数学 - 線型代数 - 線型写像 - 線型写像の像と核(行列で定義される線型写像、像と核、階数と退化次数、基底)

線型代数入門

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• 参考書籍
  • Pythonからはじめる数学入門
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  • インタラクティブ・データビジュアライゼーション

線型代数入門(松坂 和夫(著)、岩波書店)の第3章(線型写像)、9(線型写像の像と核)、問題7.を取り組んでみる。

7. 1. 
      $\begin{array}{l}L\left(e_i\right)=a_i\left(i=1,2\right)\\ L\left(e_1\right)\\ =L\left(\left(\begin{array}{c}1\\ 0\end{array}\right)\right)\\ =\left(\begin{array}{cc}1& 1\\ 1& -1\\ 0& 1\end{array}\right)\left(\begin{array}{c}1\\ 0\end{array}\right)\\ =\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)\\ \\ L\left(e_2\right)\\ =L\left(\left(\begin{array}{c}0\\ 1\end{array}\right)\right)\\ =\left(\begin{array}{cc}1& 1\\ 1& -1\\ 0& 1\end{array}\right)\left(\begin{array}{c}0\\ 1\end{array}\right)\\ =\left(\begin{array}{c}1\\ -1\\ 1\end{array}\right)\\ \\ a_1=\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)\\ a_2=\left(\begin{array}{c}1\\ -1\\ 1\end{array}\right)\\ rankL=2\end{array}$

      Lの階数は2。

      $\begin{array}{l}L\left(\left(\begin{array}{l}x\\ y\end{array}\right)\right)=0\\ \left(\begin{array}{cc}1& 1\\ 1& -1\\ 0& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=0\\ \left(\begin{array}{c}x+y\\ x-y\\ y\end{array}\right)=0\\ x+y=0\\ x-y=0\\ y=0\\ x=y=0\\ nullityL=0\end{array}$

      Lの退化次数は0となる。

      Lの像、Im Lの基底。

      $\left\{\left(\begin{array}{l}1\\ 1\\ 0\end{array}\right),\left(\begin{array}{l}1\\ -1\\ 1\end{array}\right)\right\}$

      Lの核、Ker L の基底。

      $\left\{\right\}$
   2. 
      $\begin{array}{l}L\left(e_i\right)=a_i\left(i=1,2,3\right)\\ L\left(e_1\right)\\ =L\left(\left(\begin{array}{c}1\\ \begin{array}{l}0\\ 0\end{array}\end{array}\right)\right)\\ =\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 1& 0\end{array}\right)\left(\begin{array}{c}1\\ \begin{array}{l}0\\ 0\end{array}\end{array}\right)\\ =\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\\ \\ L\left(e_2\right)\\ =L\left(\left(\begin{array}{c}0\\ \begin{array}{l}1\\ 0\end{array}\end{array}\right)\right)\\ =\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 1& 0\end{array}\right)\left(\begin{array}{c}0\\ \begin{array}{l}1\\ 0\end{array}\end{array}\right)\\ =\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\\ \\ L\left(e_3\right)\\ =L\left(\left(\begin{array}{c}0\\ \begin{array}{l}0\\ 1\end{array}\end{array}\right)\right)\\ =\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 1& 0\end{array}\right)\left(\begin{array}{c}0\\ \begin{array}{l}0\\ 1\end{array}\end{array}\right)\\ =\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)\\ \\ a_1=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\\ a_2=\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\\ a_3=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)\\ \mathrm{Im}L=\langle a_1,a_2\rangle \\ rankL=2\end{array}$

      よってLの階数は2。

      $\begin{array}{l}L\left(\left(\begin{array}{l}x\\ y\\ z\end{array}\right)\right)=0\\ \left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 1& 0\end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=0\\ \left(\begin{array}{l}x\\ 0\\ y\end{array}\right)=0\\ x=0,y=0\\ KerL=\langle \left(\begin{array}{l}0\\ 0\\ 1\end{array}\right)\rangle \end{array}$

      よってLの退化次数は1。

      Lの像の基底。

      $\left\{\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\right\}$

      Lの核、Ker Lの基底。

      $\left\{\left(\begin{array}{l}0\\ 0\\ 1\end{array}\right)\right\}$
   3. 
      $\begin{array}{l}L\left(e_i\right)=a_i\left(i=1,2,3\right)\\ a_1=\left(\begin{array}{c}1\\ 1\end{array}\right)\\ a_2=\left(\begin{array}{c}0\\ -2\end{array}\right)\\ a_3=\left(\begin{array}{c}1\\ 0\end{array}\right)\\ \\ a_1=a_3-\frac{1}{2}{a}_2\\ \mathrm{Im}L=\langle a_2,a_3\rangle \\ rankL=2\end{array}$

      よって、Lの階数は2。

      $\begin{array}{l}L\left(\left(\begin{array}{l}x\\ y\\ z\end{array}\right)\right)\\ =\left(\begin{array}{ccc}1& 0& 1\\ 1& -2& 0\end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)\\ =\left(\begin{array}{c}x+z\\ x-2y\end{array}\right)\\ x+z=0\\ x-2y=0\\ z=-x\\ y=\frac{1}{2}x\\ KerL=\langle \left(\begin{array}{c}2\\ 1\\ -2\end{array}\right)\rangle \end{array}$

      よってLの退化次数は1。

      Lの像、Im Lの基底。

      $\left\{\left(\begin{array}{c}0\\ -2\end{array}\right),\left(\begin{array}{c}1\\ 0\end{array}\right)\right\}$

      Lの核、Ker Lの基底。

      $\left\{\left(\begin{array}{c}2\\ 1\\ -2\end{array}\right)\right\}$