線形代数-1-6 〜正則の分割〜

どうも、いもけんぴぃです。

今回で第1章は最終節となります！

(1)$ \begin{eqnarray} \left( \begin{array}{cc} O & I \\ I & O \end{array} \right) \end{eqnarray} $$ \begin{eqnarray} \left( \begin{array}{cc} X & Y \\ Z & W \end{array} \right) \end{eqnarray} $$ $$ $$ $$ $$ $(2)$ \begin{eqnarray} \left( \begin{array}{cc} O & B \\ I & O \end{array} \right) \end{eqnarray} $$ \begin{eqnarray} \left( \begin{array}{cc} X & Y \\ Z & W \end{array} \right) \end{eqnarray} $

(3)$ \begin{eqnarray} \left( \begin{array}{cc} I & O \\ C & I \end{array} \right) \end{eqnarray} $$ \begin{eqnarray} \left( \begin{array}{cc} I & O \\ -C & I \end{array} \right) \end{eqnarray} $$ $$ $$ $$ $(4)$ \begin{eqnarray} \left( \begin{array}{cc} A & O \\ O & I \end{array} \right) \end{eqnarray} $$ \begin{eqnarray} \left( \begin{array}{cc} I & O \\ C & I \end{array} \right) \end{eqnarray} $$ \begin{eqnarray} \left( \begin{array}{cc} I & O \\ O & D \end{array} \right) \end{eqnarray} $

<解説>
定義通りに計算していきます.

(1)$ \begin{eqnarray} \left( \begin{array}{cc} O & I \\ I & O \end{array} \right) \end{eqnarray} $$ \begin{eqnarray} \left( \begin{array}{cc} X & Y \\ Z & W \end{array} \right) \end{eqnarray} $

$=$$ \begin{eqnarray} \left( \begin{array}{cc} O\times X + I\times Z & O\times Y + I\times W \\ I\times X + O\times Z & I\times Y + O\times W \\ \end{array} \right) \end{eqnarray} $

$=$$ \begin{eqnarray} \left( \begin{array}{cc} Z & W \\ X & Y \end{array} \right) \end{eqnarray} $

(2)$ \begin{eqnarray} \left( \begin{array}{cc} O & B \\ I & O \end{array} \right) \end{eqnarray} $$ \begin{eqnarray} \left( \begin{array}{cc} X & Y \\ Z & W \end{array} \right) \end{eqnarray} $

$=$$ \begin{eqnarray} \left( \begin{array}{cc} O\times X + B\times Z & O\times Y + B\times W \\ I\times X + O\times Z & I\times Y + O\times W \\ \end{array} \right) \end{eqnarray} $

$=$$ \begin{eqnarray} \left( \begin{array}{cc} BZ & BW \\ X & Y \end{array} \right) \end{eqnarray} $

(3)$ \begin{eqnarray} \left( \begin{array}{cc} I & O \\ C & I \end{array} \right) \end{eqnarray} $$ \begin{eqnarray} \left( \begin{array}{cc} I & O \\ -C & I \end{array} \right) \end{eqnarray} $

$=$$ \begin{eqnarray} \left( \begin{array}{cc} I\times I + O\times (-C) & I\times O + O\times I \\ C\times I + I\times (-C) & C\times O + I\times I \\ \end{array} \right) \end{eqnarray} $

$=$$ \begin{eqnarray} \left( \begin{array}{cc} I & O \\ O & I \end{array} \right) \end{eqnarray} $

(4)$ \begin{eqnarray} \left( \begin{array}{cc} A & O \\ O & I \end{array} \right) \end{eqnarray} $$ \begin{eqnarray} \left( \begin{array}{cc} I & O \\ C & I \end{array} \right) \end{eqnarray} $$ \begin{eqnarray} \left( \begin{array}{cc} I & O \\ O & D \end{array} \right) \end{eqnarray} $

$=$$ \begin{eqnarray} \left( \begin{array}{cc} A\times I + O\times C & A\times O + O\times I \\ O\times I + I\times C & O\times O + I\times I \\ \end{array} \right) \end{eqnarray} $$ \begin{eqnarray} \left( \begin{array}{cc} I & O \\ O & D \end{array} \right) \end{eqnarray} $

$=$$ \begin{eqnarray} \left( \begin{array}{cc} A & O \\ C & I \\ \end{array} \right) \end{eqnarray} $$ \begin{eqnarray} \left( \begin{array}{cc} I & O \\ O & D \end{array} \right) \end{eqnarray} $

$=$$ \begin{eqnarray} \left( \begin{array}{cc} A\times I + O\times O & A\times O + O\times D \\ C\times I + I\times O & C\times O + I\times D \\ \end{array} \right) \end{eqnarray} $

$=$$ \begin{eqnarray} \left( \begin{array}{cc} A & O \\ C & D \\ \end{array} \right) \end{eqnarray} $

問16

$m\times n$行列$A=(a_{1},a_{2},\cdots,a_{n})$と$n$次列ベクトル$x=\begin{eqnarray} \left( \begin{array}{cc} x_{1} \\ x_{2} \\ \vdots \\ x_{n} \\ \end{array} \right) \end{eqnarray} $に対して,
$Ax=x_{1}a_{1}+x_{2}a_{2}+\cdots+x_{n}a_{n}$
となることを確かめよ.

<解説>
パッとみて「$A$が$m\times n$行列ってどういうこと!?」と思った方もいるかも知れませんが,$a_{i}$は$A$を列分割した$m\times1$行列で,
$a_{i}=\begin{eqnarray} \left( \begin{array}{cc} a_{1i} \\ a_{2i} \\ \vdots \\ a_{mi} \\ \end{array} \right) \end{eqnarray}$のような行列です。

まず$A$は$m\times n$行列,$x$は$n\times1$行列ですので積$Ax$は定義されます。

$Ax$$=$$\begin{eqnarray} \left( \begin{array}{cc} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn}\\ \end{array} \right) \end{eqnarray} $$\begin{eqnarray} \left( \begin{array}{cc} x_{1} \\ x_{2} \\ \vdots \\ x_{n} \\ \end{array} \right) \end{eqnarray}$

$=$$\begin{eqnarray} \left( \begin{array}{cc} a_{11}x_{1} + a_{12}x_{2} + \cdots + a_{1n}x_{n}\\ a_{21}x_{1} + a_{22}x_{2} + \cdots + a_{2n}x_{n} \\ \vdots \\ a_{m1}x_{1} + a_{m2}x_{2} + \cdots + a_{mn}x_{n}\\ \end{array} \right) \end{eqnarray} $

$=$$x_{1}\begin{eqnarray} \left( \begin{array}{cc} a_{11} \\ a_{21} \\ \vdots \\ a_{m1} \\ \end{array} \right) \end{eqnarray}$$+$$x_{2}\begin{eqnarray} \left( \begin{array}{cc} a_{12} \\ a_{22} \\ \vdots \\ a_{m2} \\ \end{array} \right) \end{eqnarray}$$+\cdots+$$x_{n}\begin{eqnarray} \left( \begin{array}{cc} a_{1n} \\ a_{2n} \\ \vdots \\ a_{mn} \\ \end{array} \right) \end{eqnarray}$

$=x_{1}a_{1}+x_{2}a_{2}+\cdots+x_{n}a_{n}$

となり,証明完了です.