和達三樹《物理用數學》習題2.1.1，向量加法的結合律

==問題==

證明結合律：$\bf A+(B+C) = (A+B)+C$。

==解答==

向量加法有兩種方法：平行四邊形法、三角形首尾相連法。證明關鍵在於使用三角形首尾相連法。如果用平行四邊形法會很難做。

$\bf A+(B+C)$

$\bf (A+B)+C$

由以上兩圖可知$\bf A+(B+C) = (A+B)+C$。

(解答結束)

2階Hermite矩陣的形式

==共軛矩陣的定義==

設${\rm A} = \left[ \begin{array}{cccc} z_{11} & z_{12} & \cdots & z_{1n} \\ z_{21} & z_{22} & \cdots & z_{2n} \\ \cdots & \cdots & \cdots & \cdots \\ z_{m1} & z_{m2} & \cdots & z_{mn} \end{array} \right] \in \mathbb{C}^{m \times n}$。定義${\rm A}$的共軛矩陣為${\rm A}^*$，
$${\rm A}^* = \left[ \begin{array}{cccc} \bar{z_{11}} & \bar{z_{12}} & \cdots & \bar{z_{1n}} \\ \bar{z_{21}} & \bar{z_{22}} & \cdots & \bar{z_{2n}} \\ \cdots & \cdots & \cdots & \cdots \\ \bar{z_{m1}} & \bar{z_{m2}} & \cdots & \bar{z_{mn}} \end{array} \right].$$

例子：${\rm A} = \left[ \begin{array}{cc} 1 & 1-2i \\ 1+2i & -5 \end{array} \right]$，則${\rm A}^* = \left[ \begin{array}{cc} \bar{1} & \overline{1-2i} \\ \overline{1+2i} & \overline{-5} \end{array} \right] = \left[ \begin{array}{cc} 1 & 1+2i \\ 1-2i & -5 \end{array} \right]$。

==Hermite共軛矩陣的定義==

設${\rm A}$為n階方陣，${\rm A} = \left[ \begin{array}{cccc} z_{11} & z_{12} & \cdots & z_{1n} \\ z_{21} & z_{22} & \cdots & z_{2n} \\ \cdots & \cdots & \cdots & \cdots \\ z_{n1} & z_{n2} & \cdots & z_{nn} \end{array} \right] \in \mathbb{C}^{n \times n}$。定義${\rm A}$的Hermite共軛為${\rm A}^{\dagger}$，

$${\rm A}^{\dagger} = \left( {\rm A}^* \right)^{\rm T} = \left( \left[ \begin{array}{cccc} \bar{z_{11}} & \bar{z_{12}} & \cdots & \bar{z_{1n}} \\ \bar{z_{21}} & \bar{z_{22}} & \cdots & \bar{z_{2n}} \\ \cdots & \cdots & \cdots & \cdots \\ \bar{z_{n1}} & \bar{z_{n2}} & \cdots & \bar{z_{nn}} \end{array} \right] \right)^{\rm T} = \left[ \begin{array}{cccc} \bar{z_{11}} & \bar{z_{21}} & \cdots & \bar{z_{n1}} \\ \bar{z_{12}} & \bar{z_{22}} & \cdots & \bar{z_{n2}} \\ \cdots & \cdots & \cdots & \cdots \\ \bar{z_{1n}} & \bar{z_{2n}} & \cdots & \bar{z_{nn}} \end{array} \right].$$

例子：${\rm A} = \left[ \begin{array}{cc} 1 & 1-2i \\ 1+2i & -5 \end{array} \right]$，則${\rm A}^{\dagger} = \left[ \begin{array}{cc} 1 & 1+2i \\ 1-2i & -5 \end{array} \right]^{\rm T} = \left[ \begin{array}{cc} 1 & 1-2i \\ 1+2i & -5 \end{array} \right]$。

==Hermite矩陣(Hermitian matrix)的定義==

設${\rm A}$為n階複方陣。若${\rm A}$滿足

$${\rm A}^{\dagger} = {\rm A},$$

則稱${\rm A}$為Hermite矩陣(Hermitian matrix)。

例子：${\rm A} = \left[ \begin{array}{cc} 1 & 1-2i \\ 1+2i & -5 \end{array} \right]$是Hermite矩陣。

==2階Hermite矩陣的形式==

設${\rm A} = \left[ \begin{array}{cc} z_{11} & z_{12} \\ z_{21} & z_{22} \end{array} \right]$，其中$z_{11}, z_{12}, z_{21}, z_{22} \in \mathbb{C}$。

若${\rm A}$為Hermite矩陣，則意味著${\rm A}$滿足
$${\rm A}^{\dagger} = {\rm A}.$$
其中${\rm A}^{\dagger} = \left( \left[ \begin{array}{cc} z_{11} & z_{12} \\ z_{21} & z_{22} \end{array} \right]^* \right)^{\rm T} = \left( \left[ \begin{array}{cc} \bar{z_{11}} & \bar{z_{12}} \\ \bar{z_{21}} & \bar{z_{22}} \end{array} \right] \right)^{\rm T} = \left[ \begin{array}{cc} \bar{z_{11}} & \bar{z_{21}} \\ \bar{z_{12}} & \bar{z_{22}} \end{array} \right]$，於是有
$$\left[ \begin{array}{cc} \bar{z_{11}} & \bar{z_{21}} \\ \bar{z_{12}} & \bar{z_{22}} \end{array} \right] = \left[ \begin{array}{cc} z_{11} & z_{12} \\ z_{21} & z_{22} \end{array} \right].$$
從而得到
$$\bar{z_{11}} = z_{11}, \bar{z_{21}} = z_{12}, \bar{z_{12}} = z_{21}, \bar{z_{22}} = z_{22}.$$
所以$z_{11}, z_{22}$皆為實數，而$z_{12}$與$z_{21}$互為共軛複數(因此實部相同)。

總結上述，我們可以得到：${\rm A} = \left[ \begin{array}{cc} z_{11} & z_{12} \\ z_{21} & z_{22} \end{array} \right]$為Hermite矩陣，若且唯若${\rm A}$形如
$${\rm A} = \left[ \begin{array}{cc} a & p+qi \\ p-qi & b \end{array} \right], a, b, p, q \in \mathbb{R}.$$
(主對角線元素為實數，副對角線元素為共軛複數對)

(證明終了)