[線性代數] 淺論座標

令 $V$ 為 $n$ 維向量空間，則我們知道 $V$ 有基底 (basis) $S$ 且其元素為 $n$ 維向量。現在我們定義 $S :=\{{\bf v}_1,...{\bf v}_n\}$ 為向量空間 $V$ 的一組有序基底 (ordered basis) 則任意向量 ${\bf v} \in V$ 可由上述有序基底唯一表示成以下的線性組合形式：
$${\bf v} = a_1 {\bf v}_1 + a_2 {\bf v}_2 + ... + a_n {\bf v}_n$$ 其中 $a_1,...a_n \in \mathbb{R}^1$

Definition: Coordinate Vector
定義 ${\bf v}$ 對應於有序基底 $S$ 的座標向量 (coordinate vector) 為
$${[{\bf{v}}]_S}: = \left[ \begin{array}{l} {a_1}\\ {a_2}\\  \vdots \\ {a_n} \end{array} \right]$$ 且其中 $[{\bf v}]_S$ 的元素 $a_i$ 稱之為 ${\bf v}$ 對應於有序基底的座標。

Example 1:
考慮向量空間 $$V:= P_2 := \{p(t) = a_2t^2 + a_1t + a_0: a_2,a_1,a_0 \in \mathbb{R}^2\}$$ 且令基底 $S= \{t^2, t, 1\}$ 現考慮 ${\bf v}:= p(t) = \alpha t^2 + \alpha t^1 + \alpha$ 求 $[{\bf v}]_S = ?$

Solution
注意到 ${\bf v}:= p(t) = \alpha t^2 + \alpha t^1 + \alpha$，暫稱此式為 $(*)$ 又因為 ${\bf v} \in V$ 故由 $\bf v$ 可由 $S$ 的有序基底 $\{ t^2, t,1\}$ 透過線性組合唯一表示：也就是說
$${\bf{v}} = p\left( t \right) \in {P_2} \Leftrightarrow {\bf v} = {a_2}{t^2} + {a_1}t + {a_0} \;\;\;\;\; (\star)$$
故比較 $(*)$ 與 $(\star)$ 兩式
$${a_2}{t^2} + {a_1}t + {a_0} = \alpha {t^2} + \alpha t + \alpha$$ 可得 $a_2 = \alpha$， $a_1 = \alpha$ 與 $a_0 = \alpha$ 故
$${[{\bf{v}}]_S}: = \left[ \begin{array}{l} {a_2}\\ {a_1}\\ {a_0} \end{array} \right] = \left[ \begin{array}{l} \alpha \\ \alpha \\ \alpha \end{array} \right]$$

Example 2:
考慮向量空間 $$V:= P_2 := \{p(t) = a_2t^2 + a_1t + a_0: a_2,a_1,a_0 \in \mathbb{R}^2\}$$ 且令基底 $S= \{t^2-t+1, t+1, 1^2+1\}$ 現考慮 ${\bf v}:= p(t) = 4 t^2 -2 t^1 + 3$ 求 $[{\bf v}]_S = ?$

Solution:
令 $${\left[ {\bf{v}} \right]_S} = \left[ \begin{array}{l} {a_1}\\ {a_2}\\ {a_3} \end{array} \right]$$ 且 $\bf v$ 可透過基底 $S$ 做線性組合
$$\begin{array}{l} {\bf{v}} = {a_1}{{\bf{v}}_1} + {a_2}{{\bf{v}}_2} + {a_3}{{\bf{v}}_3}\\ \begin{array}{*{20}{c}} {}&{} \end{array} = {a_1}\left( {{t^2} - t + 1} \right) + {a_2}\left( {t + 1} \right) + {a_3}\left( {{t^2} + 1} \right)\\ \begin{array}{*{20}{c}} {}&{} \end{array} = \left( {{a_1} + {a_3}} \right){t^2} + \left( {{a_2} - {a_1}} \right)t + \left( {{a_1} + {a_2} + {a_3}} \right)\;\;\;\; (*) \end{array}$$ 又因為
$${\bf v} = 4t^2 -2t +3 \;\;\;\; (\star)$$ 故比較 $(\star)$ 與 $(*)$ 係數可得
$$\begin{gathered}   4{t^2} - 2t + 3 = \left( {{a_1} + {a_3}} \right){t^2} + \left( {{a_2} - {a_1}} \right)t + \left( {{a_1} + {a_2} + {a_3}} \right) \hfill \\    \Rightarrow \left\{ \begin{gathered}   {a_1} + {a_3} = 4 \hfill \\   {a_2} - {a_1} =  - 2 \hfill \\   {a_1} + {a_2} + {a_3} = 3 \hfill \\ \end{gathered}  \right. \hfill \\ \end{gathered}$$
將上式改寫成矩陣求解 $a_1, a_2, a_3$如下
$$\begin{gathered}    \left[ {\begin{array}{*{20}{c}}   1&0&1 \\   { - 1}&1&0 \\   1&1&1 \end{array}} \right]\left[ \begin{gathered}   {a_1} \hfill \\   {a_2} \hfill \\   {a_3} \hfill \\ \end{gathered}  \right] = \left[ \begin{gathered}   4 \hfill \\    - 2 \hfill \\   3 \hfill \\ \end{gathered}  \right] \hfill \\    \Rightarrow \left\{ \begin{gathered}   {a_1} = 1 \hfill \\   {a_2} =  - 1 \hfill \\   {a_3} = 3 \hfill \\ \end{gathered}  \right. \hfill \\ \end{gathered}$$