_{m}\right\}$.我们现在定义$nm$个基向量$\vec{e}_{i} \otimes \vec{f}_{j}$,其中$i=1,\cdots,n$和$j=1,\cdots,m$.$\otimes$具体运算下面将会讲明，可以看到$\vec{e}_i\otimes\vec{e}_j$有两个指标，这也是为什么直积也叫张量积的原因。
张量积是双线性的，即在$V$中是线性的，在$W$中也是线性的. 如果有多于两个线性空间，则叫多重线性的。这意味着$\vec{v} \otimes \vec{w}=\left(\sum_{i}^{n} v_{i} \vec{e}_{i}\right) \otimes\left(\sum_{j}^{m} w_{j} \vec{f}_{j}\right)=\sum_{i}^{n} \sum_{j}^{m} v_{i} w_{j}\left(\vec{e}_{i} \otimes \vec{f}_{j}\right)$. 换句话说由向量基底$\vec{e}_i\otimes\vec{f}_i$生成的线性空间$V\otimes W$中的向量由这组基$\vec{v}_i\otimes\vec{w}_j$线性表出，系数为$v_iw_j$
一种使它非常明确的方法是把它写出来。例如假设$n=2,m=3$。张量积空间的维数为$nm=6$新的基底分别是$\vec{e}_{1} \otimes \vec{f}_{1}, \vec{e}_{1} \otimes \vec{f}_{2}, \vec{e}_{1} \otimes \vec{f}_{3}, \vec{e}_{2} \otimes \vec{f}_{1}, \vec{e}_{2} \otimes \vec{f}_{2}, \vec{e}_{2} \otimes \vec{f}_{3}$. 写出这6个基底对应的列向量
\begin{equation}
\begin{split}
&\vec{e}_{1} \otimes \vec{f}_{1}=\left(\begin{array}{c}
1 \\
0 \\
0 \\
\hline 0 \\
0 \\
0
\end{array}\right), \quad \vec{e}_{1} \otimes \vec{f}_{2}=\left(\begin{array}{c}
0 \\
1 \\
0 \\
0 \\
0 \\
0
\end{array}\right), \quad \vec{e}_{1} \otimes \vec{f}_{3}=\left(\begin{array}{c}
0 \\
0 \\
1 \\
\hline 0 \\
0 \\
0
\end{array}\right)\\
&\vec{e}_{2} \otimes \vec{f}_{1}=\left(\begin{array}{c}
0 \\
0 \\
0 \\
\hline 1 \\
0 \\
0
\end{array}\right), \quad \vec{e}_{2} \otimes \vec{f}_{2}=\left(\begin{array}{c}
0 \\
0 \\
0 \\
\hline 0 \\
1 \\
0
\end{array}\right), \quad \vec{e}_{2} \otimes \vec{f}_{3}=\left(\begin{array}{c}
0 \\
0 \\
0 \\
\hline 0 \\
0 \\
1
\end{array}\right)
\end{split}
\end{equation}这里横线把基向量分割成了两个不同的子集，第一部分由$\vec{e}_1$生成，第二部分由$\vec{e}_2$生成。对于一般的向量$\vec{v}\in V,\vec{w}\in W$，张量积为
\begin{equation}
\vec{v} \otimes \vec{w}=\left(\begin{array}{l}
v_{1} w_{1} \\
v_{1} w_{2} \\
v_{1} w_{3} \\
\hline v_{2} w_{1} \\
v_{2} w_{2} \\
v_{2} w_{3}
\end{array}\right)
\end{equation}在数学上矩阵是一个线性空间到另一个线性空间的线性映射。同样的我们只关注量子力学感兴趣的$A:V\mapsto V$
或者说
\begin{equation}
\left(\begin{array}{c}
v_{1} \\
v_{2} \\
\vdots \\
v_{n}
\end{array}\right) \longmapsto\left(\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1 n} \\
a_{22} & a_{22} & \cdots & a_{2 n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{n 1} & a_{n 2} & \cdots & a_{n n}
\end{array}\right)\left(\begin{array}{c}
v_{1} \\
v_{2} \\
\vdots \\
v_{n}
\end{array}\right)_{V}
\end{equation}在张量积空间上，相同的矩阵仍然可以作用在矢量上，使得$\vec{v} \mapsto A \vec{v}$但$\vec{w} \mapsto \vec{w}$保持不变。这个矩阵可以写成$A\otimes I$，其中$I$是恒等矩阵。在之前的$n=2,m=3$例子里，矩阵$A$是$2\times2$维矩阵，显然$A\otimes I$是$6\times6$维矩阵
\begin{equation}
A \otimes I=\left(\begin{array}{ccc|ccc}
a_{11} & 0 & 0 & a_{12} & 0 & 0 \\
0 & a_{11} & 0 & 0 & a_{12} & 0 \\
0 & 0 & a_{11} & 0 & 0 & a_{12} \\
\hline a_{21} & 0 & 0 & a_{22} & 0 & 0 \\
0 & a_{21} & 0 & 0 & a_{22} & 0 \\
0 & 0 & a_{21} & 0 & 0 & a_{22}
\end{array}\right)
\end{equation}把这个矩阵作用在$\vec{v}\otimes\vec{w}$上就知道了之所以这样构造的原因
\begin{equation}
\begin{aligned}
(A \otimes I)(\vec{v} \otimes \vec{w}) &=\left(\begin{array}{ccc|ccc}
a_{11} & 0 & 0 & a_{12} & 0 & 0 \\
0 & a_{11} & 0 & 0 & a_{12} & 0 \\
0 & 0 & a_{11} & 0 & 0 & a_{12} \\
\hline a_{21} & 0 & 0 & a_{22} & 0 & 0 \\
0 & a_{21} & 0 & 0 & a_{22} & 0 \\
0 & 0 & a_{21} & 0 & 0 & a_{22}
\end{array}\right)\left(\begin{array}{c}
v_{1} w_{1} \\
v_{1} w_{2} \\
v_{1} w_{3} \\
\hline v_{2} w_{1} \\
v_{2} w_{2} \\
v_{2} w_{3}
\end{array}\right) \\
=&\left(\begin{array}{c}
\left(a_{11} v_{1}+a_{12} v_{2}\right) w_{1} \\
\left(a_{11} v_{1}+a_{12} v_{2}\right) w_{2} \\
\left(a_{11} v_{1}+a_{12} v_{2}\right) w_{3} \\
\hline\left(a_{21} v_{1}+a_{22} v_{2}\right) w_{1} \\
\left(a_{21} v_{1}+a_{22} v_{2}\right) w_{2} \\
\left(a_{21} v_{1}+a_{22} v_{2}\right) w_{3}
\end{array}\right) &=(A \vec{v}) \otimes \vec{w}
\end{aligned}
\end{equation}显然矩阵$A$仅仅只作用在$\vec{v}\in V$上并且保持$\vec{w}\in W$不变。
类似的对于矩阵$B:W\mapsto W$, $\vec{w}\mapsto B\vec{w}$. 同样的也能用$I\otimes B$作用在$V\otimes W$中的矢量。其中
\begin{equation}
I \otimes B=\left(\begin{array}{ccc|ccc}
b_{11} & b_{12} & b_{13} & 0 & 0 & 0 \\
b_{21} & b_{22} & b_{23} & 0 & 0 & 0 \\
b_{31} & b_{32} & b_{33} & 0 & 0 & 0 \\
\hline 0 & 0 & 0 & b_{11} & b_{12} & b_{13} \\
0 & 0 & 0 & b_{21} & b_{22} & b_{23} \\
0 & 0 & 0 & b_{31} & b_{32} & b_{33}
\end{array}\right)
\end{equation}作用在$\vec{v}\otimes\vec{w}$为
\begin{equation}
\begin{aligned}
(I \otimes B)(\vec{v} \otimes \vec{w}) &=\left(\begin{array}{ccc|ccc}
b_{11} & b_{12} & b_{13} & 0 & 0 & 0 \\
b_{21} & b_{22} & b_{23} & 0 & 0 & 0 \\
b_{31} & b_{32} & b_{33} & 0 & 0 & 0 \\
\hline 0 & 0 & 0 & b_{11} & b_{12} & b_{13} \\
0 & 0 & 0 & b_{21} & b_{22} & b_{23} \\
0 & 0 & 0 & b_{31} & b_{32} & b_{33}
\end{array}\right)\left(\begin{array}{c}
v_{1} w_{1} \\
v_{1} w_{2} \\
v_{1} w_{3} \\
\hline v_{2} w_{1} \\
v_{2} w_{2} \\
v_{2} w_{3}
\end{array}\right) \\
=&\left(\begin{array}{c}
v_{1}\left(b_{11} w_{1}+b_{12} w_{2}+b_{13} w_{3}\right) \\
v_{1}\left(b_{21} w_{1}+b_{22} w_{2}+b_{23} w_{3}\right) \\
v_{1}\left(b_{31} w_{1}+b_{32} w_{2}+b_{33} w_{3}\right) \\
\hline v_{2}\left(b_{11} w_{1}+b_{12} w_{2}+b_{13} w_{3}\right) \\
v_{2}\left(b_{21} w_{1}+b_{22} w_{2}+b_{23} w_{3}\right) \\
v_{2}\left(b_{31} w_{1}+b_{32} w_{2}+b_{33} w_{3}\right)
\end{array}\right) &=\vec{v} \otimes(B \vec{w})
\end{aligned}
\end{equation}一般而言
\begin{equation}
\begin{split}
(A \otimes I)(\vec{v} \otimes \vec{w})&=(A \vec{v}) \otimes \vec{w}\\
(I \otimes B)(\vec{v} \otimes \vec{w})&=\vec{v} \otimes(B \vec{w})
\end{split}
\end{equation}利用上面式子很容易证明下列式子
\begin{equation}
\begin{array}{l}
\left(A_{1} \otimes I\right)\left(A_{2} \otimes I\right)=\left(A_{1} A_{2}\right) \otimes I \\
\left(I \otimes B_{1}\right)\left(I \otimes B_{2}\right)=I \otimes\left(B_{1} B_{2}\right) \\
\left(A_{1} \otimes B_{1}\right)\left(A_{2} \otimes B_{2}\right)=\left(A_{1}A_{2}\right) \otimes\left(B_{1} B_{2}\right) \\
(A \otimes I)(I \otimes B)=(I \otimes B)(A \otimes I)=(A \otimes B)
\end{array}
\end{equation}证明：
\begin{equation}
(A_1\otimes I)(A_2\otimes I)(\vec{v}\otimes\vec{w})=(A_1\otimes I)(A_2\vec{v}\otimes\vec{w})=(A_1A_2\vec{v}\otimes\vec{w})=((A_1 A_2)\otimes I)(\vec{v}\otimes\vec{w})
\end{equation}第二三式类似证明，不再赘述
\begin{equation}
(A\otimes I)(I\otimes B)(\vec{v}\otimes\vec{w})=(A\otimes I)(v\otimes B\vec{w})=(A\vec{v})\otimes(B\vec{w})=(A\otimes B)(\vec{v}\otimes\vec{w})=(I \otimes B)(A \otimes I)(\vec{v}\otimes\vec{w})
\end{equation}最后一式定义了$A\otimes B$，也可以写出矩阵形式
\begin{equation}
\begin{split}
A \otimes B&=(A\otimes I)(I\otimes B)\\
&=\left(\begin{array}{ccc|ccc}
a_{11} & 0 & 0 & a_{12} & 0 & 0 \\
0 & a_{11} & 0 & 0 & a_{12} & 0 \\
0 & 0 & a_{11} & 0 & 0 & a_{12} \\
\hline a_{21} & 0 & 0 & a_{22} & 0 & 0 \\
0 & a_{21} & 0 & 0 & a_{22} & 0 \\
0 & 0 & a_{21} & 0 & 0 & a_{22}
\end{array}\right)\left(\begin{array}{ccc|ccc}
b_{11} & b_{12} & b_{13} & 0 & 0 & 0 \\
b_{21} & b_{22} & b_{23} & 0 & 0 & 0 \\
b_{31} & b_{32} & b_{33} & 0 & 0 & 0 \\
\hline 0 & 0 & 0 & b_{11} & b_{12} & b_{13} \\
0 & 0 & 0 & b_{21} & b_{22} & b_{23} \\
0 & 0 & 0 & b_{31} & b_{32} & b_{33}
\end{array}\right)\\
&=\left(\begin{array}{ccc|ccc}
a_{11} b_{11} & a_{11} b_{12} & a_{11} b_{13} & a_{12} b_{11} & a_{12} b_{12} & a_{12} b_{13} \\
a_{11} b_{21} & a_{11} b_{22} & a_{11} b_{23} & a_{12} b_{21} & a_{12} b_{22} & a_{12} b_{23} \\
a_{11} b_{31} & a_{11} b_{32} & a_{11} b_{33} & a_{12} b_{31} & a_{12} b_{32} & a_{12} b_{33} \\
\hline a_{21} b_{11} & a_{21} b_{12} & a_{21} b_{13} & a_{22} b_{11} & a_{22} b_{12} & a_{22} b_{13} \\
a_{21} b_{21} & a_{21} b_{22} & a_{21} b_{23} & a_{22} b_{21} & a_{22} b_{22} & a_{22} b_{23} \\
a_{21} b_{31} & a_{21} b_{32} & a_{21} b_{33} & a_{22} b_{31} & a_{22} b_{32} & a_{22} b_{33}
\end{array}\right)
\end{split}
\end{equation}值得注意，张量积空间中不是所有矩阵都能写成两个子空间矩阵张量积的形式。此外还有两个常用公式
\begin{equation}
\begin{array}{l}
\operatorname{det}(A \otimes B)=(\operatorname{det} A)^{m}(\operatorname{det} B)^{n} \\
\operatorname{Tr}(A \otimes B)=(\operatorname{Tr} A)(\operatorname{Tr} B)
\end{array}
\end{equation}