基底の変換

$3$\displaystyle{n}$ 次元実ベクトル空間 $3$\displaystyle{{\displaystyle{R}}^{n}}$ の任意のベクトル $3$\displaystyle{\displaystyle{a}}$ が， $3$\displaystyle{{\displaystyle{R}}^{n}}$ のある基底 $3$\displaystyle{ \{ \hspace{1}{\displaystyle{v}}_{1},\text{\hspace{1}}\hspace{1}{\displaystyle{v}}_{2},\text{\hspace{1}}\cdots ,\text{\hspace{1}}\hspace{1}{\displaystyle{v}}_{n} \} }$ を用いて

$3$\displaystyle{ \displaystyle{a}=\hspace{1}{a}_{1}\hspace{1}{\displaystyle{v}}_{1}+\hspace{1}{a}_{2}\hspace{1}{\displaystyle{v}}_{2}+\cdots +\hspace{1}{a}_{n}\hspace{1}{\displaystyle{v}}_{n} }$

と表されるとする．このベクトル $3$\displaystyle{\displaystyle{a}}$ が， $3$\displaystyle{{\displaystyle{R}}^{n}}$ の別の基底 $3$\displaystyle{ \{ \displaystyle{v}_{1}^{\prime },\text{\hspace{1}}\displaystyle{v}_{2}^{\prime },\text{\hspace{1}}\cdots ,\text{\hspace{1}}\displaystyle{v}_{n}^{\prime } \} }$ を用いて，

$3$\displaystyle{ \displaystyle{a}=a_{1}^{\prime }\displaystyle{v}_{1}^{\prime }+a_{2}^{\prime }\displaystyle{v}_{2}^{\prime }+\cdots +a_{n}^{\prime }\displaystyle{v}_{n}^{\prime } }$

と表されるとき，行列 $3$\displaystyle{ T={ $ \begin{array} \hspace{1}{\displaystyle{v}}_{1} & \hspace{1}{\displaystyle{v}}_{2} &\cdots & \hspace{1}{\displaystyle{v}}_{n} & \vspace{6}\\ \end{array} $ }^{ - 1 }\text{\hspace{1}} $ \begin{array} \displaystyle{v}_{1}^{\prime } & \displaystyle{v}_{2}^{\prime } &\cdots & \displaystyle{v}_{n}^{\prime } & \vspace{6}\\ \end{array} $ }$ とすると，これらの係数の間には1次変換

$3$\displaystyle{ $ \begin{array} a_{1}^{\prime } & \vspace{6}\\ a_{2}^{\prime } & \vspace{6}\\ \vdots & \vspace{6}\\ a_{n}^{\prime } & \vspace{6}\\ \end{array} $ ={T}^{ - 1 }\text{\hspace{1}} $ \begin{array} \hspace{1}{a}_{1} & \vspace{6}\\ \hspace{1}{a}_{2} & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{a}_{n} & \vspace{6}\\ \end{array} $ }$

の関係がある．また，線形写像 $3$\displaystyle{f:{\displaystyle{R}}^{n}\rightarrow {\displaystyle{R}}^{n}}$ によって $3$\displaystyle{\displaystyle{b}=f$\displaystyle{a}$}$ と変換されるとき，基底 $3$\displaystyle{ \{ \hspace{1}{\displaystyle{v}}_{1},\text{\hspace{1}}\hspace{1}{\displaystyle{v}}_{2},\text{\hspace{1}}\cdots ,\text{\hspace{1}}\hspace{1}{\displaystyle{v}}_{n} \} }$ における $3$\displaystyle{f}$ の表現行列を $3$\displaystyle{A}$ とすると，

$3$\displaystyle{ \displaystyle{b}=\hspace{1}{b}_{1}\hspace{1}{\displaystyle{v}}_{1}+\hspace{1}{b}_{2}\hspace{1}{\displaystyle{v}}_{2}+\cdots +\hspace{1}{b}_{n}\hspace{1}{\displaystyle{v}}_{n} }$ $3$\displaystyle{ =b_{1}^{\prime }\displaystyle{v}_{1}^{\prime }+b_{2}^{\prime }\displaystyle{v}_{2}^{\prime }+\cdots +b_{n}^{\prime }\displaystyle{v}_{n}^{\prime } }$

$3$\displaystyle{ $ \begin{array} \hspace{1}{b}_{1} & \vspace{6}\\ \hspace{1}{b}_{2} & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{b}_{n} & \vspace{6}\\ \end{array} $ =A\text{\hspace{1}} $ \begin{array} \hspace{1}{a}_{1} & \vspace{6}\\ \hspace{1}{a}_{2} & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{a}_{n} & \vspace{6}\\ \end{array} $ }$ ⇔ $3$\displaystyle{ $ \begin{array} b_{1}^{\prime } & \vspace{6}\\ b_{2}^{\prime } & \vspace{6}\\ \vdots & \vspace{6}\\ b_{n}^{\prime } & \vspace{6}\\ \end{array} $ ={T}^{ - 1 }AT\text{\hspace{1}} $ \begin{array} a_{1}^{\prime } & \vspace{6}\\ a_{2}^{\prime } & \vspace{6}\\ \vdots & \vspace{6}\\ a_{n}^{\prime } & \vspace{6}\\ \end{array} $ }$

の関係がある．

以上のことから，基底の変換

$3$\displaystyle{ $ \begin{array} \displaystyle{v}_{1}^{\prime } & \displaystyle{v}_{2}^{\prime } &\cdots & \displaystyle{v}_{n}^{\prime } & \vspace{6}\\ \end{array} $ = $ \begin{array} \hspace{1}{\displaystyle{v}}_{1} & \hspace{1}{\displaystyle{v}}_{2} &\cdots & \hspace{1}{\displaystyle{v}}_{n} & \vspace{6}\\ \end{array} $ \text{\hspace{1}}T }$

により，基底 $3$\displaystyle{ \{ \hspace{1}{\displaystyle{v}}_{1},\text{\hspace{1}}\hspace{1}{\displaystyle{v}}_{2},\text{\hspace{1}}\cdots ,\text{\hspace{1}}\hspace{1}{\displaystyle{v}}_{n} \} }$ において成分表示で $3$\displaystyle{ $ \begin{array} \hspace{1}{a}_{1} & \vspace{6}\\ \hspace{1}{a}_{2} & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{a}_{n} & \vspace{6}\\ \end{array} $ }$ と表されるベクトルは，別の基底 $3$\displaystyle{ \{ \displaystyle{v}_{1}^{\prime },\text{\hspace{1}}\displaystyle{v}_{2}^{\prime },\text{\hspace{1}}\cdots ,\text{\hspace{1}}\displaystyle{v}_{n}^{\prime } \} }$ においては成分表示で $3$\displaystyle{ {T}^{ - 1 }\text{\hspace{1}} $ \begin{array} \hspace{1}{a}_{1} & \vspace{6}\\ \hspace{1}{a}_{2} & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{a}_{n} & \vspace{6}\\ \end{array} $ }$ と表され，基底 $3$\displaystyle{ \{ \hspace{1}{\displaystyle{v}}_{1},\text{\hspace{1}}\hspace{1}{\displaystyle{v}}_{2},\text{\hspace{1}}\cdots ,\text{\hspace{1}}\hspace{1}{\displaystyle{v}}_{n} \} }$ における線形写像の表現行列 $3$\displaystyle{A}$ は，基底 $3$\displaystyle{ \{ \displaystyle{v}_{1}^{\prime },\text{\hspace{1}}\displaystyle{v}_{2}^{\prime },\text{\hspace{1}}\cdots ,\text{\hspace{1}}\displaystyle{v}_{n}^{\prime } \} }$ において $3$\displaystyle{{T}^{ - 1 }AT}$ と表されることがわかる．

基底	$3$\displaystyle{ \{ \hspace{1}{\displaystyle{v}}_{1},\text{\hspace{1}}\hspace{1}{\displaystyle{v}}_{2},\text{\hspace{1}}\cdots ,\text{\hspace{1}}\hspace{1}{\displaystyle{v}}_{n} \} }$	$3$\displaystyle{ \{ \displaystyle{v}_{1}^{\prime },\text{\hspace{1}}\displaystyle{v}_{2}^{\prime },\text{\hspace{1}}\cdots ,\text{\hspace{1}}\displaystyle{v}_{n}^{\prime } \} }$
変換行列	$3$\displaystyle{ T={ $ \begin{array} \hspace{1}{\displaystyle{v}}_{1} & \hspace{1}{\displaystyle{v}}_{2} &\cdots & \hspace{1}{\displaystyle{v}}_{n} & \vspace{6}\\ \end{array} $ }^{ - 1 }\text{\hspace{1}} $ \begin{array} \displaystyle{v}_{1}^{\prime } & \displaystyle{v}_{2}^{\prime } &\cdots & \displaystyle{v}_{n}^{\prime } & \vspace{6}\\ \end{array} $ }$
ベクトルの成分表示	$3$\displaystyle{ \displaystyle{a}= $ \begin{array} \hspace{1}{a}_{1} & \vspace{6}\\ \hspace{1}{a}_{2} & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{a}_{n} & \vspace{6}\\ \end{array} $ }$	$3$\displaystyle{ \displaystyle{a}= $ \begin{array} a_{1}^{\prime } & \vspace{6}\\ a_{2}^{\prime } & \vspace{6}\\ \vdots & \vspace{6}\\ a_{n}^{\prime } & \vspace{6}\\ \end{array} $ ={T}^{ - 1 }\text{\hspace{1}} $ \begin{array} \hspace{1}{a}_{1} & \vspace{6}\\ \hspace{1}{a}_{2} & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{a}_{n} & \vspace{6}\\ \end{array} $ }$
1次変換	$3$\displaystyle{ $ \begin{array} \hspace{1}{b}_{1} & \vspace{6}\\ \hspace{1}{b}_{2} & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{b}_{n} & \vspace{6}\\ \end{array} $ =A\text{\hspace{1}} $ \begin{array} \hspace{1}{a}_{1} & \vspace{6}\\ \hspace{1}{a}_{2} & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{a}_{n} & \vspace{6}\\ \end{array} $ }$	$3$\displaystyle{ $ \begin{array} b_{1}^{\prime } & \vspace{6}\\ b_{2}^{\prime } & \vspace{6}\\ \vdots & \vspace{6}\\ b_{n}^{\prime } & \vspace{6}\\ \end{array} $ ={T}^{ - 1 }AT\text{\hspace{1}} $ \begin{array} a_{1}^{\prime } & \vspace{6}\\ a_{2}^{\prime } & \vspace{6}\\ \vdots & \vspace{6}\\ a_{n}^{\prime } & \vspace{6}\\ \end{array} $ }$

詳しい解説

ホーム>>カテゴリー分類>>行列>>線形代数>>基底の変換

最終更新日： 2023年2月9日

基底	$3$\displaystyle{ \{ \hspace{1}{\displaystyle{v}}_{1},\text{\hspace{1}}\hspace{1}{\displaystyle{v}}_{2},\text{\hspace{1}}\cdots ,\text{\hspace{1}}\hspace{1}{\displaystyle{v}}_{n} \} }$	$3$\displaystyle{ \{ \displaystyle{v}_{1}^{\prime },\text{\hspace{1}}\displaystyle{v}_{2}^{\prime },\text{\hspace{1}}\cdots ,\text{\hspace{1}}\displaystyle{v}_{n}^{\prime } \} }$
変換行列	$3$\displaystyle{ T={ \( \begin{array} \hspace{1}{\displaystyle{v}}_{1} & \hspace{1}{\displaystyle{v}}_{2} &\cdots & \hspace{1}{\displaystyle{v}}_{n} & \vspace{6}\\ \end{array} \) }^{ - 1 }\text{\hspace{1}} \( \begin{array} \displaystyle{v}_{1}^{\prime } & \displaystyle{v}_{2}^{\prime } &\cdots & \displaystyle{v}_{n}^{\prime } & \vspace{6}\\ \end{array} \) }$
ベクトルの成分表示	$3$\displaystyle{ \displaystyle{a}= \( \begin{array} \hspace{1}{a}_{1} & \vspace{6}\\ \hspace{1}{a}_{2} & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{a}_{n} & \vspace{6}\\ \end{array} \) }$	$3$\displaystyle{ \displaystyle{a}= \( \begin{array} a_{1}^{\prime } & \vspace{6}\\ a_{2}^{\prime } & \vspace{6}\\ \vdots & \vspace{6}\\ a_{n}^{\prime } & \vspace{6}\\ \end{array} \) ={T}^{ - 1 }\text{\hspace{1}} \( \begin{array} \hspace{1}{a}_{1} & \vspace{6}\\ \hspace{1}{a}_{2} & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{a}_{n} & \vspace{6}\\ \end{array} \) }$
1次変換	$3$\displaystyle{ \( \begin{array} \hspace{1}{b}_{1} & \vspace{6}\\ \hspace{1}{b}_{2} & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{b}_{n} & \vspace{6}\\ \end{array} \) =A\text{\hspace{1}} \( \begin{array} \hspace{1}{a}_{1} & \vspace{6}\\ \hspace{1}{a}_{2} & \vspace{6}\\ \vdots & \vspace{6}\\ \hspace{1}{a}_{n} & \vspace{6}\\ \end{array} \) }$	$3$\displaystyle{ \( \begin{array} b_{1}^{\prime } & \vspace{6}\\ b_{2}^{\prime } & \vspace{6}\\ \vdots & \vspace{6}\\ b_{n}^{\prime } & \vspace{6}\\ \end{array} \) ={T}^{ - 1 }AT\text{\hspace{1}} \( \begin{array} a_{1}^{\prime } & \vspace{6}\\ a_{2}^{\prime } & \vspace{6}\\ \vdots & \vspace{6}\\ a_{n}^{\prime } & \vspace{6}\\ \end{array} \) }$