[線性代數] 座標轉換矩陣

考慮 $V$ 為 $n$ 維向量空間，且 ${\bf v} \in V$。現在考慮兩組 有序基底 (ordered basis)

$$\begin{array}{l} S: = \{ {{\bf{v}}_1},{{\bf{v}}_2},...,{{\bf{v}}_n}\} \\ T: = \{ {{\bf{w}}_1},{{\bf{w}}_2},...,{{\bf{w}}_n}\} \end{array}$$

則 我們可將 ${\bf v}$ 用上述有序基底做線性組合唯一表示，比如說

$${\bf v} = c_1 { {\bf w}_1} + ... c_n {\bf w}_n$$ 且其對應於  $T$ 基底的 座標向量 (coordinate vector) 與 對 $S$ 基底的 coordinate vector 我們定義如下 

$${\left[ {\bf{v}} \right]_T} := \left[ {\begin{array}{*{20}{l}} {{c_1}}\\ {{c_2}}\\  \vdots \\ {{c_n}} \end{array}} \right]; \;\;{[{\bf{v}}]_S} := \left[ \begin{array}{l} {a_1}\\ {a_2}\\  \vdots \\ {a_n} \end{array} \right]$$ 注意到事實上 上述 對 $T$ 基底的 coordinate vector 可看成是函數，比如說令 $L: V \to \mathbb{R}^n$ 滿足
$$L({\bf v}) = [{\bf v}]_T$$ 同理，對 $S$ 基底的 coordinate vector 令 $L': V \to \mathbb{R}^n$ 滿足
$$L'({\bf v}) = [{\bf v}]_S$$
Comment:
1. 上述 $L, L'$ 統稱為 coordinate mapping
2. Coordinate mapping 為 bijective linear transformation 或稱 isomorphism。

現在若我們想建構 對於 $S$ 基底的座標向量 與 $T$ 基底座標向量之間兩者的關係，利用  coordinate mapping $[{\bf v}]_S$ 為 linear transformation 性質 ，我們有
$$\begin{array}{l} {[{\bf{v}}]_S} = {[{c_1}{{\bf{w}}_1} + ...{c_n}{{\bf{w}}_n}]_S}\\ \begin{array}{*{20}{c}} {}&{} \end{array} = {[{c_1}{{\bf{w}}_1}]_S} + ... + {[{c_n}{{\bf{w}}_n}]_S}\\ \begin{array}{*{20}{c}} {}&{} \end{array} = {c_1}{[{{\bf{w}}_1}]_S} + ... + {c_n}{[{{\bf{w}}_n}]_S}\\ \begin{array}{*{20}{c}} {}&{} \end{array} = \left[ {\begin{array}{*{20}{c}} |&|&{}&|\\ {{{[{{\bf{w}}_1}]}_S}}&{{{[{{\bf{w}}_2}]}_S}}&{...}&{{{[{{\bf{w}}_n}]}_S}}\\ |&|&{}&| \end{array}} \right]\left[ \begin{array}{l} {c_1}\\ {c_2}\\  \vdots \\ {c_n} \end{array} \right]\\ \begin{array}{*{20}{c}} {}&{} \end{array} = {P_{S \leftarrow T}}\left[ \begin{array}{l} {c_1}\\ {c_2}\\  \vdots \\ {c_n} \end{array} \right] \end{array}$$ 其中 ${P_{S \leftarrow T}}: = \left[ {\begin{array}{*{20}{c}} |&|&{}&|\\ {{{[{{\bf{w}}_1}]}_S}}&{{{[{{\bf{w}}_2}]}_S}}&{...}&{{{[{{\bf{w}}_n}]}_S}}\\ |&|&{}&| \end{array}} \right]$ 稱為 從 $T$ 基底到 $S$ 基底的座標轉換矩陣  (Transition Matrix from T-basis to the S basis )且對於個別的 coordinate vector; e.g., $[{\bf w}_j]_S$ 我們有
$${[{{\bf{w}}_j}]_S} = \left[ \begin{array}{l} {a_{1j}}\\ {a_{2j}}\\  \vdots \\ {a_{nj}} \end{array} \right]$$
故我們有以下結果
$$[{\bf v}]_S = P_{S \leftarrow T} [{\bf v}]_T$$

以下是一些關於 Transition Matrix 的性質：
令 $S,T$ 為向量空間的兩組 ordered basis 則
FACT 1. $P_{T \leftarrow T} = I$
FACT 2. $P_{S \leftarrow T}$ 為 nonsingular

現在看幾個例子：

Example 1.
令 $$S: = \left\{ {{{\bf{v}}_1},{{\bf{v}}_2}} \right\} = \left\{ {\left[ \begin{array}{l} 1\\ 2 \end{array} \right],\left[ \begin{array}{l} 0\\ 1 \end{array} \right]} \right\};T: = \left\{ {{{\bf{w}}_1},{{\bf{w}}_2}} \right\} = \left\{ {\left[ \begin{array}{l} 1\\ 1 \end{array} \right],\left[ \begin{array}{l} 2\\ 3 \end{array} \right]} \right\}$$ 為 ordered bases for $\mathbb{R}^2$。現在令 ${\bf v} = [1 \;\; 5]^T$ 與 ${\bf w} := [5 \;\; 4]^T$ 。
(a) 試求 coordinate vectors $[{\bf v}]_T$ 與 $[{\bf w}]_T$
(b) 試求 $P_{S \leftarrow T}$
(c) 試求 coordinate vectors $[{\bf v}]_S$ 與 $[{\bf w}]_S$

Solution(a)
由 $[{\bf v}]_T$ 的定義可知 ${\left[ {\bf{v}} \right]_T} = \left[ \begin{array}{l} {a_1}\\ {a_2} \end{array} \right]$且
$$\begin{array}{l} \left[ \begin{array}{l} 1\\ 5 \end{array} \right] = {a_1}\left[ \begin{array}{l} 1\\ 1 \end{array} \right] + {a_2}\left[ \begin{array}{l} 2\\ 3 \end{array} \right]\\  \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&2\\ 1&3 \end{array}} \right]\left[ \begin{array}{l} {a_1}\\ {a_2} \end{array} \right] = \left[ \begin{array}{l} 1\\ 5 \end{array} \right]\\  \Rightarrow \left\{ \begin{array}{l} {a_1} =  - 7\\ {a_2} = 4 \end{array} \right. \end{array}$$ 同理 ${\left[ {\bf{w}} \right]_T} = \left[ \begin{array}{l} {b_1}\\ {b_2} \end{array} \right]$
$$\begin{array}{l} \underbrace {\left[ \begin{array}{l} 5\\ 4 \end{array} \right]}_{ = {\bf{w}}} = {b_1}\left[ \begin{array}{l} 1\\ 1 \end{array} \right] + {b_2}\left[ \begin{array}{l} 2\\ 3 \end{array} \right]\\  \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&2\\ 1&3 \end{array}} \right]\left[ \begin{array}{l} {b_1}\\ {b_2} \end{array} \right] = \left[ \begin{array}{l} 5\\ 4 \end{array} \right]\\  \Rightarrow \left\{ \begin{array}{l} {b_1} = 7\\ {b_2} =  - 1 \end{array} \right. \end{array}$$

Solution (b)
我們要求 $P_{S \leftarrow T}$，由 part (a) 可知我們有 $[{\bf v}]_T = [-7\;\;4]$ 故
$$\begin{array}{l} {[{\bf{v}}]_S} = {\left[ { - 7{{\bf{w}}_1} + 4{{\bf{w}}_2}} \right]_S}\\ \begin{array}{*{20}{c}} {}&{} \end{array} =  - 7{\left[ {{{\bf{w}}_1}} \right]_S} + 4{\left[ {{{\bf{w}}_2}} \right]_S}\\ \begin{array}{*{20}{c}} {}&{} \end{array} = \underbrace {\left[ {\begin{array}{*{20}{c}} |&|\\ {{{\left[ {{{\bf{w}}_1}} \right]}_S}}&{{{\left[ {{{\bf{w}}_2}} \right]}_S}}\\ |&| \end{array}} \right]}_{ = {P_{S \leftarrow T}}}\underbrace {\left[ \begin{array}{l}  - 7\\ 4 \end{array} \right]}_{ = {{[{\bf{v}}]}_T}}\\ \begin{array}{*{20}{c}} {}&{} \end{array} = \underbrace {\left[ {\begin{array}{*{20}{c}} {{b_{11}}}&{{b_{12}}}\\ {{b_{21}}}&{{b_{22}}} \end{array}} \right]}_{{P_{S \leftarrow T}}}\left[ \begin{array}{l}  - 7\\ 4 \end{array} \right] \end{array}$$
現在我們分別求取 $[{\bf w}_1]_S$ 與 $[{\bf w}_2]_S$如下:
$$\begin{array}{l} {\left[ {{{\bf{w}}_1}} \right]_S} = \left[ \begin{array}{l} {b_{11}}\\ {b_{21}} \end{array} \right]\\  \Rightarrow {b_{11}}\left[ \begin{array}{l} 1\\ 2 \end{array} \right] + {b_{21}}\left[ \begin{array}{l} 0\\ 1 \end{array} \right] = \underbrace {\left[ \begin{array}{l} 1\\ 1 \end{array} \right]}_{ = {{\bf{w}}_1}}\\  \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&0\\ 2&1 \end{array}} \right]\left[ \begin{array}{l} {b_{11}}\\ {b_{21}} \end{array} \right] = \left[ \begin{array}{l} 1\\ 1 \end{array} \right]\\  \Rightarrow \left\{ \begin{array}{l} {b_{11}} = 1\\ {b_{21}} =  - 1 \end{array} \right. \end{array}$$
且
$$\begin{array}{l} {\left[ {{{\bf{w}}_2}} \right]_S} = \left[ \begin{array}{l} {b_{12}}\\ {b_{22}} \end{array} \right]\\  \Rightarrow {b_{12}}\left[ \begin{array}{l} 1\\ 2 \end{array} \right] + {b_{22}}\left[ \begin{array}{l} 0\\ 1 \end{array} \right] = \underbrace {\left[ \begin{array}{l} 2\\ 3 \end{array} \right]}_{ = {{\bf{w}}_2}}\\  \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&0\\ 2&1 \end{array}} \right]\left[ \begin{array}{l} {b_{11}}\\ {b_{21}} \end{array} \right] = \left[ \begin{array}{l} 2\\ 3 \end{array} \right]\\  \Rightarrow \left\{ \begin{array}{l} {b_{12}} = 2\\ {b_{21}} =  - 1 \end{array} \right. \end{array}$$ 故
$${P_{S \leftarrow T}} = \left[ {\begin{array}{*{20}{c}} 1&2\\ { - 1}&{ - 1} \end{array}} \right]$$
Solution (c)
$$\begin{array}{l} {[{\bf{v}}]_S} = {P_{S \leftarrow T}}{[{\bf{v}}]_T}\\ \begin{array}{*{20}{c}} {}&{} \end{array} = \left[ {\begin{array}{*{20}{c}} 1&2\\ { - 1}&{ - 1} \end{array}} \right]\left[ \begin{array}{l}  - 7\\ 4 \end{array} \right] = \left[ \begin{array}{l} 1\\ 3 \end{array} \right]\\ {[{\bf{w}}]_S} = {P_{S \leftarrow T}}{[{\bf{w}}]_T}\\ \begin{array}{*{20}{c}} {}&{} \end{array} = \left[ {\begin{array}{*{20}{c}} 1&2\\ { - 1}&{ - 1} \end{array}} \right]\left[ \begin{array}{l} 7\\  - 1 \end{array} \right] = \left[ \begin{array}{l} 5\\  - 6 \end{array} \right] \end{array}$$

Example 2.
令 $V:= \mathbb{R^3}$ 且令 $S:=\{{\bf v}_1,{\bf v}_2, {\bf v}_3\}$ 且 $T = \{{\bf w}_1, {\bf w}_2, {\bf w}_3\}$ 為  $\mathbb{R}^3$ 的ordered basis，其中
$$\begin{array}{l} {{\bf{v}}_1} = \left[ \begin{array}{l} 2\\ 0\\ 1 \end{array} \right];{{\bf{v}}_2} = \left[ \begin{array}{l} 1\\ 2\\ 0 \end{array} \right];{{\bf{v}}_3} = \left[ \begin{array}{l} 1\\ 1\\ 1 \end{array} \right]\\ {{\bf{w}}_1} = \left[ \begin{array}{l} 6\\ 3\\ 3 \end{array} \right];{{\bf{w}}_2} = \left[ \begin{array}{l} 4\\  - 1\\ 3 \end{array} \right];{{\bf{w}}_3} = \left[ \begin{array}{l} 5\\ 5\\ 2 \end{array} \right] \end{array}$$
(a) 試計算 $P_{S \leftarrow T}$
(Exercise) 令 ${\bf v} = [4 \;\; -9 \;\; 5]$ 驗證 $[{\bf v}]_S = P_{S \leftarrow T} [{\bf v}]_T$

Solution (a):
由前面討論可知
$${P_{S \leftarrow T}}: = \left[ {\begin{array}{*{20}{c}} |&|&|\\ {{{[{{\bf{w}}_1}]}_S}}&{{{[{{\bf{w}}_2}]}_S}}&{{{[{{\bf{w}}_3}]}_S}}\\ |&|&| \end{array}} \right]$$ 故我們需分別求出 $[{\bf w}_1]_S, [{\bf w}_2]_S$ 與 $[{\bf w}_3]_S$：

首先求 ${[{{\bf{w}}_1}]_S} = \left[ \begin{array}{l} {a_{11}}\\ {a_{21}}\\ {a_{31}} \end{array} \right]$ 如下：由於 ${\bf w}_1$ 可用 $S$ 有序基底作唯一線性組合表示，故
$$\begin{array}{l} \left[ \begin{array}{l} 6\\ 3\\ 3 \end{array} \right] = {a_{11}}\left[ \begin{array}{l} 2\\ 0\\ 1 \end{array} \right] + {a_{21}}\left[ \begin{array}{l} 1\\ 2\\ 0 \end{array} \right] + {a_{31}}\left[ \begin{array}{l} 1\\ 1\\ 1 \end{array} \right]\\  \Rightarrow \left[ {\begin{array}{*{20}{c}} 2&1&1\\ 0&2&1\\ 1&0&1 \end{array}} \right]\left[ \begin{array}{l} {a_{11}}\\ {a_{21}}\\ {a_{31}} \end{array} \right] = \left[ \begin{array}{l} 6\\ 3\\ 3 \end{array} \right] \Rightarrow \left\{ \begin{array}{l} {a_{11}} = 2\\ {a_{21}} = 1\\ {a_{31}} = 1 \end{array} \right. \end{array}$$
接著我們求 ${[{{\bf{w}}_2}]_S} = \left[ \begin{array}{l} {a_{12}}\\ {a_{22}}\\ {a_{32}} \end{array} \right]$ ，由 ${\bf w}_2$ 可用 $S$ 有序基底作唯一線性組合表示，我們可得
$$\begin{array}{l} \left[ \begin{array}{l} 4\\  - 1\\ 3 \end{array} \right] = {a_{12}}\left[ \begin{array}{l} 2\\ 0\\ 1 \end{array} \right] + {a_{22}}\left[ \begin{array}{l} 1\\ 2\\ 0 \end{array} \right] + {a_{32}}\left[ \begin{array}{l} 1\\ 1\\ 1 \end{array} \right]\\  \Rightarrow \left[ {\begin{array}{*{20}{c}} 2&1&1\\ 0&2&1\\ 1&0&1 \end{array}} \right]\left[ \begin{array}{l} {a_{11}}\\ {a_{21}}\\ {a_{31}} \end{array} \right] = \left[ \begin{array}{l} 4\\  - 1\\ 3 \end{array} \right] \Rightarrow \left\{ \begin{array}{l} {a_{12}} = 2\\ {a_{22}} =  - 1\\ {a_{32}} = 1 \end{array} \right. \end{array}$$
最後求 ${[{{\bf{w}}_3}]_S} = \left[ \begin{array}{l} {a_{13}}\\ {a_{23}}\\ {a_{33}} \end{array} \right]$，同前述方法，利用 ${\bf w}_3$ 可透過 $S$ 有序基底作唯一線性組合表示，我們可得
$$\begin{array}{l} \left[ \begin{array}{l} 5\\ 5\\ 2 \end{array} \right] = {a_{13}}\left[ \begin{array}{l} 2\\ 0\\ 1 \end{array} \right] + {a_{23}}\left[ \begin{array}{l} 1\\ 2\\ 0 \end{array} \right] + {a_{33}}\left[ \begin{array}{l} 1\\ 1\\ 1 \end{array} \right]\\  \Rightarrow \left[ {\begin{array}{*{20}{c}} 2&1&1\\ 0&2&1\\ 1&0&1 \end{array}} \right]\left[ \begin{array}{l} {a_{11}}\\ {a_{21}}\\ {a_{31}} \end{array} \right] = \left[ \begin{array}{l} 5\\ 5\\ 2 \end{array} \right] \Rightarrow \left\{ \begin{array}{l} {a_{13}} = 1\\ {a_{23}} = 2\\ {a_{33}} = 1 \end{array} \right. \end{array}$$
$$\begin{array}{l} {P_{S \leftarrow T}}: = \left[ {\begin{array}{*{20}{c}} |&|&|\\ {{{[{{\bf{w}}_1}]}_S}}&{{{[{{\bf{w}}_2}]}_S}}&{{{[{{\bf{w}}_3}]}_S}}\\ |&|&| \end{array}} \right]\\ \begin{array}{*{20}{c}} {}&{} \end{array} = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\ {{a_{12}}}&{{a_{22}}}&{{a_{23}}}\\ {{a_{13}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2&2&1\\ 1&{ - 1}&2\\ 1&1&1 \end{array}} \right] \end{array}$$