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行列の対角化を考えるときに固有値や固有ベクトルを求めれば良い理由

Last updated at Posted at 2020-01-02

$A$ を対角化可能な $n$ 次正方行列とすると,ある $n$ 次正方行列 $P$ が存在して

P^{-1}AP=
\begin{pmatrix}
\lambda_1 &  & O \\
 & \ddots &  \\
O &  & \lambda_n
\end{pmatrix}

が成り立つ.このとき

AP=P
\begin{pmatrix}
\lambda_1 &  & O \\
 & \ddots &  \\
O &  & \lambda_n
\end{pmatrix}

である.$p_1,\cdots,p_n$ を $n$ 次列ベクトルとして $P=(p_1,\cdots,p_n)$ と書き表すと

A(p_1,\cdots,p_n)=(p_1,\cdots,p_n)
\begin{pmatrix}
\lambda_1 &  & O \\
 & \ddots &  \\
O &  & \lambda_n
\end{pmatrix}

となり,これより各 $k=1,\cdots,n$ に対して

Ap_k=\lambda_kp_k

が得られる.すなわち,$\lambda_k$ は $A$ の固有値であり,$p_k$ は $\lambda_k$ に対応する $A$ の固有ベクトルである.従って,$A$ の対角化を実現するには $A$ の固有値と固有ベクトルを求めればよい.

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