# 行列の対角化

## ■問題

$A=\left(\begin{array}{cc}4& -3\\ 2& -1\end{array}\right)$ を行列$P$ を用いて対角化せよ．

## ■答

$P=\left(\begin{array}{cc}1& 3\\ 1& 2\end{array}\right)$ のとき ${P}^{-1}AP=\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)$

$P=\left(\begin{array}{cc}3& 1\\ 2& 1\end{array}\right)$ のとき ${P}^{-1}AP=\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)$

## ■計算

$Ax=\lambda x$ より，$\left(A-\lambda E\right)x=0$  ･･････(1)

ここで

$Τ=A-\lambda E$ $=\left(\begin{array}{cc}4& -3\\ 2& -1\end{array}\right)-\left(\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right)$ $=\left(\begin{array}{cc}4-\lambda & -3\\ 2& -1-\lambda \end{array}\right)$

とおく．

$|Τ|=|A-\lambda E|$ $=|\begin{array}{cc}4-\lambda & -3\\ 2& -1-\lambda \end{array}|=0$

$\left(4-\lambda \right)\left(-1-\lambda \right)+6=0$ ， ${\lambda }^{2}-3\lambda +2=0$

$\left(\lambda -1\right)\left(\lambda -2\right)=0$

$\therefore \lambda ＝1,2$

(i)$\lambda =1$ のとき，固有ベクトル${x}_{1}=\left(\begin{array}{c}{\alpha }_{1}\\ {\alpha }_{2}\end{array}\right)$ とおくと，(1)は

$\left(\begin{array}{cc}3& -3\\ 2& -2\end{array}\right)\left(\begin{array}{c}{\alpha }_{1}\\ {\alpha }_{2}\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$

となり

${\alpha }_{1}-{\alpha }_{2}=0$${\alpha }_{1}={\alpha }_{2}$

となる．

${\alpha }_{1}={\alpha }_{2}={k}_{1}$ とおくと

${x}_{1}=\left(\begin{array}{c}{\alpha }_{1}\\ {\alpha }_{2}\end{array}\right)={k}_{1}\left(\begin{array}{c}1\\ 1\end{array}\right)$

(ii)$\lambda =2$ のとき，固有ベクトル${x}_{2}=\left(\begin{array}{c}{\beta }_{1}\\ {\beta }_{2}\end{array}\right)$とおくと，(1)は

$\left(\begin{array}{cc}2& -3\\ 2& -3\end{array}\right)\left(\begin{array}{c}{\beta }_{1}\\ {\beta }_{2}\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$

となり

$2{\beta }_{1}-3{\beta }_{2}=0$$2{\beta }_{1}=3{\beta }_{2}$

となる．

${\beta }_{1}=3{k}_{2}$とおくと,   ${\beta }_{2}=2{k}_{2}$

${x}_{2}=\left(\begin{array}{c}{\beta }_{1}\\ {\beta }_{2}\end{array}\right)={k}_{2}\left(\begin{array}{c}3\\ 2\end{array}\right)$

(i)(ii)より，${x}_{1}\prime =\left(\begin{array}{c}1\\ 1\end{array}\right)$ は固有値${\lambda }_{1}=1$ の固有ベクトル，${x}_{2}\prime =\left(\begin{array}{c}3\\ 2\end{array}\right)$ は固有値${\lambda }_{2}=2$ の固有ベクトルである．

$P=\left(\begin{array}{cc}{x}_{1}\prime & {x}_{2}\prime \end{array}\right)=\left(\begin{array}{cc}1& 3\\ 1& 2\end{array}\right)$

で対角化可能で

${P}^{-1}AP=\left(\begin{array}{cc}{\lambda }_{1}& 0\\ 0& {\lambda }_{2}\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)$

となる.

$P=\left(\begin{array}{cc}{x}_{2}\prime & {x}_{1}\prime \end{array}\right)=\left(\begin{array}{cc}3& 1\\ 2& 1\end{array}\right)$

とすると

${P}^{-1}AP=\left(\begin{array}{cc}{\lambda }_{2}& 0\\ 0& {\lambda }_{1}\end{array}\right)=\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)$

となる.

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